Comparing More Revision and Fixed-Point Theories of Truth

Abstract

Kremer presented three approaches of comparing fixed-point and revision theories of truth in Kremer (Journal of Philosophical Logic, 38(4), 363–403, 2009). Using these approaches, he established the relationships among ten fixed-point theories suggested by Kripke in (Journal of Philosophy, 72(19), 690–716, 1975) and three revision theories presented by Gupta and Belnap in (1993). This paper continues Kremer’s work. We add five other revision theories to the comparisons, including the theory proposed by Gupta in (Journal of Philosophical Logic, 11(1), 1–60, 1982), the theory proposed by Herzberger in (Journal of Philosophical Logic, 11(1), 61–102, 1982), the theory based on fully-varied revision sequences (and stability) proposed by Gupta and Belnap in (1993), the theory proposed by Yaqūb in (1993), and the theory based on weakly consistent revision sequences (and stability). We show that, the notion of Thomason model defined by Belnap’s limit rule is not equivalent to the one defined by Gupta’s limit rule, and that the theory based on fully-varied revision sequences is ≤₂-equivalent to the one based on the greatest intrinsic fixed point of σ.

Appendix: Proof of Lemma 23

Proof

Besides T and quote names, let the truth language ${{\mathscr{L}}}^{+}$ have only the following nonlogical symbols: countably many nonquote names $\overline {\alpha }(\alpha \in 3\omega )$, and a_n,α(n ∈ ω,α ∈ 3ω); three 1-place predicate symbols E, O, and S; four 2-place predicate symbols ≤, ≤_S, V_e and V_o; one 3-place predicate symbol P; and three 1-place function symbols F₁, F₂ and F_ω. Define formulas with exactly 1 free variable φ₁(x), φ₂(x), φ₃(x), φ₄(x) and φ₅(x), and sentences φ₆, φ_c, φ_others and A_n,α(n ∈ ω,α ∈ 3ω) as follows.

φ₁(x) is the conjunction of the following four formulas:¹

• (1) Sx;

• (2) ∀y(Sy ∧ F_ωy < F_ωx →¬Ty);

• (3) $\forall y(Sy\land y<_{S}x \land F_{\omega } y=F_{\omega } x \to (V_{e} yx\leftrightarrow (\boldsymbol {T} y \leftrightarrow \boldsymbol {T} x)))$;

• (4) $\forall y(Sy\land y\ge _{S}x \to (\boldsymbol {T} y \leftrightarrow EF_{1} y))$.

φ₂(x) is the conjunction of the following five formulas:²

• (1) Sx;

• (2) $F_{1} x=\overline {0} \land F_{2} x\not = \overline {0}\land F_{2} x\not = \overline {\omega }\land F_{2} x\not = \overline {2\omega }$;

• (3) ∀y(Sy ∧ F_ωy < F_ωx →¬Ty);

• (4) $\forall y(Sy\land y<_{S}x \land F_{\omega } y=F_{\omega } x \to (V_{o} yx\leftrightarrow (\boldsymbol {T} y \leftrightarrow \boldsymbol {T} x)))$;

• (5) $\forall y(Sy\land y\ge _{S}x \to (\boldsymbol {T} y \leftrightarrow EF_{1} y))$.

φ₃(x) is the conjunction of the following five formulas:

• (1) Sx;

• (2) x = a_0,0 ∨ x = a_0,ω ∨ x = a_0,2ω;

• (3) ∀y(Sy ∧ F_ωy < F_ωx →¬Ty);

• (4) $\forall y(Sy \land F_{\omega } y=F_{\omega } x \to (\boldsymbol {T} y \leftrightarrow OF_{2} y))$;

• (5) $\forall y(Sy \land F_{\omega } y>F_{\omega } x \to (\boldsymbol {T} y \leftrightarrow EF_{1} y))$.

φ₄(x) is the conjunction of the following five formulas:

• (1) Sx;

• (2) $F_{1} x=\overline {0}$;

• (3) ∀y(Sy ∧ F_ωy < F_ωx →¬Ty);

• (4) $\forall y(Sy \land F_{\omega } y=F_{\omega } x \to (F_{2} y<F_{2} x\to (\boldsymbol {T} y \leftrightarrow \boldsymbol {T} x) )\land (F_{2} y\ge F_{2} x\to (\boldsymbol {T} y \leftrightarrow EF_{2} y)))$;

• (5) $\forall y(Sy \land F_{\omega } y>F_{\omega } x \to (\boldsymbol {T} y \leftrightarrow EF_{1} y))$.

φ₅(x) is the conjunction of the following four formulas:

• (1) $x=\overline {1}\lor x=\overline {2}\lor x=\overline {3}$;

• (2) ∀y(Sy ∧ F_ωy < x →¬Ty);

• (3) ∀y(Sy ∧ F_ωy = x →Ty);

• (4) $\forall y(Sy \land F_{\omega } y> x \to (\boldsymbol {T} y \leftrightarrow EF_{1} y))$.

• φ₆ = _df∀x(Sx →¬Tx).

• $\varphi _{c}=_{df}\forall x\forall y\forall z(Pxyz\to ((\boldsymbol {T} x \leftrightarrow \boldsymbol {T} y)\leftrightarrow \boldsymbol {T} z))$.

• φ_others = _df¬(φ_c ∧ (∃xφ₁(x) ∨∃xφ₂(x) ∨∃xφ₃(x) ∨∃xφ₄(x) ∨∃xφ₅(x) ∨ φ₆)).

For every n ∈ ω and every α ∈ 3ω, A_n,α is the disjunction of the following eleven formulas:

• (1) φ_c∧∃x(φ₁(x)∧a_n,α ≤_Sx∧F_ωa_n,α = F_ωx∧((EF₁x∧V_oa_n,αx)∨(OF₁x∧V_ea_n,αx)));

• (2) $\varphi _{c} \land \exists x(\varphi _{1}(x)\land a_{n,\alpha } >_{S} x \land E\overline {n})$;

• (3) φ_c ∧∃x(φ₂(x) ∧ a_n,α <_Sx ∧ F_ωa_n,α = F_ωx ∧ V_oa_n,αx);

• (4) $\varphi _{c} \land \exists x(\varphi _{2}(x)\land a_{n,\alpha } >_{S} x \land E\overline {n})$;

• (5) $\varphi _{c} \land \exists x(\varphi _{3}(x)\land F_{\omega } a_{n,\alpha }= F_{\omega } x\land \overline {\alpha }\neq F_{2} x\land E\overline {\alpha })$;

• (6) $\varphi _{c} \land \exists x(\varphi _{3}(x)\land F_{\omega } a_{n,\alpha }> F_{\omega } x\land E\overline {n})$;

• (7) $\varphi _{c} \land \exists x(\varphi _{4}(x) \land F_{\omega } a_{n,\alpha }= F_{\omega } x\land \overline {\alpha }\le F_{2} x\land OF_{2} x)$;

• (8) $\varphi _{c} \land \exists x(\varphi _{4}(x) \land F_{\omega } a_{n,\alpha }= F_{\omega } x\land \overline {\alpha }>F_{2} x\land E\overline {\alpha })$;

• (9) $\varphi _{c} \land \exists x(\varphi _{4}(x) \land F_{\omega } a_{n,\alpha }> F_{\omega } x\land E\overline {n})$;

• (10) $\varphi _{c} \land \exists x(\varphi _{5}(x) \land F_{\omega } a_{n,\alpha }> x\land E\overline {n}\land (\overline {n}=\overline {0}\land x=\overline {1}\to \overline {\alpha } \neq \overline {\omega })\land (\overline {n}=\overline {0}\land x=\overline {2}\to \overline {\alpha } \neq \overline {2\omega }))$;

• (11) $\varphi _{others}\land E\overline {n}$.

Let ${\mathscr{M}}=\langle D,I\rangle$ be the ground model of ${{\mathscr{L}}}^{+}$ with

• $D=Sent({{\mathscr{L}}}^{+})\cup 3\omega$;

• for every α ∈ 3ω, $I(\overline {\alpha })=\alpha$;

• for every n ∈ ω and every α ∈ 3ω, I(a_n,α) = A_n,α;

• I(E) = {α ∈ 3ω∣α = ω × m + n,m,n ∈ ω,n is even};

• I(O) = {α ∈ 3ω∣α = ω × m + n,m,n ∈ ω,n is odd};

• I(S) = {A_n,α∣n ∈ ω,α ∈ 3ω};

• I(≤) = {(α,β)∣α,β ∈ 3ω,α ≤ β};³

• I(≤_S) = {(A_n,α,A_m,β)∣m,n ∈ ω,α,β ∈ 3ω,α < β ∨ (α = β ∧ n ≤ m)};

• I(V_e) = {(A_n,α,A_m,β)∣m,n ∈ ω,α,β ∈ 3ω,∃k ∈ ω(α = β + 2k ∨ β = α + 2k)};

• I(V_o) = {(A_n,α,A_m,β))∣m,n ∈ ω,α,β ∈ 3ω,∃k ∈ ω(α = β + 2k + 1 ∨ β = α + 2k + 1)};

• $I(P)=\{(A_{n,\alpha },A_{m,\beta },A_{n,\alpha }\leftrightarrow A_{m,\beta })\mid m,n \in \omega ,\alpha ,\beta \in 3\omega \}$; and

• I(F₁), I(F₂) and I(F_ω) are 1-place functions on D f₁, f₂ and f_ω respectively.

f₁ is defined as follows: for every d ∈ D,

$$ f_{1}(d)=\left\{\begin{array}{ll} n & \text{ if there is an }n \in \omega \text{ and an } \alpha \in 3\omega \text{ such that } d=A_{n,\alpha }\text{,}\\ 0& \text{ otherwise.} \end{array}\right. $$

f₂ is defined as follows: for every d ∈ D,

$$ f_{2}(d)=\left\{\begin{array}{ll} \alpha & \text{ if there is an }n \in \omega \text{ and an }\alpha \in 3\omega \text{ such that } d=A_{n,\alpha }\text{,}\\ 0& \text{ otherwise.} \end{array}\right. $$

f_ω is defined as follows: for every d ∈ D,

$$ f_{\omega }(d) = \!\left\{\!\!\begin{array}{ll} 1 & \text{ if there is an }n \in \omega \text{ and an }\alpha \in 3\omega \text{ such that } d=A_{n,\alpha }\text{ and }\alpha <\omega \text{,}\\ 2 & \text{ if there is an }n \in \omega \text{ and an }\alpha \in 3\omega \text{ such that } d = A_{n,\alpha }\text{ and }\omega \le \alpha <2\omega\text{,} \\ 3 & \text{ if there is an }n \!\in\! \omega \text{ and an }\alpha \!\in \!3\omega \text{ such that } d = A_{n,\alpha }\text{ and }2\omega \le \alpha <3\omega\text{,}\\ 0& \text{ otherwise.} \end{array}\right. $$

The restriction of I(≤_S) to I(S) is a linear order. If n≠m or α≠β, then A_n,α≠A_m,β. Hence, f₁, f₂ and f_ω are 1-place functions on D. For every n ∈ ω and every α ∈ 3ω, we say that A_n,α is in plane k iff f_ω(A_n,α) = k (of course, k ∈{1, 2, 3}); say that A_n,α is in line α; say that A_n,α is the n th sentence of line α; and say that line α is line t of plane k + 1 if α = ω × k + t and t ∈ ω. Call n the abscissa of A_n,α and α the ordinate of A_n,α. The relation I(V_e) holds between A_n,α and A_m,β iff, they are in the same line, or they are in the same plane and there are an odd number of lines between them. The relation I(V_o) holds between A_n,α and A_m,β iff, they are in the same plane, but not in the same line, and there are an even number of lines between them. Note that for every maximally consistent classical hypothesis h, ${\mathscr{M}}+h\models \varphi _{c}$.

We can use the following table to represent the sentences in I(S):

In the table above, a sentence in some plane is less than every sentence in its next plane (if exists); a sentence in some line is less than every sentence in its next line; and for any two different sentences in the same line, the one on the left is less than the one on the right.⁴ We say that A_n,α is before A_m,β, or A_m,β is afterA_n,α iff, A_n,α is less than A_m,β. Similarly, we say that a sentence is before (after) some line (plane), say that a line is before (after) another line, say that a plane is before (after) another plane, and so on.

Let h be a classical hypothesis that it declares the sentences in line α as follows: h declares the 0th sentence of line α true, h declares the 1th sentence of line α false, h declares the 2th sentence of line α true, h declares the 3th sentence of line α false, and so on. We can use the following table to represent the declared truth value of line α by h:⁵

For every classical hypothesis and every α ∈ 3ω, if the declared truth value of line α by this hypothesis enters a loop from some sentence, then we can use a table of the same kind to represent the declared truth value of line α by this hypothesis. Similarly, for every classical hypothesis and every k ∈{1, 2, 3}, if the declared truth value of each line of plane k by this hypothesis enters a loop from some sentence, and the declared truth value of plane k by this hypothesis enters a loop from some line, then we can use a table to represent the declared truth value of plane k by this hypothesis. For example, let $h^{\prime }$ be a classical hypothesis that it declares the sentences in plane k (where k ∈{1, 2, 3}) as follows: $h^{\prime }$ declares the sentences in line 0 of plane k true; $h^{\prime }$ declares the sentences in line 1 of plane k false; $h^{\prime }$ declares the sentences in line 2 of plane k with even abscissas true, and others false; and the declared truth value of each of the other lines of plane k by $h^{\prime }$ is the same as that of line 2 of plane k by $h^{\prime }$. We can use the following table to represent the declared truth value of plane k by $h^{\prime }$:

Conventions in this proof: if h is a classical hypothesis and φ is a sentence, then h⊧φ is an abbreviation for ${\mathscr{M}}+h\models \varphi$, and we say that h satisfies φ iff h⊧φ; when we mention a sentence A_n,α or nonquote name a_n,α, n ∈ ω and α ∈ 3ω by default; and when we mention plane k, k ∈{1, 2, 3} by default.

Claim 1. Let h be a classical hypothesis, and s be a function from the set of variables to D.

• (1) If h⊧∃xφ₁(x), then there is a unique element A_n,α in I(S) such that ${\mathscr{M}}+h,s(x\mid A_{n,\alpha })\models \varphi _{1}(x)$.⁶

• (2) If h⊧∃xφ₂(x), then there is a unique element A_0,α(α≠ 0,ω, 2ω) in I(S) such that ${\mathscr{M}}+h,s(x\mid A_{0,\alpha })\models \varphi _{2}(x)$.

• (3) If h⊧∃xφ₃(x), then there is a unique element A_0,α(α = 0,ω, 2ω) in I(S) such that ${\mathscr{M}}+h,s(x\mid A_{0,\alpha })\models \varphi _{3}(x)$.

• (4) If h⊧∃xφ₄(x), then there is a unique element A_0,α in I(S) such that ${\mathscr{M}}+h,s(x\mid A_{0,\alpha })\models \varphi _{4}(x)$.

• (5) If h⊧∃xφ₅(x), then there is a unique element k(k = 1, 2, 3) in D such that ${\mathscr{M}}+h,s(x\mid k)\models \varphi _{5}(x)$.

Proof of (1). If h⊧∃xφ₁(x), then there is an element d ∈ D such that ${\mathscr{M}}+h,s(x\mid d)\models \varphi _{1}(x)$. According to the first conjunct of φ₁(x), d is an element in I(S). Let d be A_n,α. According to the second conjunct of φ₁(x), all sentences in a plane before A_n,α are declared false by h. According to the fourth conjunct of φ₁(x), the sentences after A_n,α and A_n,α are declared true by h iff their abscissas are even. In particular, the sentences in a plane after A_n,α are declared true by h iff their abscissas are even. According to the third conjunct of φ₁(x), for each of sentences that is in the same plane with A_n,α and that is before A_n,α, its declared truth value by h is the same as that of A_n,α by h iff, it is in the same line with A_n,α, or there are an odd number of lines between it and A_n,α. Suppose that A_n,α is in plane k. If n is even, then the declared truth value of plane k by h is as follows:⁷

If n is odd, then the declared truth value of plane k by h is as follows:⁸

Obviously, for any d≠A_n,α, ${\mathscr{M}}+h,s(x\mid d)\not \models \varphi _{1}(x)$. So (1) holds.

Similarly, we can prove (2). By the definition of φ₂(x), we can easily prove that there is a unique element A_0,α(α≠ 0,ω, 2ω) in I(S) such that ${\mathscr{M}}+h,s(x\mid A_{0,\alpha })\models \varphi _{2}(x)$. Suppose that A_0,α is in plane k. All sentences in a plane before plane k are declared false by h. The sentences in a plane after plane k are declared true by h iff their abscissas are even. And the declared truth value of plane k by h is as follows:⁹

Similarly, we can prove (3). By the definition of φ₃(x), we can easily prove that there is a unique element A_0,α(α = 0,ω, 2ω) in I(S) such that ${\mathscr{M}}+h,s(x\mid A_{0,\alpha })\models \varphi _{3}(x)$. Suppose that A_0,α is in plane k. All sentences in a plane before plane k are declared false by h. The sentences in a plane after plane k are declared true by h iff their abscissas are even. And the declared truth value of plane k by h is as follows:¹⁰

Similarly, we can prove (4). By the definition of φ₄(x), we can easily prove that there is a unique element A_0,α in I(S) such that ${\mathscr{M}}+h,s(x\mid A_{0,\alpha })\models \varphi _{4}(x)$. Suppose that A_0,α is in plane k. All sentences in a plane before plane k are declared false by h. The sentences in a plane after plane k are declared true by h iff their abscissas are even. If α = ω × (k − 1) + n and n is even, then the declared truth value of plane k by h is as follows:¹¹

If α = ω × (k − 1) + n and n is odd, then the declared truth value of plane k by h is as follows:¹²

Similarly, we can prove (5). By the definition of φ₅(x), we can easily prove that there is a unique element k(k = 1, 2, 3) in D such that ${\mathscr{M}}+h,s(x\mid k)\models \varphi _{5}(x)$. All sentences in a plane before plane k are declared false by h. The sentences in a plane after plane k are declared true by h iff their abscissas are even. And all sentences in plane k are declared true by h.

Claim 2. Let U be the set {φ_c ∧∃xφ₁(x),φ_c ∧∃xφ₂(x),φ_c ∧∃xφ₃(x),φ_c ∧∃xφ₄(x),φ_c ∧∃xφ₅(x),φ_c ∧ φ₆,φ_others}. Let h be a classical hypothesis. Then there is a unique sentence φ ∈ U such that h satisfies φ.

The proof of Claim 2 is as follows. Note that if a classical hypothesis satisfies φ₆, then this hypothesis declares all sentences in I(S) false. From this and the proof of Claim 1, we can know that if there is an i ∈{1, 2, 3, 4, 5, 6} such that h satisfies ∃xφ_i(x), then for every j ∈{1, 2, 3, 4, 5, 6}, if j≠i then h does not satisfy ∃xφ_j(x).¹³ Then by the definition of φ_others, we can see that Claim 2 holds.

Claim 3. Suppose that h is a classical hypothesis.

• (1) For every n ∈ ω and α ∈ 3ω, if h⊧φ_c ∧ φ₁(a_n,α), then $\tau _{{\mathscr{M}}}(h)\models \varphi _{1}(a_{n+1,\alpha })$.¹⁴

• (2) For every α ∈ 3ω such that α≠ 0,ω, 2ω, if h⊧φ_c ∧ φ₂(a_0,α), then $\tau _{{\mathscr{M}}}(h)\models \varphi _{1}(a_{1,\alpha })$.

• (3) For α = 0,ω, 2ω, if h⊧φ_c ∧ φ₃(a_0,α), then $\tau _{{\mathscr{M}}}(h)\models \varphi _{4}(a_{0,\alpha +1})$.

• (4) For every α ∈ 3ω, if h⊧φ_c ∧ φ₄(a_0,α), then $\tau _{{\mathscr{M}}}(h)\models \varphi _{4}(a_{0,\alpha +1})$.

• (5) For k = 1, 2, if $h\models \varphi _{c} \land \varphi _{5}(\overline {k})$, then $\tau _{{\mathscr{M}}}(h)\models \varphi _{1}(a_{1,\omega \times k})$. If $h\models \varphi _{c} \land \varphi _{5}(\overline {3})$, then $\tau _{{\mathscr{M}}}(h)\models \varphi _{6}$.

• (6) If h⊧φ_c ∧ φ₆, then $\tau _{{\mathscr{M}}}(h)\models \varphi _{6}$.

• (7) If h⊧φ_others, then $\tau _{{\mathscr{M}}}(h)\models \varphi _{1}(a_{0,0})$.

Proof of (1). Suppose that h⊧φ_c ∧ φ₁(a_n,α). By Claim 1, for any A_m,β≠A_n,α, h⊮φ₁(a_m,β). By Claim 2, for any i ∈{2, 3, 4, 5, 6}, h⊮∃xφ_i(x), and h⊮φ_others. Suppose that A_n,α is in plane k. Obviously, $\tau _{{\mathscr{M}}}(h)$ satisfies the first conjunct of φ₁(a_n+ 1,α). Let A_m,β be any sentence in I(S) with f_ω(A_m,β) < k. By the definition of A_m,β, we know that h⊮A_m,β. So $\tau _{{\mathscr{M}}}(h)(A_{m,\beta })={\mathbf {f}}$. Hence $\tau _{{\mathscr{M}}}(h)$ satisfies the second conjunct of φ₁(a_n+ 1,α). Let A_m,β be any sentence in I(S) that is greater than or equal to A_n+ 1,α. By the definition of A_m,β (the second disjunct), we know that h⊧A_m,β iff $h\models E(\overline {m})$. So $\tau _{{\mathscr{M}}}(h)(A_{m,\beta })={\mathbf {t}}$ iff m is even. Hence $\tau _{{\mathscr{M}}}(h)$ satisfies the fourth conjunct of φ₁(a_n+ 1,α). In particular, $\tau _{{\mathscr{M}}}(h)(A_{n+1,\alpha })={\mathbf {t}}$ iff $h\models E(\overline {n+1})$. By the definition of A_n,α (the first disjunct), we know that h⊧A_n,α iff $h\models O(\overline {n})$. So $\tau _{{\mathscr{M}}}(h)$ declares A_n,α and A_n+ 1,α the same truth value. Let A_m,β be any sentence in I(S) that is less than A_n,α and such that f_ω(A_m,β) = k. By the definition of A_m,β (the first disjunct), h⊧A_m,β iff, $h\models E(\overline {n})\land V_{o} a_{m,\beta } a_{n,\alpha }$ or $h\models O(\overline {n})\land V_{e} a_{m,\beta } a_{n,\alpha }$. Note that h⊧V_ea_m,βa_n,α iff h⊮V_oa_m,βa_n,α, and that $h\models E(\overline {n})$ iff $h\not \models O(\overline {n})$. So if h⊧V_ea_m,βa_n,α, then h⊧A_m,β iff $h\models O(\overline {n})$; and if h⊮V_ea_m,βa_n,α, then h⊧A_m,β iff $h\not \models O(\overline {n})$. Hence, if h⊧V_ea_m,βa_n,α, then h⊧A_m,β iff h⊧A_n,α; and if h⊮V_ea_m,βa_n,α, then h⊧A_m,β iff h⊮A_n,α. Therefore, the relation I(V_e) holds between A_m,β and A_n,α iff, $\tau _{{\mathscr{M}}}(h)$ declares them the same truth value. Note that $\tau _{{\mathscr{M}}}(h)$ declares A_n,α and A_n+ 1,α the same truth value. Hence $\tau _{{\mathscr{M}}}(h)$ satisfies the third conjunct of φ₁(a_n+ 1,α). Therefore, (1) holds.

Similarly, we can prove (2)–(7).

Claim 4. (1) For every classical hypothesis h, $\tau _{{\mathscr{M}}}(h)$ satisfies φ₁(a_0,0), φ₁(a_n,α)(n ≥ 1), φ₄(a_0,α)(α≠ 0,ω, 2ω) or φ₆.¹⁵

(2) If a classical hypothesis h is a fixed point of $\tau _{{\mathscr{M}}}$, then h satisfies φ₆.

Claim 4 is a corollary of Claim 1, Claim 2 and Claim 3.

Claim 5. For every classical hypothesis h, if there is an i ∈{1, 2, 3, 4, 5, 6} such that h satisfies ∃xφ_i(x), then h has the following six properties.

(I) h declares each line , , , or , where sequences , , , and are defined as follows:

• (II) The total number of occurrences of or is less than or equal to 1.¹⁶

• (III) only occurs after , , and .¹⁷

• (IV) and only occur after and ; if one line is , then its previous line (if exists) is ; and if one line is , then its previous line (if exists) is .¹⁸

• (V) For every plane k, if there are successive occurrences of or successive occurrences of in plane k, then plane k is one of the following four cases:¹⁹

  

• (VI) If some plane is not all- , then the previous plane of it (if exists) is all- .

We can see that Claim 5 holds from Claim 1 and its proof and the definition of φ₆.

Claim 6. For every classical hypothesis h, if h has all of the properties (I)–(VI) in Claim 5, then there is an i ∈{1, 2, 3, 4, 5, 6} such that h satisfies ∃xφ_i(x).

We prove Claim 6 in two cases. (1) Suppose that h declares each plane all- or all- . Since h has the property (III), must occur after . So there are four possibilities: the three planes are all- ; the first two planes are all- , and the third plane is all- ; the first plane is all- , and the last two planes are all- ; and the three planes are all- . In these four cases, h satisfies φ₆, φ₁(a_0,2ω), φ₁(a_0,ω) and φ₁(a_0,0), respectively.

(2) Suppose that there is some plane that h declares neither all- nor all- . Since h has the property (VI), if some plane is neither all- nor all- , then the planes before it are all- , and the planes after it are all- . So there is at most one such plane. Let plane k be the plane that h declares neither all- nor all- . Next, we prove in two subcases that there is an i ∈{1, 2, 3, 4, 5, 6} such that h satisfies ∃xφ_i(x).

(2.1) Suppose that there is no occurrence of , or in plane k. Since h has the property (I), only and occur in plane k. Since h has the property (III), must occur before . And since h has the property (V), there are no successive occurrences of in plane k. So the declared truth values of lines in plane k are as follows: line 0 of plane k is , and the other lines of plane k are . Hence h satisfies φ₁(a_{0,ω×(k− 1)+ 1}).

(2.2) Suppose that there is an occurrence of , or in plane k. (2.2.1) Suppose that there is an occurrence of in plane k. Since h has the property (II), there is no occurrence of in plane k, and occurs only once (in plane k). Suppose that it is line t of plane k that is . And since h has the properties (IV) and (I), we know that: the lines before line t of plane k must be or ; the lines after line t of plane k must be ; and if t≠ 0, then line t − 1 of plane k is . Since h has the property (V), if t ≥ 2, then in plane k, there are neither successive occurrences of nor successive occurrences of before . So, if t ≥ 2, then in plane k, and alternately occur before . In either case (t ≤ 1 or t ≥ 2), h satisfies ∃xφ₁(x). (2.2.2) Suppose that there is an occurrence of in plane k. Similar to (2.2.1), we can prove that h satisfies ∃xφ₁(x). (2.2.3) Suppose that there is no occurrence of or in plane k. Then there is an occurrence of in plane k. (2.2.3.1) Suppose that there is no occurrence of in plane k. Since h has the properties (V) and (I), and alternately occur in plane k, or plane k is one of the other three cases besides all- shown in the property (V). So h satisfies φ₃(a_{0,ω×(k− 1)}), ∃xφ₄(x) or $\varphi _{5}(\overline {k})$. (2.2.3.2) Suppose that there is an occurrence of in plane k. Since h has the property (III), must occur after and . Suppose that the earliest occurrence of is in line 1 of plane k. Line 0 of plane k can only be , and so h satisfies φ₂(a_{0,ω×(k− 1)+ 1}). Suppose that the earliest occurrence of is after line 1 of plane k. Since h has the properties (V) and (I), in plane k, and alternately occur before the earliest occurrence of . So h satisfies ∃xφ₁(x) or ∃xφ₂(x).

In conclusion, Claim 6 holds.

Claim 7. For every classical hypothesis h, if h satisfies φ_c ∧ φ_others, then h does not have at least one of the six properties (I)–(VI).

By Claim 6 and the definition of φ_others, we can easily see that Claim 7 holds.

Claim 8. Let h₀ be the empty hypothesis, i.e., h₀(d) = n for every d ∈ D. Suppose that h is a maximally consistent classical hypothesis, and that h⊧φ_others. Then $h\not \ge {\sigma _{2}}_{{\mathscr{M}}}(h_{0})$.

The proof of Claim 8 is as follows. By Claim 7, h does not have at least one of the six properties (I)–(VI). It suffices to prove that no matter which of the six properties h does not have, we can find a sentence that is declared true by h and that is declared false by ${\sigma _{2}}_{{\mathscr{M}}}(h_{0})$. We prove this proposition in six cases.

• (1) Suppose that h does not have the property (I). Then there is at least one line such that the declared truth value of this line by h is different from , , , and . Let line α be such a line. So there is a least natural number n such that the restriction to n of the declared truth value of line α by h is different from $\upharpoonright n$, $\upharpoonright n$, $\upharpoonright n$, $\upharpoonright n$ and $\upharpoonright n$. Define the sets S₁ and S₂, and sentence ϕ as follows:

  $$ \begin{array}{@{}rcl@{}} &&S_{1}={~}_{df}\{ A_{m,\alpha} \mid m\le n, h (A_{m,\alpha} )=\mathbf{t} \}\\ &&S_{2}={~}_{df}\{ A_{m,\alpha} \mid m\le n,h (A_{m,\alpha} )={\mathbf{f}} \}\\ &&\phi={~}_{df}\bigwedge (S_{1}\cup \{\neg A_{m,\alpha} \mid A_{m,\alpha} \in S_{2}\}) \end{array} $$

  Since h is maximally consistent, h declares ϕ true. By Claim 4 and the definitions of φ₁(x), φ₄(x) and φ₆, we know that for any classical hypothesis $h^{\prime }$, the restriction to n of the declared truth value of line α by $\tau _{{\mathscr{M}}}(h^{\prime })$ is different from the restriction to n of the declared truth value of line α by h. So, it can not be the case that $\tau _{{\mathscr{M}}}(h^{\prime })$ declares all sentences in S₁ true and declares all sentences in S₂ false. Therefore, it can not be the case that $\tau _{{\mathscr{M}}}(h^{\prime })$ declares ϕ true. Hence ${\sigma _{2}}_{{\mathscr{M}}}(h_{0})(\phi )={\mathbf {f}}$.

• (2) Suppose that h does not have the property (II). Then the total number of occurrences of or is greater than or equal to 2. (2.1) Suppose that both and occur. Let line α be some line that is declared by h, and let line β be some line that is declared by h. Of course, α≠β. Let n be the least natural number such that the n th sentence of line α (i.e., A_n,α) is declared true by h. Let B₁ be the set consisting of the first n + 1 sentences of line α. Let m be the least natural number such that the m th sentence of line β (i.e., A_m,β) is declared false by h. Let B₂ be the set consisting of the first m + 1 sentences of line β. We can represent B₁ and B₂ as follows:

  

  Define the sets S₁ and S₂, and sentence ϕ as follows:

  $$ \begin{array}{@{}rcl@{}} &&S_{1}={~}_{df}\{ A_{m,\alpha} \mid A_{m,\alpha} \in B_{1} \cup B_{2}, h (A_{m,\alpha} )=\mathbf{t} \}\\ &&S_{2}={~}_{df}\{ A_{m,\alpha} \mid A_{m,\alpha} \in B_{1} \cup B_{2},h (A_{m,\alpha} )={\mathbf{f}} \}\\ &&\phi={~}_{df}\bigwedge (S_{1}\cup \{\neg A_{m,\alpha} \mid A_{m,\alpha} \in S_{2}\}) \end{array} $$

  Since h is maximally consistent, h declares ϕ true. By Claim 4 and the definitions of φ₁(x), φ₄(x) and φ₆, we know that for any classical hypothesis $h^{\prime }$, it can not be the case that $\tau _{{\mathscr{M}}}(h^{\prime })$ declares all sentences in S₁ true and declares all sentences in S₂ false. So ${\sigma _{2}}_{{\mathscr{M}}}(h_{0})(\phi )={\mathbf {f}}$. (2.2) Suppose that the number of occurrences of is greater than or equal to 2. Choose two lines that are declared by h. Let the line before the other be line α, and let the other line be line β. Of course, α < β. Let n be the least natural number such that the n th sentence of line α is declared true by h. Let B₁ be the set consisting of the first n + 1 sentences of line α. Let B₂ be the set consisting of the first two sentences of line β. Similar to (2.1), B₁ and B₂ determine a sentence that h declares true and that ${\sigma _{2}}_{{\mathscr{M}}}(h_{0})$ declares false. (2.3) Suppose that the number of occurrences of is greater than or equal to 2. Choose two lines that are declared by h. Let the line before the other be line α, and let the other line be line β. Of course, α < β. Let n be the least natural number such that the n th sentence of line α is declared false by h. Let B₁ be the set consisting of the first n + 1 sentences of line α. Let B₂ be the set consisting of the first two sentences of line β. Similar to (2.1), B₁ and B₂ determine a sentence that h declares true and that ${\sigma _{2}}_{{\mathscr{M}}}(h_{0})$ declares false.

• (3) Suppose that h does not have the property (III). Then there is a line such that this line is and there is a line after this line such that it is , , or . Let line α be such a line. And let line β be a line after line α that is , , or . Of course, α < β. Let B₁ be the set consisting of the first two sentences of line α. Let B₂ be the set consisting of the first two sentences of line β. Similar to (2.1), B₁ and B₂ determine a sentence that h declares true and that ${\sigma _{2}}_{{\mathscr{M}}}(h_{0})$ declares false.

• (4) Suppose that h does not have the property (IV). Then there are three possibilities: (4.1) there is a line such that this line is or and there is a line after this line such that it is or ; (4.2) there is a line that is , but its previous line is not ; (4.3) there is a line that is , but its previous line is not . In case (4.1), let line α be a line that is or and such that there is a line after line α that is or . And let line β be a line that is after line α and that is or . α < β. Let n be the greatest natural number such that the first n sentences of line α are declared the same truth value by h. Let B₁ be the set consisting of A_n− 1,α and A_n,α. Let B₂ be the set consisting of the first two sentences of line β. Similar to (2.1), B₁ and B₂ determine a sentence that h declares true and that ${\sigma _{2}}_{{\mathscr{M}}}(h_{0})$ declares false. In case (4.2), because of case (2) and case (3), we only need to consider the following case: there is a line that is , but its previous line is . Let line α + 1 be such a line. Its previous line, i.e., line α, is . Define sets B₁ and B₂ as follows:

  

  Similar to (2.1), B₁ and B₂ determine a sentence that h declares true and that ${\sigma _{2}}_{{\mathscr{M}}}(h_{0})$ declares false. In case (4.3), because of case (2) and case (3), we only need to consider the following case: there is a line that is , but its previous line is . Let line α + 1 be such a line. Its previous line, i.e., line α, is . Define sets B₁ and B₂ as follows:

  

  Similar to (2.1), B₁ and B₂ determine a sentence that h declares true and that ${\sigma _{2}}_{{\mathscr{M}}}(h_{0})$ declares false.

• (5) Suppose that h does not have the property (V). Then there is a plane such that there are successive occurrences of or successive occurrences of in this plane and this plane is not one of the four cases shown in the property (V). Let plane k be such a plane. (5.1) Suppose that there are successive occurrences of in plane k. Obviously, plane k is not one of the following two cases:

  

  Choose two adjacent lines in plane k that are declared by h. Let them be line α and line α + 1 respectively. Then ω × (k − 1) ≤ α < ω × k. We consider in four cases: (5.1.1) in plane k, there is an occurrence of , or ; (5.1.2) in plane k, there is no occurrence of , or , and there is a line before line α that is ; (5.1.3) in plane k, there is no occurrence of , or , and all lines before line α are , and there are successive occurrences of after line α + 1; (5.1.4) in plane k, there is no occurrence of , or , and all lines before line α are , and there are no successive occurrences of after line α + 1.

  In case (5.1.1), because of case (3) and case (4), we only need to consider the following case: in plane k, there is an occurrence of , or after line α + 1. Let line β be a line in plane k that is , or and that is after line α + 1. Then α + 1 < β < ω × k. Let B₁ be the set consisting of A_0,α. Let B₂ be the set consisting of the first two sentences of line α + 1. Choose a sentence in line β that is declared true by h, and another sentence in line β that is declared false by h. Let B₃ be the set consisting of these two sentences. Similar to (2.1), B₁, B₂ and B₃ determine a sentence that h declares true and that ${\sigma _{2}}_{{\mathscr{M}}}(h_{0})$ declares false.

  In case (5.1.2), let line β be a line in plane k that is and that is before line α. Then ω × (k − 1) ≤ β < α. Let B₁ be the set consisting of A_0,β. Let B₂ be the set consisting of A_0,α. Let B₃ be the set consisting of the first two sentences of line α + 1. Similar to (2.1), B₁, B₂ and B₃ determine a sentence that h declares true and that ${\sigma _{2}}_{{\mathscr{M}}}(h_{0})$ declares false.

  In case (5.1.3), let line β and line β + 1 be lines in plane k that are declared by h. Then α + 1 < β < ω × k. Let B₁ be the set consisting of A_0,α. Let B₂ be the set consisting of A_0,α+ 1. Let B₃ be the set consisting of A_0,β. Let B₄ be the set consisting of A_0,β+ 1. Similar to (2.1), B₁, B₂, B₃ and B₄ determine a sentence that h declares true and that ${\sigma _{2}}_{{\mathscr{M}}}(h_{0})$ declares false.

  In case (5.1.4), there is a unique line in plane k such that, in plane k, all lines before it are , and it is , and in plane k, starting with this line, and alternately occur. Let line β be this line. Then α + 1 < β < ω × k. Since h does not have the property (V), there is not an odd number n greater than or equal to 3 such that β = ω × (k − 1) + n. Then there is an even number n such that β = ω × (k − 1) + n. Let B₁ be the set consisting of A_0,α. Let B₂ be the set consisting of A_0,α+ 1. Let B₃ be the set consisting of A_0,β. Similar to (2.1), B₁, B₂ and B₃ determine a sentence that h declares true and that ${\sigma _{2}}_{{\mathscr{M}}}(h_{0})$ declares false.

  (5.2) Suppose that there are successive occurrences of in plane k. The proof is similar to (5.1).

• (6) Suppose that h does not have the property (VI). Then there is a plane that is not all- and its previous plane is not all- . Let plane k be such a plane. Then k ∈{2, 3}. Since plane k is not all- , then there is a sentence in plane k such that either its abscissa is even and it is declared false by h, or its abscissa is odd and it is declared true by h. Choose such a sentence, and let B₁ be the set consisting of it. Since plane k − 1 is not all- , then there is a sentence in plane k − 1 that is declared true by h. Choose such a sentence, and let B₂ be the set consisting of it. Similar to (2.1), B₁ and B₂ determine a sentence that h declares true and that ${\sigma _{2}}_{{\mathscr{M}}}(h_{0})$ declares false.

In conclusion, Claim 8 holds.

Claim 9. ${\mathbf {T}}^{lfp,\sigma _{2}}$ dictates that truth behaves like a classical concept in ${\mathscr{M}}$.

The proof of Claim 9 is as follows. Let h₀ be the empty hypothesis. Let ${\sigma _{2}}_{{\mathscr{M}}}^{\alpha }(h_{0})(\alpha \in On)$ be the unique ordinal-length sequence of hypotheses obtained in the construction of the least fixed point of ${\sigma _{2}}_{{\mathscr{M}}}$ according to Kripke’s inductive construction, beginning with h₀. To be specific, ${\sigma _{2}}_{{\mathscr{M}}}^{\alpha }(h_{0})(\alpha \in On)$ is defined as follows: if α = 0, then ${\sigma _{2}}_{{\mathscr{M}}}^{\alpha }(h_{0})=h_{0}$; if α = β + 1, then ${\sigma _{2}}_{{\mathscr{M}}}^{\alpha }(h_{0})={\sigma _{2}}_{{\mathscr{M}}}({\sigma _{2}}_{{\mathscr{M}}}^{\beta }(h_{0}))$; and if α is a limit ordinal, then ${\sigma _{2}}_{{\mathscr{M}}}^{\alpha }(h_{0})=\bigcup _{\beta < \alpha }{\sigma _{2}}_{{\mathscr{M}}}^{\beta }(h_{0})$. We know that for all ordinals α and β, if β ≤ α, then ${\sigma _{2}}_{{\mathscr{M}}}^{\beta }(h_{0})\le {\sigma _{2}}_{{\mathscr{M}}}^{\alpha }(h_{0})$; that for every ordinal α, ${\sigma _{2}}_{{\mathscr{M}}}^{\alpha }(h_{0})$ is strongly consistent; and that for every ordinal α, there is a maximally consistent classical hypothesis such that it is greater than or equal to ${\sigma _{2}}_{{\mathscr{M}}}^{\alpha }(h_{0})$.

By the claims above, the following propositions hold.

• (1) By Claim 8, any maximally consistent classical hypothesis greater than or equal to ${\sigma _{2}}_{{\mathscr{M}}}^{1}(h_{0})$ does not satisfy φ_others.

• (2) By (1), Claim 2 and Claim 1, every maximally consistent classical hypothesis greater than or equal to ${\sigma _{2}}_{{\mathscr{M}}}^{1}(h_{0})$ satisfies φ₁(a_n,α), φ₂(a_0,β)(β≠ 0,ω, 2ω), φ₃(a_0,γ)(γ = 0,ω, 2ω), φ₄(a_0,λ), $\varphi _{5}(\overline {k})(k=1,2,3)$ or φ₆.

• (3) By (2) and Claim 3 (and the definitions of φ₁(x), φ₄(x) and φ₆), for every maximally consistent classical hypothesis $h\ge {\sigma _{2}}_{{\mathscr{M}}}^{1}(h_{0})$, $\tau _{{\mathscr{M}}}(h)(A_{0,0})=\tau _{{\mathscr{M}}}(h)(A_{1,0})$, so $h\models A_{0,0}\leftrightarrow A_{1,0}$. Hence ${\sigma _{2}}_{{\mathscr{M}}}^{2}(h_{0})(A_{0,0}\leftrightarrow A_{1,0})=\mathbf {t}$. But every maximally consistent classical hypothesis that satisfies φ₁(a_0,0) declares $A_{0,0}\leftrightarrow A_{1,0}$ false. Therefore, any maximally consistent classical hypothesis greater than or equal to ${\sigma _{2}}_{{\mathscr{M}}}^{2}(h_{0})$ does not satisfy φ₁(a_0,0).

• (4) By (2), (3) and Claim 3, for every maximally consistent classical hypothesis $h\ge {\sigma _{2}}_{{\mathscr{M}}}^{2}(h_{0})$, $h\models A_{1,0}\leftrightarrow A_{2,0}$. Hence ${\sigma _{2}}_{{\mathscr{M}}}^{3}(h_{0})(A_{1,0}\leftrightarrow A_{2,0})=\mathbf {t}$. Therefore, any maximally consistent classical hypothesis greater than or equal to ${\sigma _{2}}_{{\mathscr{M}}}^{3}(h_{0})$ does not satisfy φ₁(a_1,0).

• (5) Similar to (4), for every natural number n ≥ 2, any maximally consistent classical hypothesis greater than or equal to ${\sigma _{2}}_{{\mathscr{M}}}^{n+2}(h_{0})$ does not satisfy φ₁(a_n,0).

• (6) By (2)–(5), every maximally consistent classical hypothesis greater than or equal to ${\sigma _{2}}_{{\mathscr{M}}}^{\omega }(h_{0})$ satisfies φ₁(a_n,α)(α ≥ 1), φ₂(a_0,β)(β≠ 0,ω, 2ω), φ₃(a_0,γ)(γ = 0,ω, 2ω), φ₄(a_0,λ), $\varphi _{5}(\overline {k})(k=1,2,3)$ or φ₆.

• (7) By (6) and Claim 3, for every maximally consistent classical hypothesis $h\ge {\sigma _{2}}_{{\mathscr{M}}}^{\omega }(h_{0})$, $h\models A_{0,1}\leftrightarrow A_{1,1}$. So ${\sigma _{2}}_{{\mathscr{M}}}^{\omega +1}(h_{0})(A_{0,1}\leftrightarrow A_{1,1})=\mathbf {t}$. Hence, any maximally consistent classical hypothesis greater than or equal to ${\sigma _{2}}_{{\mathscr{M}}}^{\omega +1}(h_{0})$ satisfies neither φ₁(a_0,1) nor φ₂(a_0,1).

• (8) Similar to (4)–(5), for every natural number n ≥ 1, any maximally consistent classical hypothesis greater than or equal to ${\sigma _{2}}_{{\mathscr{M}}}^{\omega +n+1}(h_{0})$ does not satisfy φ₁(a_n,1).

• (9) By (6)–(8), every maximally consistent classical hypothesis greater than or equal to ${\sigma _{2}}_{{\mathscr{M}}}^{2\omega }(h_{0})$ satisfies φ₁(a_n,α)(α ≥ 2), φ₂(a_0,β)(β ≥ 2,β≠ω, 2ω), φ₃(a_0,γ)(γ = 0,ω, 2ω), φ₄(a_0,λ), $\varphi _{5}(\overline {k})(k=1,2,3)$ or φ₆.

• (10) Similar to (7)–(9), for every natural number m ≥ 2 and every natural number i, any maximally consistent classical hypothesis greater than or equal to ${\sigma _{2}}_{{\mathscr{M}}}^{\omega \times m+1}(h_{0})$ does not satisfy φ₂(a_0,m); any maximally consistent classical hypothesis greater than or equal to ${\sigma _{2}}_{{\mathscr{M}}}^{\omega \times m+i+1}(h_{0})$ does not satisfy φ₁(a_i,m); and every maximally consistent classical hypothesis greater than or equal to ${\sigma _{2}}_{{\mathscr{M}}}^{\omega \times (m+1)}(h_{0})$ satisfies φ₁(a_n,α)(α ≥ m + 1), φ₂(a_0,β)(β ≥ m + 1,β≠ω, 2ω), φ₃(a_0,γ)(γ = 0,ω, 2ω), φ₄(a_0,λ), $\varphi _{5}(\overline {k})(k=1,2,3)$ or φ₆.

• (11) By (10), every maximally consistent classical hypothesis greater than or equal to ${\sigma _{2}}_{{\mathscr{M}}}^{\omega ^{2}}(h_{0})$ satisfies φ₁(a_n,α)(α ≥ ω), φ₂(a_0,β)(β > ω,β≠ 2ω), φ₃(a_0,γ)(γ = 0,ω, 2ω), φ₄(a_0,λ), $\varphi _{5}(\overline {k})(k=1,2,3)$ or φ₆.

• (12) By (11) and Claim 3, ${\sigma _{2}}_{{\mathscr{M}}}^{\omega ^{2}+1}(h_{0})(A_{0,0}\leftrightarrow A_{0,1})=\mathbf {t}$, so any maximally consistent classical hypothesis greater than or equal to ${\sigma _{2}}_{{\mathscr{M}}}^{\omega ^{2}+1}(h_{0})$ satisfies neither φ₃(a_0,0) nor φ₄(a_0,0).

• (13) Similar to (12), for every natural number n ≥ 1, ${\sigma _{2}}_{{\mathscr{M}}}^{\omega ^{2}+n+1}(h_{0})(A_{0,n}\leftrightarrow A_{0,n+1})=\mathbf {t}$, so any maximally consistent classical hypothesis greater than or equal to ${\sigma _{2}}_{{\mathscr{M}}}^{\omega ^{2}+n+1}(h_{0})$ does not satisfy φ₄(a_0,n).

• (14) By (11)–(13), every maximally consistent classical hypothesis greater than or equal to ${\sigma _{2}}_{{\mathscr{M}}}^{\omega ^{2}+\omega }(h_{0})$ satisfies φ₁(a_n,α)(α ≥ ω), φ₂(a_0,β)(β > ω,β≠ 2ω), φ₃(a_0,γ)(γ = ω, 2ω), φ₄(a_0,λ)(λ ≥ ω), $\varphi _{5}(\overline {k})(k=1,2,3)$ or φ₆.

• (15) By (14) and Claim 3, for every maximally consistent classical hypothesis $h\ge {\sigma _{2}}_{{\mathscr{M}}}^{\omega ^{2}+\omega }(h_{0})$, $h\models A_{0,\omega }\leftrightarrow A_{1,\omega }$. So any maximally consistent classical hypothesis greater than or equal to ${\sigma _{2}}_{{\mathscr{M}}}^{\omega ^{2}+\omega +1}(h_{0})$ satisfies neither $\varphi _{5}(\overline {1})$ nor φ₁(a_0,ω).

• (16) Similar to (4)–(5), for every natural number n ≥ 1, any maximally consistent classical hypothesis greater than or equal to ${\sigma _{2}}_{{\mathscr{M}}}^{\omega ^{2}+\omega +n+1}(h_{0})$ does not satisfy φ₁(a_n,ω).

• (17) By (14)–(16), every maximally consistent classical hypothesis greater than or equal to ${\sigma _{2}}_{{\mathscr{M}}}^{\omega ^{2}+\omega +\omega }(h_{0})$, i.e., ${\sigma _{2}}_{{\mathscr{M}}}^{\omega ^{2}+2\omega }(h_{0})$, satisfies φ₁(a_n,α)(α ≥ ω + 1), φ₂(a_0,β)(β > ω,β≠ 2ω), φ₃(a_0,γ)(γ = ω, 2ω), φ₄(a_0,λ)(λ ≥ ω), $\varphi _{5}(\overline {k})(k=2,3)$ or φ₆.

• (18) Similar to (7)–(11), every maximally consistent classical hypothesis greater than or equal to ${\sigma _{2}}_{{\mathscr{M}}}^{\omega ^{2}+\omega +\omega ^{2}}(h_{0})$, i.e., ${\sigma _{2}}_{{\mathscr{M}}}^{2\omega ^{2}}(h_{0})$, satisfies φ₁(a_n,α)(α ≥ 2ω), φ₂(a_0,β)(β > 2ω), φ₃(a_0,γ)(γ = ω, 2ω), φ₄(a_0,λ)(λ ≥ ω), $\varphi _{5}(\overline {k})(k=2,3)$ or φ₆.

• (19) Similar to (12)–(14), every maximally consistent classical hypothesis greater than or equal to ${\sigma _{2}}_{{\mathscr{M}}}^{2\omega ^{2}+\omega }(h_{0})$ satisfies φ₁(a_n,α)(α ≥ 2ω), φ₂(a_0,β)(β > 2ω), φ₃(a_0,2ω), φ₄(a_0,λ)(λ ≥ 2ω), $\varphi _{5}(\overline {k})(k=2,3)$ or φ₆.

• (20) Similar to (15), any maximally consistent classical hypothesis greater than or equal to ${\sigma _{2}}_{{\mathscr{M}}}^{2\omega ^{2}+\omega +1}(h_{0})$ satisfies neither $\varphi _{5}(\overline {2})$ nor φ₁(a_0,2ω).

• (21) Similar to (16)–(18), every maximally consistent classical hypothesis greater than or equal to ${\sigma _{2}}_{{\mathscr{M}}}^{2\omega ^{2}+\omega +\omega ^{2}}(h_{0})$, i.e., ${\sigma _{2}}_{{\mathscr{M}}}^{3\omega ^{2}}(h_{0})$, satisfies φ₃(a_0,2ω), φ₄(a_0,λ)(λ ≥ 2ω), $\varphi _{5}(\overline {3})$ or φ₆.

• (22) Similar to (19), every maximally consistent classical hypothesis greater than or equal to ${\sigma _{2}}_{{\mathscr{M}}}^{3\omega ^{2}+\omega }(h_{0})$ satisfies $\varphi _{5}(\overline {3})$ or φ₆.

• (23) By (22) and Claim 3, for any two sentences A_n,α and A_m,β in I(S), ${\sigma _{2}}_{{\mathscr{M}}}^{3\omega ^{2}+\omega +1}(h_{0})(A_{n,\alpha })={\mathbf {f}}$ and ${\sigma _{2}}_{{\mathscr{M}}}^{3\omega ^{2}+\omega +1}(h_{0})(A_{n,\alpha }\leftrightarrow A_{m,\beta })=\mathbf {t}$.

Hence for any two sentences A_n,α and A_m,β in I(S), $lfp({\sigma _{2}}_{{\mathscr{M}}})(A_{n,\alpha })={\mathbf {f}}$ and $lfp({\sigma _{2}}_{{\mathscr{M}}})(A_{n,\alpha }\leftrightarrow A_{m,\beta })=\mathbf {t}$. Let Y be the set $I(S)\cup \{A_{n,\alpha }\leftrightarrow A_{m,\beta }\mid A_{n,\alpha }, A_{m,\beta }\in I(S) \}$. Then $Y \subseteq \{ A \in Sent({{\mathscr{L}}}^{+})\mid A\in {\mathbf {V}}_{{\mathscr{M}}}^{lfp,\sigma _{2}}$ or $\neg A\in {\mathbf {V}}_{{\mathscr{M}}}^{lfp,\sigma _{2}}\}$. ${\mathscr{M}}$ is $(Sent({{\mathscr{L}}}^{+})\setminus Y)$-neutral. By Theorem 4, we know ${\mathbf {T}}^{lfp,\sigma _{2}} \le _3{\mathbf {T}}^{lfp,\sigma _{2}}$. So ${\mathbf {T}}^{lfp,\sigma _{2}}$ dictates that truth behaves like a classical concept in ${\mathscr{M}}$.

Let ${\mathscr{S}}$ be a sequence of classical hypotheses, and $lh({\mathscr{S}})$ be a limit ordinal or the class of all ordinals On. We say that line β is stably in ${\mathscr{S}}$ iff, the sentences in line β with even abscissas are stably true in ${\mathscr{S}}$, and the sentences in line β with odd abscissas are stably false in ${\mathscr{S}}$. We say that line β is unstable in ${\mathscr{S}}$ iff each sentence in line β is unstable in ${\mathscr{S}}$. For every limit ordinal $\alpha \le lh({\mathscr{S}})$, we say that line β is unstable up to α in ${\mathscr{S}}$ iff each sentence in line β is unstable up to α in ${\mathscr{S}}$. We say that plane k is stably all- in ${\mathscr{S}}$ iff, the sentences in plane k with even abscissas are stably true in ${\mathscr{S}}$, and the sentences in plane k with odd abscissas are stably false in ${\mathscr{S}}$. We say that plane k is unstable in ${\mathscr{S}}$ iff each sentence in plane k is unstable in ${\mathscr{S}}$. For every limit ordinal $\alpha \le lh({\mathscr{S}})$, we say that plane k is unstable up to α in ${\mathscr{S}}$ iff each sentence in plane k is unstable up to α in ${\mathscr{S}}$.

Claim 10. Let λ be a limit ordinal, and ${\mathscr{S}}$ be a λ-length revision sequence.²⁰ Suppose that there is no hypothesis in ${\mathscr{S}}$ that satisfies φ_c ∧ φ₆. Then if there is a sentence A_n,α ∈ I(S) such that A_n,α is stable in ${\mathscr{S}}$ and all planes after A_n,α are stably all- in ${\mathscr{S}}$, then α ≥ 1, and line α and all lines after line α are stably in ${\mathscr{S}}$.

The proof of Claim 10 is as follows. Let δ be the supremum of the set {β∣A_n,α or some sentence in a plane after A_n,α is stable in ${\mathscr{S}}$ from stage β}. Let γ = δ + 1. Note that ${\mathscr{S}}_{\gamma }$ does not satisfy φ_c ∧ φ₆. By Claim 3 and Claim 4, α ≥ 1 and ${\mathscr{S}}_{\gamma }$ satisfies φ₁(a_p,𝜖)(p ≥ 1,𝜖 < α), φ₁(a_0,0) or φ₄(a_0,ε)(ε < ω × (f_ω(A_n,α) − 1), ε≠ 0,ω), otherwise the declared truth value of A_n,α would change after a finite number of applications of $\tau _{{\mathscr{M}}}$ to ${\mathscr{S}}_{\gamma }$.²¹ Next, we prove by transfinite induction that for every γ ≤ β < λ, ${\mathscr{S}}_{\beta }$ declares line α and all lines after line α . For β = γ, the proposition holds, because ${\mathscr{S}}_{\gamma }$ satisfies φ₁(a_p,𝜖)(p ≥ 1,𝜖 < α), φ₁(a_0,0) or φ₄(a_0,ε)(ε < ω × (f_ω(A_n,α) − 1), ε≠ 0,ω). The argument for the successor case is similar to the argument for the case β = γ. The limit case is trivial.

Claim 11. Let λ be a limit ordinal, and ${\mathscr{S}}$ be a λ-length revision sequence. Suppose that there is no hypothesis in ${\mathscr{S}}$ that satisfies φ_c ∧ φ₆. Let δ be a limit ordinal such that δ > λ. Let $X\subseteq Sent({{\mathscr{L}}}^{+})$. Basing on ${\mathscr{S}}$ and using X as the bootstrapper in the interval [λ,δ), we get a unique δ-length revision sequence ${\mathscr{S}}^{\prime }$. Suppose further that X declares plane k neither all- nor all- ,²² and that all planes after plane k are stably all- in ${\mathscr{S}}$. Then in the interval [λ,δ), plane k would be declared neither all- nor all- , and the declared truth value of each of the planes after plane k would not change. I.e., for every λ ≤ β < δ, ${\mathscr{S}}^{\prime }_{\beta }$ declares plane k neither all- nor all- , and ${\mathscr{S}}^{\prime }_{\beta }$ declares all planes after plane k all- .

The proof of Claim 11 is as follows. We prove by transfinite induction that for every λ ≤ β < δ, (1) ${\mathscr{S}}^{\prime }_{\beta }$ declares plane k neither all- nor all- ; and (2) ${\mathscr{S}}^{\prime }_{\beta }$ declares all planes after plane k all- . For β = γ, (2) holds trivially. If there is some sentence A_n,α in plane k such that A_n,α is stable in ${\mathscr{S}}$, then by Claim 10, line α is stably in ${\mathscr{S}}$. Obviously, ${\mathscr{S}}^{\prime }_{\beta }$ declares plane k neither all- nor all- . Assume that there is no sentence in plane k that is stable in ${\mathscr{S}}$. Since X declares plane k neither all- nor all- , ${\mathscr{S}}^{\prime }_{\beta }$ declares plane k neither all- nor all- , too. So (1) holds for β. Let β be a successor ordinal. By the induction hypothesis and Claim 3, (1) and (2) hold for β. The argument for the limit case is similar to the argument for the case β = γ.

Claim 12. Let λ be a limit ordinal, and ${\mathscr{S}}$ be a λ-length revision sequence. Let δ be a limit ordinal such that δ > λ. Let $X\subseteq Sent({{\mathscr{L}}}^{+})$. Basing on ${\mathscr{S}}$ and using X as the bootstrapper in the interval [λ,δ), we get a unique δ-length revision sequence ${\mathscr{S}}^{\prime }$. Suppose further that X declares plane k either all- or all- , where k = 2 or 3, and that plane k and the plane after plane k (if exists) are stably all- in ${\mathscr{S}}$. Then the declared truth value of each of line ω × (k − 1) + 2 and the lines after line ω × (k − 1) + 2 would not change. I.e., for every λ ≤ β < δ, ${\mathscr{S}}^{\prime }_{\beta }$ declares line ω × (k − 1) + 2 and all lines after line ω × (k − 1) + 2 .

The proof of Claim 12 is as follows. We prove by transfinite induction that for every λ ≤ β < δ, (1) it is not the case that ${\mathscr{S}}^{\prime }_{\beta }$ declares one of line ω × (k − 1) and line ω × (k − 1) + 1 , and the other ; and (2) ${\mathscr{S}}^{\prime }_{\beta }$ declares line ω × (k − 1) + 2 and all lines after line ω × (k − 1) + 2 . The case β = γ is trivial. Let β = 𝜖 + 1. By the induction hypothesis, ${\mathscr{S}}^{\prime }_{\epsilon }$ satisfies φ_c ∧ φ₁(a_n,α)(α ≤ ω × (k − 1) + 1), φ_c ∧ φ₂(a_0,ε)(ε ≤ ω × (k − 1) + 1,ε≠ 0,ω, 2ω), φ_c ∧ φ₃(a_0,ζ)(ζ < ω × (k − 1), ζ = 0,ω), φ_c ∧ φ₄(a_0,η)(η < ω × (k − 1)), $\varphi _{c}\land \varphi _{5}(\overline {i})(i<k,i=1,2)$ or φ_others. Then by Claim 3, ${\mathscr{S}}^{\prime }_{\beta }$ satisfies φ₁(a_n+ 1,α), φ₁(a_1,ε), φ₄(a_0,ζ+ 1), φ₄(a_0,η+ 1), φ₁(a_1,ω×i) or φ₁(a_0,0). Hence the proposition holds for β. Let β be a limit ordinal. By the induction hypothesis, (2) holds for β. By the induction hypothesis, there is no hypothesis in ${\mathscr{S}}^{\prime } \upharpoonright \beta$ that satisfies φ_c ∧ φ₆. By Claim 10, in ${\mathscr{S}}^{\prime } \upharpoonright \beta$, there are three possibilities: line ω × (k − 1) and line ω × (k − 1) + 1 are unstable; line ω × (k − 1) is unstable and line ω × (k − 1) + 1 is stably ; and line ω × (k − 1) and line ω × (k − 1) + 1 are stably . Note that X declares plane k either all- or all- . So in each case, (1) holds for β.

Claim 13. Let λ be a limit ordinal, and ${\mathscr{S}}$ be a λ-length revision sequence. Suppose that plane 2 and plane 3 are stably all- in ${\mathscr{S}}$. Let δ be the least cardinal that is larger than $\arrowvert \mathcal {P}(Sent({{\mathscr{L}}}^{+})) \arrowvert$. Let $X\subseteq Sent({{\mathscr{L}}}^{+})$. Basing on ${\mathscr{S}}$ and using X as the bootstrapper in the interval [λ,λ + δ), we get a unique (λ + δ)-length revision sequence ${\mathscr{S}}^{\prime }$. Suppose further that X declares plane 1 either all- or all- . Then plane 2 is stably all- in ${\mathscr{S}}^{\prime }$.

The proof of Claim 13 is as follows. First, we prove that there is a hypothesis in the interval [λ,λ + δ) (of ${\mathscr{S}}^{\prime }$, of course) that satisfies φ_others. We prove this by reduction to absurdity. Suppose that there is no hypothesis in the interval [λ,λ + δ) that satisfies φ_others. In particular, ${\mathscr{S}}^{\prime }_{\lambda }$ does not satisfy φ_others. (1) Suppose that there is some sentence in plane 1 that is stable in ${\mathscr{S}}$. Since ${\mathscr{S}}^{\prime }_{\lambda }$ does not satisfy φ_others, by Claim 10 and the premise that X declares plane 1 either all- or all- , line 0 of plane 1 is unstable in ${\mathscr{S}}$, and other lines of plane 1 are stably in ${\mathscr{S}}$. Then ${\mathscr{S}}^{\prime }_{\lambda }$ satisfies φ_c ∧ φ₁(a_0,1) or φ_c ∧ φ₂(a_0,1). By Claim 3, ${\mathscr{S}}^{\prime }_{\lambda +\omega }$ satisfies φ_others, a contradiction. (2) Suppose that there is no sentence in plane 1 that is stable in ${\mathscr{S}}$. Then ${\mathscr{S}}^{\prime }_{\lambda }$ satisfies either φ_c ∧ φ₁(a_0,ω) or $\varphi _{c}\land \varphi _{5}(\overline {1})$. By Claim 3, line 0 of plane 2 is unstable up to λ + ω in ${\mathscr{S}}^{\prime }$, and all sentences of the form $A_{n,\omega }\leftrightarrow A_{m,\omega }$, where n,m ∈ ω, are stably true up to λ + ω in ${\mathscr{S}}^{\prime }$. By supposition ${\mathscr{S}}^{\prime }_{\lambda +\omega }$ does not satisfy φ_others. Then ${\mathscr{S}}^{\prime }_{\lambda +\omega }$ must satisfy φ_c. Hence X declares line 0 of plane 2 either or , and ${\mathscr{S}}^{\prime }_{\lambda +\omega }$ satisfies φ_c ∧ φ₁(a_0,ω+ 1) or φ_c ∧ φ₂(a_0,ω+ 1). By Claim 3, we know that: the first two lines of plane 2 are unstable up to λ + 2ω in ${\mathscr{S}}^{\prime }$; all sentences of the form $A_{n,\alpha }\leftrightarrow A_{m,\alpha }$, where n,m ∈ ω and α = ω or ω + 1, are stably true up to λ + 2ω in ${\mathscr{S}}^{\prime }$; and all sentences of the form $A_{p,\omega }\leftrightarrow A_{q,\omega +1}$, where p,q ∈ ω, are stably false up to λ + 2ω in ${\mathscr{S}}^{\prime }$. Since ${\mathscr{S}}^{\prime }_{\lambda +2\omega }$ does not satisfy φ_others, X declares line 1 of plane 2 either or , and the declared truth value of line 1 of plane 2 by X is different from the declared truth value of line 0 of plane 2 by X, and ${\mathscr{S}}^{\prime }_{\lambda +2\omega }$ satisfies φ_c ∧ φ₁(a_0,ω+ 2) or φ_c ∧ φ₂(a_0,ω+ 2). Similarly, for every natural number t ≥ 2, X declares line t of plane 2 either or , and the declared truth value of line t of plane 2 by X is different from the declared truth value of line t − 1 of plane 2 by X, and ${\mathscr{S}}^{\prime }_{\lambda +\omega \times (t+1)}$ satisfies φ_c ∧ φ₁(a_0,ω+t+ 1) or φ_c ∧ φ₂(a_0,ω+t+ 1). So there are two possibilities: X declares all even lines of plane 2 , and all odd lines of plane 2 ; and X declares all even lines of plane 2 , and all odd lines of plane 2 .²³ Plane 2 is unstable up to λ + ω² in ${\mathscr{S}}^{\prime }$. Since ${\mathscr{S}}^{\prime }_{\lambda +\omega ^{2}}$ does not satisfy φ_others, ${\mathscr{S}}^{\prime }_{\lambda +\omega ^{2}}$ satisfies either φ_c ∧ φ₃(a_0,ω) or φ_c ∧ φ₄(a_0,ω). By Claim 3, plane 2 is unstable up to λ + ω² + ω in ${\mathscr{S}}^{\prime }$; and the sentence $A_{0,\omega }\leftrightarrow A_{0,\omega +1}$ is stably true up to λ + ω² + ω in ${\mathscr{S}}^{\prime }$. Hence ${\mathscr{S}}^{\prime }_{\lambda +\omega ^{2}+\omega }$ must not satisfy φ_c. Then ${\mathscr{S}}^{\prime }_{\lambda +\omega ^{2}+\omega }$ satisfies φ_others, a contradiction. We conclude that there is a hypothesis in the interval [λ,λ + δ) that satisfies φ_others. Let ${\mathscr{S}}^{\prime }_{\gamma }$ be such a hypothesis (where λ ≤ γ < λ + δ).

Next, we prove by transfinite induction that for every γ + 1 ≤ β < λ + δ, ${\mathscr{S}}^{\prime }_{\beta }$ declares line 2 and all lines after line 2 . Let β = γ + 1. Since ${\mathscr{S}}^{\prime }_{\gamma }$ satisfies φ_others, by Claim 3, the proposition holds for β. The limit case is trivial. Let β = 𝜖 + 1. Suppose that 𝜖 is a limit ordinal. By Claim 10, in ${\mathscr{S}}^{\prime }$, there are two possibilities: line 0 and line 1 are unstable up to 𝜖; and line 0 is unstable up to 𝜖 and line 1 is stably up to 𝜖. Hence ${\mathscr{S}}^{\prime }_{\epsilon }$ satisfies φ_c ∧ φ₁(a_0,1), φ_c ∧ φ₂(a_0,1) or φ_others. By Claim 3, ${\mathscr{S}}^{\prime }_{\beta }$ satisfies either φ₁(a_1,1) or φ₁(a_0,0). So ${\mathscr{S}}^{\prime }_{\beta }$ declares line 2 and all lines after line 2 . Suppose that 𝜖 is a successor ordinal. By Claim 4 and the induction hypothesis, ${\mathscr{S}}^{\prime }_{\epsilon }$ satisfies either φ₁(a_p,q)(p ≥ 1,q = 0, 1) or φ₁(a_0,0). By Claim 3, ${\mathscr{S}}^{\prime }_{\beta }$ satisfies either φ₁(a_p+ 1,q) or φ₁(a_1,0). So ${\mathscr{S}}^{\prime }_{\beta }$ declares line 2 and all lines after line 2 . Therefore the proposition holds for β.

Claim 14. Let λ be a limit ordinal, and ${\mathscr{S}}$ be a λ-length revision sequence. Suppose that plane 2 and plane 3 are stably all- in ${\mathscr{S}}$. Let δ be the least cardinal that is larger than $\arrowvert \mathcal {P}(Sent({{\mathscr{L}}}^{+})) \arrowvert$. Let C = {A_n,α ∈ I(S)∣n is even }. Let $X,Y,Z\subseteq Sent({{\mathscr{L}}}^{+})$. We get a unique (λ + δ × ω × 2)-length revision sequence ${\mathscr{S}}^{\prime }$ basing on ${\mathscr{S}}$ and using X, Y, Z and C as the bootstrappers in the interval [λ,λ + δ × ω × 2) in the following ways: in the intervals of the form [λ + δ × (2m + 1),λ + δ × (2m + 2)), where m ∈ ω, use X as the bootstrapper; in the intervals of the form [λ + δ × (2m + 2),λ + δ × (2m + 3)), where m ∈ ω, use Y as the bootstrapper; in the interval [λ + δ × ω,λ + δ × ω + δ) use Z as the bootstrapper; and use C as the bootstrapper at the other limit stages. Then plane 2 and plane 3 are stably all- in ${\mathscr{S}}^{\prime }$.

The proof of Claim 14 is as follows. First, we prove that there is no hypothesis in the interval [λ,λ + δ × ω × 2) (of ${\mathscr{S}}^{\prime }$, of course) that declares plane 3 either all- or all- . (1) In the interval [λ,λ + δ), C is used as the bootstrapper. Note that C declares plane 1 neither all- nor all- . By Claim 11 (k = 1), in the interval [λ,λ + δ), the declared truth value of each of plane 2 and plane 3 would not change. (2) In the interval [λ + δ,λ + δ × ω), alternately use X and Y as the bootstrappers. Next we prove in four cases that in the interval [λ + δ,λ + δ × ω), the declared truth value of plane 3 would not change, i.e., each hypothesis in the interval [λ + δ,λ + δ × ω) declares plane 3 all- . (2.1) Suppose that both X and Y declare plane 1 neither all- nor all- . By Claim 11 (k = 1), in the interval [λ + δ,λ + δ × ω), the declared truth value of each of plane 2 and plane 3 would not change. (2.2) Suppose that X declares plane 1 either all- or all- , and that Y declares plane 1 neither all- nor all- . By Claim 13, plane 2 is stably all- up to λ + 2δ in ${\mathscr{S}}^{\prime }$. By Claim 11 (k = 2) and Claim 12 (k = 2), in the interval [λ + δ,λ + 2δ), the declared truth value of plane 3 would not change. Since Y declares plane 1 neither all- nor all- , by Claim 11 (k = 1), in the interval [λ + 2δ,λ + 3δ) the declared truth value of each of plane 2 and plane 3 would not change. In this case, we can see that in the interval [λ + δ,λ + δ × ω), the declared truth value of plane 3 would not change. (2.3) Suppose that X declares plane 1 neither all- nor all- , and that Y declares plane 1 either all- or all- . The argument is similar to the argument for (2.2). (2.4) Suppose that both X and Y declare plane 1 either all- or all- . By Claim 13, plane 2 is stably all- up to λ + 2δ in ${\mathscr{S}}^{\prime }$. By Claim 11 (k = 2) and Claim 12 (k = 2), in the interval [λ + δ,λ + 2δ), the declared truth value of plane 3 would not change. Similarly, plane 2 is stably all- up to λ + 3δ in ${\mathscr{S}}^{\prime }$, and the declared truth value of plane 3 would not change in the interval [λ + 2δ,λ + 3δ). In this case, we can see that in the interval [λ + δ,λ + δ × ω), the declared truth value of plane 3 would not change. By (2.1)–(2.4), we conclude that the declared truth value of plane 3 would not change in the interval [λ + δ,λ + δ × ω). (3) In the interval [λ + δ × ω,λ + δ × ω + δ), Z is used as the bootstrapper. If Z declares plane 3 neither all- nor all- , by Claim 11 (k = 3), in the interval [λ + δ × ω,λ + δ × ω + δ), the declared truth value of plane 3 would be neither all- nor all- . If Z declares plane 3 either all- or all- , by Claim 12 (k = 3), each hypothesis in the interval [λ + δ × ω,λ + δ × ω + δ) declares each of line 2ω + 2 and the lines after line 2ω + 2 . We can see that in the interval [λ + δ × ω,λ + δ × ω + δ), the declared truth value of plane 3 would be neither all- nor all- . (4) In the interval [λ + δ × ω + δ,λ + δ × ω × 2), C is used as the bootstrapper. By Claim 11 (k = 3), in the interval [λ + δ × ω + δ,λ + δ × ω × 2), plane 3 would be declared neither all- nor all- . By (1)–(4), we conclude that there is no hypothesis in the interval [λ,λ + δ × ω × 2) that declares plane 3 either all- or all- .

Next, we prove that plane 2 and plane 3 are stably all- in ${\mathscr{S}}^{\prime }$. Consider the classical hypothesis ${\mathscr{S}}^{\prime }_{\lambda +\delta \times \omega +\delta +1}$. ${\mathscr{S}}^{\prime }_{\lambda +\delta \times \omega +\delta +1}$ declares plane 3 neither all- nor all- . By Claim 4, ${\mathscr{S}}^{\prime }_{\lambda +\delta \times \omega +\delta +1}$ satisfies φ₁(A_n,α)(n ≥ 1), φ₁(a_0,0) or φ₄(a_0,β)(β≠ 0,ω, 2ω). If ${\mathscr{S}}^{\prime }_{\lambda +\delta \times \omega +\delta +1}$ satisfies φ₁(A_n,α)(n ≥ 1), by Claim 3, line α is unstable up to λ + δ × ω + δ + ω in ${\mathscr{S}}^{\prime }$, and the sentence $A_{0,\alpha }\leftrightarrow A_{1,\alpha }$ is stably true up to λ + δ × ω + δ + ω in ${\mathscr{S}}^{\prime }$. Since C is the bootstrapper used at stage λ + δ × ω + δ + ω, ${\mathscr{S}}^{\prime }_{\lambda +\delta \times \omega +\delta +\omega }$ does not satisfy φ_c, and hence satisfies φ_others. If ${\mathscr{S}}^{\prime }_{\lambda +\delta \times \omega +\delta +1}$ satisfies φ₁(a_0,0), similarly, it can be proved that ${\mathscr{S}}^{\prime }_{\lambda +\delta \times \omega +\delta +\omega }$ satisfies φ_others. If ${\mathscr{S}}^{\prime }_{\lambda +\delta \times \omega +\delta +1}$ satisfies φ₄(a_0,β)(β≠ 0,ω, 2ω), by Claim 3, both A_0,β and A_1,β are unstable up to λ + δ × ω + δ + ω in ${\mathscr{S}}^{\prime }$, and the sentence $A_{0,\beta }\leftrightarrow A_{1,\beta }$ is stably true up to λ + δ × ω + δ + ω in ${\mathscr{S}}^{\prime }$. Since C is the bootstrapper used at stage λ + δ × ω + δ + ω, ${\mathscr{S}}^{\prime }_{\lambda +\delta \times \omega +\delta +\omega }$ satisfies φ_others. In each case, ${\mathscr{S}}^{\prime }_{\lambda +\delta \times \omega +\delta +\omega }$ satisfies φ_others. By Claim 3, plane 2 and plane 3 are stably all- up to λ + δ × ω + δ + 2ω in ${\mathscr{S}}^{\prime }$. Note that C declares plane 1 neither all- nor all- . By Claim 11 (k = 1), in the interval [λ + δ × ω + δ + 2ω,λ + δ × ω × 2) the declared truth value of each of plane 2 and plane 3 would not change. Hence plane 2 and plane 3 are stably all- in ${\mathscr{S}}^{\prime }$.

Claim 15. T^Y does not dictate that truth behaves like a classical concept in ${\mathscr{M}}$.

The proof of Claim 15 is as follows. Next we construct a Y-sequence, and prove that this revision sequence does not culminate in a fixed point. By Claim 3, a fixed point of $\tau _{{\mathscr{M}}}$ must satisfy φ_c ∧ φ₆. Let δ be the least cardinal that is larger than $\arrowvert \mathcal {P}(Sent({{\mathscr{L}}}^{+})) \arrowvert$. Let C = {A_n,α ∈ I(S)∣n is even }. Let $U=\{(X,Y,Z)\mid X,Y,Z\subseteq Sent({{\mathscr{L}}}^{+})\}$. $\arrowvert U \arrowvert =\arrowvert \mathcal {P}(Sent({{\mathscr{L}}}^{+})) \arrowvert <\delta$. So there is a surjection from δ to U. Let f be such a function. For every α ∈ δ, denote the first element of f(α) by X_α, the second element of f(α) by Y_α, and the third element of f(α) by Z_α. Let h be the classical hypothesis whose extension is C. h does not satisfy φ_c, and hence satisfies φ_others. Define a y-bootstrapping-policy relative to hBr as follows: for every limit ordinal λ,

$$ Br(\lambda)= \left\{\begin{array}{lll} X_{\alpha} & \text{ if there is an ordinal}~\beta~\text{and an}~\alpha<\delta~\text{and an}~m\in\omega~\text{such that }\\ & \delta^2\times\beta+\delta\times\omega\times2\times\alpha+\delta\times(2m+1)\le\lambda~ \text{and }\\ & \lambda<\delta^2\times\beta+\delta\times\omega\times2\times\alpha+\delta\times(2m+2),\\ Y_{\alpha}&\text{ if there is an ordinal}~\beta~\text{and an}~ \alpha<\delta~\text{and an}~m\in\omega~\text{such that }\\ & \delta^2\times\beta+\delta\times\omega\times2\times\alpha+\delta\times(2m+2)\le\lambda~ \text{and }\\ & \lambda<\delta^2\times\beta+\delta\times\omega\times2\times\alpha+\delta\times(2m+3),\\ Z_{\alpha} & \text{ if there is an ordinal}~\beta~\text{and an}~\alpha<\delta~\text{such that }\\ & \delta^2\times\beta+\delta\times\omega\times2\times\alpha+\delta\times\omega\le\lambda~\text{and}\\ & \lambda<\delta^2\times\beta+\delta\times\omega\times2\times\alpha+\delta\times(\omega+1),\\ C & \text{otherwise.} \end{array}\right. $$

Let ${\mathscr{S}}={\mathscr{M}}[h,Br]$. Obviously, ${\mathscr{S}}$ is a Y-sequence. We prove by transfinite induction that for every ordinal α, plane 2 and plane 3 are stably all- up to δ × ω × 2 × (α + 1) in ${\mathscr{S}}$. For α = 0, by Claim 3, plane 2 and plane 3 are stably all- up to ω in ${\mathscr{S}}$. By Claim 14, the proposition holds for α. For α = β + 1, by Claim 14 and the induction hypothesis, the proposition holds. Let α be a limit ordinal. By definition, in the interval [δ × ω × 2 × α,δ × ω × 2 × α + δ), C is used as the bootstrapper. Since C declares plane 3 neither all- nor all- , By Claim 11 (k = 3), in the interval [δ × ω × 2 × α,δ × ω × 2 × α + δ), plane 3 would be declared neither all- nor all- . By Claim 3, ${\mathscr{S}}_{\delta \times \omega \times 2\times \alpha +1}$ satisfies φ₁(a_n,η)(n ≥ 1), φ₁(a_0,0) or φ₄(a_0,ζ)(ζ≠ 0,ω, 2ω). In each case, ${\mathscr{S}}_{\delta \times \omega \times 2\times \alpha +\omega }$ satisfies φ_others. By Claim 3, plane 2 and plane 3 are stably all- up to δ × ω × 2 × α + 2ω in ${\mathscr{S}}$. By Claim 14, the proposition holds for α. We conclude that for every ordinal α, plane 2 and plane 3 are stably all- up to δ × ω × 2 × (α + 1) in ${\mathscr{S}}$. Obviously, ${\mathscr{S}}$ does not culminate in a fixed point.

By Claim 9 and Claim 15, ${\mathbf {T}}^{lfp,\sigma _{2}}\not \le _2{\mathbf {T}}^Y$. □