Equivalence, and Property Internalization and Preservation for Equational Programs

Abstract

An equational theory $\mathcal {E}=(\varSigma , E \cup B)$ is an equational program if its equations E, oriented as rewrite rules $\vec{E}$, are ground convergent modulo axioms B. Its properties are the inductive theorems of the initial algebra $\mathbb {T}_{\varSigma /E \cup B}$ defined by $(\varSigma , E \cup B)$. Since programs are structured in module hierarchies, checkable syntactic conditions are given to preserve program properties up and/or down such hierarchies. Two equational programs $\mathcal {E}=(\varSigma , E \cup B)$ and $\mathcal {E}'=(\varSigma , E' \cup B')$ are equivalent iff they define the same computable functions on the same algebraic data types. Succinct conditions to verify $\mathcal {E}$ and $\mathcal {E}'$ equivalent are given. A useful internalization method to extend an equational program $\mathcal {E}$ into an equivalent one by adding new rewrite rules or structural axioms that are inductive theorems of $\mathcal {E}$ is also given. This method can make proofs of program properties simpler and shorter, and offers a new way to prove equational theories ground convergent.

Appendices

$\text {A}\qquad$ Proofs of Theorems and Lemmas

Proof of Theorem 2.

Proof

First of all, note that, since $\mathbb {C}_{\varSigma /\vec{E},B}|_{\varOmega ^{+}} = \mathbb {C}_{\varOmega ^{+}/\vec{E}_{\varOmega ^{+}},B_{\varOmega ^{+}}}$ and $\mathbb {C}_{\varSigma '/\vec{E}',B'}|_{\varOmega '^{+}} = \mathbb {C}_{\varOmega '^{+}/\vec{E'}_{\varOmega '^{+}},B'_{\varOmega '^{+}}}$, the S-sorted function of the unique $\varSigma$-homomorphism $h: \mathbb {C}_{\varSigma /\vec{E},B} \rightarrow \mathbb {C}_{\varSigma '/\vec{E}',B'}|_{\varSigma }$ coincides with the S-sorted function of the unique $\varOmega ^{+}$-homomorphism $h: \mathbb {C}_{\varOmega ^{+}/\vec{E}_{\varOmega ^{+}},B_{\varOmega ^{+}}} \rightarrow \mathbb {C}_{\varOmega '^{+}/\vec{E}'_{\varOmega '^{+}},B'_{\varOmega '^{+}}}|_{\varOmega ^{+}}$. Therefore, in each case (1)–(3) we just need to show that, for each sort $s \in S$, the function $h_{s}: C_{\varOmega ^{+}/\vec{E}_{\varOmega ^{+}},B_{\varOmega ^{+}},s} \rightarrow C_{\varOmega '^{+}/\vec{E}'_{\varOmega '^{+}},B'_{\varOmega '^{+}},s}$ is injective, resp. surjective, resp. bijective. Furthermore, since the unique surjective $\varOmega ^{+}$-homomorphism $\mathbb {T}_{\varOmega ^{+}} \rightarrow \mathbb {C}_{\varOmega ^{+}/\vec{E}_{\varOmega ^{+}},B_{\varOmega ^{+}}}$ maps each $u \in T_{\varOmega ^{+},s}$ to $[u!_{\vec{E}_{\varOmega ^{+}},B_{\varOmega ^{+}}}] \in C_{\varOmega ^{+}/\vec{E}_{\varOmega ^{+}},B_{\varOmega ^{+}},s}$, and the unique $\varOmega ^{+}$-homomorphism $\mathbb {T}_{\varOmega ^{+}} \rightarrow \mathbb {C}_{\varOmega '^{+}/\vec{E}'_{\varOmega '^{+}},B'_{\varOmega '^{+}}}|_{\varOmega ^{+}}$ maps each $u \in T_{\varOmega ^{+},s}$ to $[u!_{\vec{E}'_{\varOmega '^{+}},B'_{\varOmega '^{+}}}] \in C_{\varOmega '^{+}/\vec{E}'_{\varOmega '^{+}},B'_{\varOmega '^{+}},s}$, by initiality of $\mathbb {T}_{\varOmega ^{+}}$ $h_{s}$ must be the function $h_{s}: C_{\varOmega ^{+}/\vec{E}_{\varOmega ^{+}},B_{\varOmega ^{+}},s} \ni [u!_{\vec{E}_{\varOmega ^{+}},B_{\varOmega ^{+}}}]_{B_{\varOmega ^{+}}} \mapsto [u!_{\vec{E}'_{\varOmega '^{+}},B'_{\varOmega '^{+}}}]_{B'_{\varOmega '^{+}}}\in C_{\varOmega '^{+}/\vec{E}'_{\varOmega '^{+}},B'_{\varOmega '^{+}},s}$.

If the inclusion is extending we need to show that for any $[u]_{B_{\varOmega ^{+}}} , [v]_{B_{\varOmega ^{+}}} \in C_{\varOmega ^{+}/\vec{E}_{\varOmega ^{+}},B_{\varOmega ^{+}},s}$, if $[u!_{\vec{E}'_{{\varOmega '^{+}},{B'_{{\varOmega '}^{+}}}}}]_{B'_{\varOmega '^{+}}}= [v!_{\vec{E}'_{\varOmega '^{+}},B'_{\varOmega '^{+}}}]_{B'_{\varOmega '^{+}}}$, then $[u]_{B_{\varOmega ^{+}}} = [v]_{B_{\varOmega ^{+}}}$. But $[u!_{\vec{E}'_{\varOmega '^{+}},B'_{\varOmega '^{+}}}]_{B'_{\varOmega '^{+}}}= [v!_{\vec{E}'_{\varOmega '^{+}},B'_{\varOmega '^{+}}}]_{B'_{\varOmega '^{+}}}$ just means $u!_{\vec{E}'_{\varOmega '^{+}},B'_{\varOmega '^{+}}}=_{B'_{\varOmega '^{+}}} v!_{\vec{E}'_{\varOmega '^{+}},B'_{\varOmega '^{+}}}$. First of all note that, $u!_{\vec{E}'_{\varOmega '^{+}},B'_{\varOmega '^{+}}}=u$ and $v!_{\vec{E}'_{\varOmega '^{+}},B'_{\varOmega '^{+}}}=v$, i.e., both terms are in $\vec{E}'_{\varOmega '^{+}},B'_{\varOmega '^{+}}$-canonical form. Let me show this for $u!_{\vec{E}'_{\varOmega '^{+}},B'_{\varOmega '^{+}}}$ since the argument is identical for $v!_{\vec{E}'_{\varOmega '^{+}},B'_{\varOmega '^{+}}}$. Suppose $u\rightarrow _{\vec{E}'_{\varOmega '^{+}},B'_{\varOmega '^{+}}}w$. This happens iff there is a term $w'$ such that $u!_{{\vec{U}_{\varOmega '^{+}}}, {B'_{A\vee C,\varOmega '^{+}}}}=u'$, $u' \rightarrow _ {{\vec{E}'_{\varOmega '^{+}}}{U_{\varOmega '^{+}}},{B'_{A \vee C,\varOmega '^{+}}}}w'$ and $w =_{B'_{\varOmega '^{+}}}w'$. But note that $u'=u!_{\vec{U}_{\varOmega '^{+}}, B'_{A\vee C,\varOmega '^{+}}}$ is actually $u'=u!_{\vec{U}_{\varOmega ^{+}}, B_{A\vee C,\varOmega ^{+}}}$. This is because, (a) u is an $\varOmega ^{+}$-term and by assumption (i), if w is an $\varOmega ^{+}$-term, then $w=_{ B'_{A\vee C,\varOmega '^{+}}}w'$ iff $w'$ is an $\varOmega ^{+}$-term and $w=_{ B_{A\vee C,\varOmega ^{+}}}w'$, and (b), also by assumption (i), for any $\varOmega ^{+}$-term $u''$ if $u''\rightarrow _{\vec{U}_{\varOmega '^{+}}, B_{A\vee C,\varOmega '^{+}}}u'''$ the rule $f(x,e)\rightarrow x$, (resp. $f(e,x)\rightarrow x$) applied to $u''$ to get $u'''$ can always be chosen to be a rule in $\vec{U}_{\varOmega ^{+}}$. This means that $u'$ is an $\varOmega ^{+}$-term and $u=_{B_{\varOmega ^{+}}}u'$. But since $[u] \in C_{\varOmega ^{+}/\vec{E}_{\varOmega ^{+}},B_{\varOmega ^{+}},s}$, $u'$ must be in $\vec{E}_{\varOmega ^{+}},B_{\varOmega ^{+}}$-canonical form. This therefore means that a rewrite $u' \rightarrow _ {{\vec{E}'_{\varOmega '^{+}}}{U_{\varOmega '^{+}}},{B'_{A \vee C,\varOmega '^{+}}}}w'$ is impossible, since by (a) above and the assumption in (1) such a rewrite must be of the form $u' \rightarrow _ {{\vec{E}_{{\varOmega ^{+}}_{U_{\varOmega ^{+}}}},B_{A \vee C,\varOmega '^{+}}}}w'$, which is impossible by $u'$ in $\vec{E}_{\varOmega ^{+}},B_{\varOmega ^{+}}$-canonical form. This shows that $u!_{\vec{E}'_{\varOmega '^{+}},B'_{\varOmega '^{+}}}=u$ and $v!_{\vec{E}'_{\varOmega '^{+}},B'_{\varOmega '^{+}}}=v$. We furthermore have $u!_{\vec{E}'_{\varOmega '^{+}},B'_{\varOmega '^{+}}}=_{B'_{\varOmega '^{+}}} v!_{\vec{E}'_{\varOmega '^{+}},B'_{\varOmega '^{+}}}$, i.e., $u =_{B'_{\varOmega '^{+}}} v$, which holds iff $u!_{\vec{U}_{\varOmega '^{+}}, B'_{A\vee C,\varOmega '^{+}}} =_{B'_{A \vee C,\varOmega '^{+}}} v!_{\vec{U}_{\varOmega '^{+}}, B'_{A\vee C,\varOmega '^{+}}}$, which, reasoning as above, holds iff $u!_{\vec{U}_{\varOmega ^{+}}, B_{A \vee C,\varOmega ^{+}}} =_{B_{A \vee C,\varOmega ^{+}}} v!_{\vec{U}_{\varOmega ^{+}}, B_{A \vee C,\varOmega ^{+}}}$, which is equivalent to $u=_{B_{\varOmega ^{+}}}v$, proving the injectivity of $h_{s}$.

If the inclusion is generated-by we need to show h surjective. Since $\mathbb {C}_{\varOmega '^{+}/\vec{E'}_{\varOmega '^{+}},B'_{\varOmega '^{+}}}$ and $\mathbb {C}_{\varOmega ^{+}/\vec{E}_{\varOmega ^{+}},B_{\varOmega ^{+}}}$ are initial $\varOmega '^{+}$-, resp. $\varOmega ^{+}$-, algebras, we have surjective $\varOmega '^{+}$-, resp. $\varOmega ^{+}$-, homomorphisms $q' : \mathbb {T}_{\varOmega '^{+}}\rightarrow \mathbb {C}_{\varOmega '^{+}/\vec{E'}_{\varOmega '^{+}},B'_{\varOmega '^{+}}}$ and $q: \mathbb {T}_{\varOmega ^{+}} \rightarrow \mathbb {C}_{\varOmega ^{+}/\vec{E}_{\varOmega ^{+}},B_{\varOmega ^{+}}}$. But conditions (a)–(b) in (2) mean that $\mathbb {T}_{\varOmega '^{+}}|_{\varOmega ^{+}} = \mathbb {T}_{\varOmega ^{+}}$, which forces the unique $\varOmega ^{+}$-homomorphism $\mathbb {T}_{\varOmega ^{+}} \rightarrow \mathbb {C}_{\varOmega '^{+}/\vec{E'}_{\varOmega '^{+}},B'_{\varOmega '^{+}}}|_{\varOmega ^{+}}$ to be $q'|_{\varOmega }$ and therefore surjective, and, by initiality of $\mathbb {T}_{\varOmega ^{+}}$, the homomorphism identity $q ; h = q'|_{\varOmega ^{+}}$, which by $q'|_{\varOmega ^{+}}$ surjective forces h to be surjective.

If the inclusion is protecting, h is injective by the requiements in (1) and surjective by those in (2), and therefore bijective. $\Box$.

Proof of Theorem 3.

Proof

To see the $(\Rightarrow )$ implication, since $\mathbb {T}_{\varSigma /E \cup G} \models E$ we have a unique $\varSigma$-homomorphism $h: \mathbb {T}_{\varSigma /E} \rightarrow \mathbb {T}_{\varSigma /E \cup G}$. And since $\mathbb {T}_{\varSigma /E}\models E \cup G$, we also have a unique $\varSigma$-homomorphism $g: \mathbb {T}_{\varSigma /E \cup G} \rightarrow \mathbb {T}_{\varSigma /E }$. But then, the initiality of $\mathbb {T}_{\varSigma /E}$ forces $h;g = id _{\mathbb {T}_{\varSigma /E}}$, and the initiality of $\mathbb {T}_{\varSigma /E \cup G}$ forces $g;h = id _{\mathbb {T}_{\varSigma /E \cup G}}$. Therefore, we have an isomorphism: $\mathbb {T}_{\varSigma /E} \cong \mathbb {T}_{\varSigma /E \cup G}$. We will be done of we prove the following lemma:

Lemma 2

Let $E,E'$ be two sets of $\varSigma$-equations such that $\mathbb {T}_{\varSigma /E} \cong \mathbb {T}_{\varSigma /E'}$. Then, $\mathbb {T}_{\varSigma /E} = \mathbb {T}_{\varSigma /E'}$.

Proof

$\mathbb {T}_{\varSigma /E}$ and $\mathbb {T}_{\varSigma /E'}$ are uniquely determined by the respective ground equality relations $=_{E} \cap T_{\varSigma }^{2}$ and $=_{E'} \cap T_{\varSigma }^{2}$. We just need to show $(=_{E} \cap T_{\varSigma }^{2}) = (=_{E'} \cap T_{\varSigma }^{2})$. Since we have a $\varSigma$-isomorphism $h: \mathbb {T}_{\varSigma /E} \rightarrow \mathbb {T}_{\varSigma /E'}$, and unique $\varSigma$-homomorphisms $[\_]_{E}: \mathbb {T}_{\varSigma } \rightarrow \mathbb {T}_{\varSigma /E}$, and $[\_]_{E'}: \mathbb {T}_{\varSigma } \rightarrow \mathbb {T}_{\varSigma /E}$, the initiality of $\mathbb {T}_{\varSigma }$ forces $[\_]_{E}; h = [\_]_{E'}$, i.e., $h_{s}([t]_{E}) = [t]_{E'}$ for each $t \in T_{\varSigma ,s}, s\in S$. Let $t \in T_{\varSigma ,s}$ and $t' \in T_{\varSigma ,s'}$ with $t =_{E} t'$. Then $[s]=[s']$ and, by h order-sorted $\varSigma$-homomorphism and $[t]_{E}=[t']_{E}$, we must have $h_{s}([t]_{E})=h_{s'}([t']_{E})$, which forces:

$$h_{s}([t]_{E})= [t]_{E'}= [t']_{E'}=h_{s'}([t']_{E}) $$

giving us the containment $(=_{E} \cap T_{\varSigma }^{2}) \subseteq (=_{E'} \cap T_{\varSigma }^{2})$. Using the inverse isomorphism $h^{-1}$ we likewise get $(=_{E'} \cap T_{\varSigma }^{2}) \subseteq (=_{E} \cap T_{\varSigma }^{2})$, giving us $(=_{E} \cap T_{\varSigma }^{2}) = (=_{E'} \cap T_{\varSigma }^{2})$, as desired. $\Box$

To see the $(\Leftarrow )$ implication, since $\mathbb {T}_{\varSigma /E} = \mathbb {T}_{\varSigma /E \cup G}$ we have $\mathbb {T}_{\varSigma /E} \models G$, which exactly means $(\varSigma ,E) \models _{ind} G$. $\Box$

Proof of Lemma 1

Proof

It follows from the assumptions on $\vec{E}_{\varDelta _{0}}$ (resp. $\vec{E}'_{\varDelta _{0}}$) and $B_{\varDelta }$, that for any $t \in T_{\varDelta _{0} \uplus \varOmega ^{+}}$, if $t\rightarrow _{\vec{E},B}t'$ (resp. $t\rightarrow _{\vec{E}',B'}t'$), then $t' \in T_{\varDelta _{0} \uplus \varOmega ^{+}}$ and $t\rightarrow _{ \vec{E}_{\varDelta _{0}} \cup \vec{E}_{\varOmega _{+}}, B_{\varDelta _{0}} \cup B_{\varOmega ^{+}}}t'$ (resp. $t\rightarrow _{\vec{E}'_{\varDelta _{0}} \cup \vec{E}'_{\varOmega _{+}}, B'_{\varDelta _{0}} \cup B'_{\varOmega ^{+}}}t'$). Therefore, both $(\varDelta _{0} \uplus \varOmega ^{+}, E_{\varDelta _{0}} \cup E_{\varOmega _{+}} \cup B_{\varDelta _{0}} \cup B_{\varOmega ^{+}})$ and $(\varDelta _{0} \uplus \varOmega ^{+}, E'_{\varDelta _{0}} \cup E'_{\varOmega _{+}} \cup B'_{\varDelta _{0}} \cup B'_{\varOmega ^{+}})$ are sufficiently complete w.r.t. $\varOmega$, ground convergent, and therefore admissible. Furthermore, for each $t \in T_{\varDelta _{0} \uplus \varOmega ^{+}}$, $t!_{\vec{E},B}=t!_{\vec{E}_{\varDelta _{0}} \cup \vec{E}_{\varOmega _{+}},B_{\varDelta _{0}} \cup B_{\varOmega ^{+}}}$ and $t!_{\vec{E}',B'}=t!_{\vec{E}'_{\varDelta _{0}} \cup \vec{E}'_{\varOmega _{+}},B'_{\varDelta _{0}} \cup B'_{\varOmega ^{+}}}$, which forces $\mathbb {C}_{\varDelta \uplus \varOmega ^{+}/\vec{E},B}|_{\varDelta _{0} \uplus \varOmega ^{+}} =\mathbb {C}_{\varDelta \uplus \varOmega ^{+}/\vec{E}_{\varDelta _{0}}\cup \vec{E}_{\varOmega _{+}},B_{\varDelta _{0}} \cup B_{\varOmega ^{+}} }$ and $\mathbb {C}_{\varDelta \uplus \varOmega ^{+}/\vec{E}',B'}|_{\varDelta _{0} \uplus \varOmega ^{+}} =\mathbb {C}_{\varDelta \uplus \varOmega ^{+}/\vec{E}'_{\varDelta _{0}}\cup \vec{E}'_{\varOmega _{+}},B'_{\varDelta _{0}}\cup B'_{\varOmega ^{+}} }$, which, by $\mathcal {E} \equiv _{ sem } \mathcal {E}'$, forces $(\varDelta _{0} \uplus \varOmega ^{+}, E_{\varDelta _{0}} \cup E_{\varOmega _{+}} \cup B_{\varDelta _{0}} \cup B_{\varOmega ^{+}}) \equiv _{ sem } (\varDelta _{0} \uplus \varOmega ^{+}, E'_{\varDelta _{0}} \cup E'_{\varOmega _{+}} \cup B'_{\varDelta _{0}} \cup B'_{\varOmega ^{+}})$. $\Box$

Proof of Theorem 4.

Proof

To see $(\Rightarrow )$, note that semantic equivalence forces (1), since $\mathbb {C}_{\varSigma /\vec{E},B}=\mathbb {C}_{\varSigma /\vec{E}',B'}$ implies that $\mathbb {C}_{\varSigma /\vec{E},B}|_{\varOmega } =\mathbb {C}_{\varOmega /\vec{E}_{\varOmega },B_{\varOmega }}= \mathbb {C}_{\varOmega /\vec{E}'_{\varOmega },B'_{\varOmega }}= \mathbb {C}_{\varSigma /\vec{E}',B'}|_{\varOmega }$; and also forces (2), since $\mathcal {E} \equiv _{ind} \mathcal {E}'$ means that $\mathbb {T}_{\varSigma /E \cup B} = \mathbb {T}_{\varSigma /E' \cup B'}$, which forces $(\varSigma ,E \cup B) \models _{ind} (E'_{\varDelta } \setminus E_{\varDelta }) \cup (B_{\varDelta }' \setminus B_{\varDelta })$.

To prove the $(\Leftarrow )$ implication we first prove:

Lemma 3

For any two canonical term algebras $\mathbb {C}_{\varSigma /\vec{E},B}$ and $\mathbb {C}_{\varSigma /\vec{E}',B'}$ with respective constructor subspecifications $(\varOmega , E_{\varOmega } \cup B_{\varOmega })$ and $(\varOmega , E'_{\varOmega } \cup B'_{\varOmega })$ and such that $\mathbb {C}_{\varOmega /\vec{E}_{\varOmega },B_{\varOmega }}= \mathbb {C}_{\varOmega /\vec{E}'_{\varOmega },B'_{\varOmega }}$, $\mathbb {C}_{\varSigma /\vec{E},B}= \mathbb {C}_{\varSigma /\vec{E}',B'}$ iff for each $\varSigma$-term t, $t ! _{\vec{E}/B} =_{B_{\varOmega }} t ! _{\vec{E'}/B'}$.

Proof

To see the $(\Rightarrow )$ implication, note that the unique $\varSigma$-homomorphism $\mathbb {T}_{\varSigma } \rightarrow \mathbb {C}_{\varSigma /\vec{E},B}$ maps each $t\in \mathbb {T}_{\varSigma }$ to $[t ! _{\vec{E}/B}]_{B_{\varOmega }}\in \mathbb {C}_{\varOmega /\vec{E}_{\varOmega },B_{\varOmega }}$. Therefore, $\mathbb {C}_{\varSigma /\vec{E},B}= \mathbb {C}_{\varSigma /\vec{E}',B'}$ means that for each $t\in \mathbb {T}_{\varSigma }$, $[t !_{\vec{E}/B}]_{B_{\varOmega }} =[t ! _{\vec{E'}/B'}]_{B'_{\varOmega }}$, and therefore that $t !_{\vec{E}/B}=_{B_{\varOmega }} t ! _{\vec{E'}/B'}$.

To see the $(\Leftarrow )$ implication, since $\mathbb {C}_{\varOmega /\vec{E}_{\varOmega },B_{\varOmega }}=\mathbb {C}_{\varOmega /\vec{E}'_{\varOmega },B'_{\varOmega }}$, $\mathbb {C}_{\varSigma /\vec{E},B}$ and $\mathbb {C}_{\varSigma /\vec{E}',B'}$ have the same underlying S-sorted set of data elements, and the same interpretation $c_{\mathbb {C}_{\varSigma /\vec{E},B}}= c_{\mathbb {C}_{\varSigma /\vec{E}',B'}}$ for each constructor operator $c \in \varOmega$. Therefore, to show $\mathbb {C}_{\varSigma /\vec{E},B}= \mathbb {C}_{\varSigma /\vec{E}',B'}$ we only need to show that for each $f\in \varDelta$ we have $f_{\mathbb {C}_{\varSigma /\vec{E},B}}= f_{\mathbb {C}_{\varSigma /\vec{E}',B'}}$. Indeed, $f_{\mathbb {C}_{\varSigma /\vec{E},B}}([u_{1}],\ldots ,[ u_{n}]) =[f(u_{1},\ldots , u_{n}) !_{\vec{E}/B}] =[f(u_{1},\ldots , u_{n}) !_{\vec{E}'/B'}]= f_{\mathbb {C}_{\varSigma /\vec{E}',B'}}([u_{1}],\ldots ,[ u_{n}])$. $\Box$

Now note that (1) (using Theorems 3 and 1) and (2) force $(\varSigma ,E \cup B) \models _{ind} E' \cup B'$, which forces $t ! _{\vec{E}/B} =_{E \cup B} t ! _{\vec{E'}/B'}$, which, by the Church-Rosser property, then forces $t ! _{\vec{E}/B} =_{B_{\varOmega } } (t ! _{\vec{E'}/B'}) !_{\vec{E}/B}$, which by (1) and Lemma 3 forces $\mathbb {C}_{\varSigma /\vec{E},B}= \mathbb {C}_{\varSigma /\vec{E}',B'}$. $\Box$

Proof of Theorem 5.

Proof

$(\varSigma ,B,\vec{E} \cup \vec{G})$ will be admissible if we prove that $(\varSigma ,B,\vec{E} \cup \vec{G})$ is locally ground confluent modulo B. Let $t,u,v \in T_{\varSigma }$ be such that $u \; _{\vec{E} \cup \vec{G}/B} \! \leftarrow t \rightarrow _{\vec{E} \cup \vec{G}/B} v$. We need to show that $u \downarrow _{\vec{E} \cup \vec{G}/B} v$. This will hold if we prove that $u \downarrow _{\vec{E} /B} v$. But since $(\varSigma ,E \cup B) \models _{ ind } G$, by Theorem 3 this forces $u =_{E \cup B} v$, which, since $\vec{E}$ is gound convergent modulo B, forces $u \downarrow _{\vec{E} /B} v$. Next we have to show that $\mathbb {C}_{\varSigma /\vec{E},B}=\mathbb {C}_{\varSigma /\vec{E}\cup \vec{G},B}$. Let us first prove that for each $t \in T_{\varSigma }$ $t!_{\vec{E},B}=_{B_{\varOmega }}t! _{\vec{E}\cup \vec{G},B}$. Since $t!_{\vec{E},B}=_{E \cup G \cup B} t! _{\vec{E}\cup \vec{G},B}$, by $(\varSigma ,E \cup B) \models _{ ind } G$ and Theorem 3, $t!_{\vec{E},B}=_{E \cup B}t! _{\vec{E}\cup \vec{G},B}$, which by the ground Church-Rosser Theorem holds iff $t!_{\vec{E},B}=_{B_{\varOmega }} (t! _{\vec{E}\cup \vec{G},B})!_{\vec{E},B}$. But, by definition, $t! _{\vec{E}\cup \vec{G},B}$ is already in $\vec{E},B$-canonical form, so that $t!_{\vec{E},B}=_{B_{\varOmega }}t! _{\vec{E}\cup \vec{G},B}$. This means that $\mathbb {C}_{\varSigma /\vec{E},B}$ and $\mathbb {C}_{\varSigma /\vec{E}\cup \vec{G},B}$ have the same S-sorted set of data elements. So, to prove $\mathbb {C}_{\varSigma /\vec{E},B}=\mathbb {C}_{\varSigma /\vec{E}\cup \vec{G},B}$ it is enough to prove that for each $f \in \varSigma$ we have the function identity $f_{\mathbb {C}_{\varSigma /\vec{E},B}}=f_{\mathbb {C}_{\varSigma /\vec{E}\cup \vec{G},B}}$. But this follows easily from $t!_{\vec{E},B}=_{B_{\varOmega }}t! _{\vec{E}\cup \vec{G},B}$, as in the proof of Lemma 3. $\Box$

Proof of Theorem 6.

Proof

To show $(\varSigma ,E \cup B \cup B')$ admissible we need to show that the rules $\vec{E}$ are locally ground confluent modulo $B \cup B'$. Let $t,u,v \in T_{\varSigma }$ be such that $u \; _{\vec{E} /B \cup B'} \! \leftarrow t \rightarrow _{\vec{E} /B \cup B'} v$. We need to show that $u \downarrow _{\vec{E} /B \cup B'} v$. This will hold if we prove $u \downarrow _{\vec{E} /B} v$. But since $(\varSigma ,E \cup B) \models _{ ind } B'$, Theorem 3 forces $u =_{E \cup B} v$, which, since $\vec{E}$ is ground confluent modulo B, forces $u \downarrow _{\vec{E} /B} v$, as desired.

We will be done if we show that $\mathbb {C}_{\varSigma /\vec{E},B}=\mathbb {C}_{\varSigma /\vec{E},B \cup B'}$. First of all note that for any ground $\varSigma$-term $t\in T_{\varSigma }$, $t!_{\vec{E},B}=_{B_{\varOmega }}t!_{\vec{E},B \cup B'}$. This is because, by Theorem 3, $t!_{\vec{E},B}=_{E \cup B}t_{\vec{E},B \cup B'}$, by the ground Church Rosser Theorem $t!_{\vec{E},B}=_{B_{\varOmega }}(t!_{\vec{E},B \cup B'})!_{\vec{E},B}$ and by $t!_{\vec{E},B \cup B'}$ already being in $\vec{E},B$-normal form $t!_{\vec{E},B}=_{B_{\varOmega }}t!_{\vec{E},B \cup B'}$. Second, $\mathbb {C}_{\varSigma /\vec{E},B}$ and $\mathbb {C}_{\varSigma /\vec{E},B \cup B'}$ will have the same undelying S-sorted set of data elements if we show that $t!_{\vec{E},B}=_{B_{\varOmega }}t!_{\vec{E},B \cup B'} \Leftrightarrow t!_{\vec{E},B}=_{B_{\varOmega } \cup B'_{\varOmega } }t!_{\vec{E},B \cup B'}$, since this shows that $[t!_{\vec{E},B}]_{B_{\varOmega }}=[t!_{\vec{E},B \cup B'}]_{B_{\varOmega } \cup B'_{\varOmega } }$. This only requires showing $t!_{\vec{E},B}=_{B_{\varOmega } \cup B'_{\varOmega } }t!_{\vec{E},B \cup B'} \Rightarrow t!_{\vec{E},B}=_{B_{\varOmega }}t!_{\vec{E},B \cup B'}$. But, by Theorem 3, $t!_{\vec{E},B}=_{B_{\varOmega } \cup B'_{\varOmega } }t!_{\vec{E},B \cup B'} \Rightarrow t!_{\vec{E},B}=_{E \cup B}t!_{\vec{E},B \cup B'}$, which, as shown above, forces $t!_{\vec{E},B}=_{B_{\varOmega }}t!_{\vec{E},B\cup B'}$, as desired. The only remaining task is to show that for each $f \in \varSigma$, $f_{\mathbb {C}_{\varSigma /\vec{E},B}} = f_{\mathbb {C}_{\varSigma /\vec{E},B \cup B'}}$. But this follows easily from $t!_{\vec{E},B}=_{B_{\varOmega }}t!_{\vec{E},B \cup B'}$, exactly as in the proof of Lemma 3. $\Box$

Proof of Theorem 7

Proof

Assuming (1), (2) and Theorem 6 (with $\varDelta = \emptyset$) yields $\mathbb {C}_{\varOmega /\vec{E}_{\varOmega },B_{\varOmega }}= \mathbb {C}_{\varOmega /\vec{E}_{\varOmega },B_{\varOmega } \cup B'_{\varOmega }}$ and $\mathbb {C}_{\varOmega /\vec{E}'_{\varOmega },B'_{\varOmega }} = \mathbb {C}_{\varOmega /\vec{E}'_{\varOmega }, B_{\varOmega } \cup B'_{\varOmega }}$. Assuming (1), (3) and Theorem 5 yields $\mathbb {C}_{\varOmega /\vec{E}_{\varOmega },B_{\varOmega } \cup B'_{\varOmega }} = \mathbb {C}_{\varOmega /\vec{E}_{\varOmega } \cup \vec{E}'_{\varOmega },B_{\varOmega } \cup B'_{\varOmega }} = \mathbb {C}_{\varOmega /\vec{E}'_{\varOmega },B_{\varOmega } \cup B'_{\varOmega }}$. Therefore,

$$\mathbb {C}_{\varOmega /\vec{E}_{\varOmega },B_{\varOmega }}= \mathbb {C}_{\varOmega /\vec{E}_{\varOmega },B_{\varOmega } \cup B'_{\varOmega }}= \mathbb {C}_{\varOmega /\vec{E}_{\varOmega } \cup \vec{E}'_{\varOmega },B_{\varOmega } \cup B'_{\varOmega }} = \mathbb {C}_{\varOmega /\vec{E}'_{\varOmega },B_{\varOmega } \cup B'_{\varOmega }} = \mathbb {C}_{\varOmega /\vec{E}'_{\varOmega },B'_{\varOmega }} $$

as desired. $\Box$

Proof of Theorem 8

Proof

Since $\mathcal {E} \equiv _{ sem } \mathcal {E}''$ whe have $\mathcal {E} \models _{ ind } E_{1} \cup B_{1}$. Since by hypothesis the rules $\vec{E} \cup \vec{E}_{1}$ are terminating modulo $B \cup B_{1} \cup B_{2}$, they are a fortiori terminating modulo $B \cup B_{1}$. Therefore, Theorem 5 applies and we have $(\varSigma ,E \cup E_{1} \cup B)$ admissible and $\mathcal {E} \equiv _{ sem } (\varSigma ,E \cup E_{1} \cup B)$, which implies $(\varSigma ,E \cup E_{1} \cup B) \models _{ ind } B_{1}$, and, since $\varSigma$ is $B \cup B_{1} \cup B_{2}$-preregular, a fortiori $\varSigma$ is $B \cup B_{1}$-preregular, so that Theorem 5 applies and we have $\mathcal {E}'$ admissible and $(\varSigma ,E \cup E_{1} \cup B) \equiv _{ sem } \mathcal {E}'$. Therefore, by symmetry and transitivity of $\equiv _{ sem }$, we get $\mathcal {E} \equiv _{ sem } \mathcal {E}' \equiv _{ sem } \mathcal {E}''$. $\Box$

$\text {B}\qquad$ NuITP Proof Scripts

To prove inductive theorems with the NuITP all modules should previously be entered in Maude after giving the Maude command set include BOOL off . This is because the BOOL module, which has several built-in features, would otherwise be added by default; but the NuITP does not expect any built-in features in the modules it proves properties about.

NuITP Proof Script for Example 3

NuITP Proof Script for Example 4

NuITP Proof Script for Example 5