A note on standard completeness for some extensions of uninorm logic

Abstract

We prove standard completeness for uninorm logic extended with knotted axioms. This is done following a proof-theoretical approach, based on the elimination of the density rule in suitable hypersequent calculi.

Appendix: Proof of the derivability of \(\alpha _1,\ldots , \alpha _k \rightarrow \alpha _1^k \vee \dots \vee \alpha _k^k\) in \(UL\)

First, we show that $\text {HUL}$ derives the hypersequent $\alpha _1,\dots , \alpha _k$ $\Rightarrow \alpha _1^k \, | \, \dots \, | \, \alpha _1, \dots , \alpha _{k} \Rightarrow \alpha _{k}^{k}$ . We proceed by induction on $k$. For $k=2$, we have

For the induction step, we assume to have a derivation $d$ of $\alpha _1,\dots , \alpha _{k-1} \Rightarrow \alpha _1^{k-1} \, | \, \dots \, | \, \alpha _1,\dots , \alpha _{k-1} \Rightarrow \alpha _{k-1}^{k-1}$ in $\text {HUL}$. First, we show that for any $\alpha _i$ with $i=\{1,\dots , k-1\}$ we have a derivation $d_i$ in $\text {HUL}$ of the hypersequent $\alpha _k,\alpha _i^{k-1} \Rightarrow \alpha _k^k \, | \, \alpha _k \Rightarrow \alpha _i$ .

Consider now the following derivation. For space reasons, we abbreviate the hypersequent $\alpha _1,\dots , \alpha _{k-1} \Rightarrow \alpha _2^{k-1} \, | \, \dots \, | \, \alpha _1,\dots , \alpha _{k-1} \Rightarrow \alpha _{k-1}^{k-1}$ with $H$.

Starting from the end-hypersequent above and repeating similar derivations for any of the $d_i$, we eventually obtain the desired derivation of

$$\begin{aligned} \alpha _1,\dots , \alpha _k \Rightarrow \alpha _1^k \, | \, \dots \, | \, \alpha _1, \dots , \alpha _{k} \Rightarrow \alpha _{k}^{k} \end{aligned}$$

From this, we get:

Finally, by the completeness of $\text {HUL}$ with respect to $\text {UL}$ (see Metcalfe and Montagna 2007), we have that $\vdash _{UL} (\alpha _1,\ldots , \alpha _k) \rightarrow (\alpha _1^k \vee \dots \vee \alpha _k^k)$. This completes the proof.