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I'm stucking in understanding the usage and soundness of the rules for the quantifiers $\forall,\exists$ in sequent calculus.

$\forall-L$: $~~~~~\dfrac{\Gamma,\phi[t]\vdash \Delta}{\Gamma,\forall x\phi[x/t]\vdash\Delta}$ $~~~~~~~~~~~~~~~~~~\forall-R$: $~~~~~\dfrac{\Gamma\vdash\phi[y],\Delta}{\Gamma\vdash\forall x\phi[x/y],\Delta}$

$\exists -L$: $~~~~~\dfrac{\Gamma,\phi[y]\vdash \Delta}{\Gamma,\exists x\phi[x/y]\vdash\Delta}$$~~~~~~~~~~~~~~~~~~\exists-R$: $~~~~~\dfrac{\Gamma\vdash\phi[t],\Delta}{\Gamma\vdash\exists x\phi[x/t],\Delta}$

Take $\forall-L$ for example, how do "$\Gamma,\phi[t]\vdash \Delta$" tautologically imply "$\Gamma,\forall x\phi[x/t]\vdash\Delta$"?

Also, why is $\exists -L$ sound? If for every model $(\mathfrak{A},\sigma)$ satisfies $\Gamma,\phi[y]$ also satisfies $\Delta$, why $\Gamma,\exists x\phi[x/y]\vDash\Delta$? I tried hard to think about it, but it's so complicated... (I'm not used to the open formula involved in $\vdash$ or $\vDash$ symbol as well.) Detailed answer is welcome. Thanks.

Taroccoesbrocco
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Eric
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1 Answers1

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Concerning the rule $\forall\text{-}L$, the intuitive meaning is the following: if under the hypotheses $\Gamma$ and $\phi[t]$ (for some term $t$) you can derive $\Delta$ (this is the premise of the rule $\forall\text{-}L$), then also under the stronger hypotheses $\Gamma$ and $\forall x \phi[x]$ you can derive $\Delta$ (this is the conclusion of the rule $\forall\text{-}L$). The hypotheses $\Gamma$ and $\forall x \phi[x]$ are stronger than the hypotheses $\Gamma$ and $\phi[t]$ (for some term $t$) because if you suppose that $\forall x \phi[x]$ is true then $\phi[x]$ is true for any value of the variable $x$, in particular when $x$ is replaced by the term $t$.

Concerning the rule $\exists\text{-}L$, first you have to recall that this rule (as well as the rule $\forall\text{-}R$) is valid only if the variable $y$ does not occur free within $\Gamma$ and $\Delta$. The intuitive meaning of the rule is the following. Suppose that under the hypotheses $\Gamma$ and $\phi[y]$ you can derive $\Delta$ (this is the premise of the rule $\exists\text{-}L$). Now, since $y$ does not occur free in $\Gamma$ and $\Delta$, this means that you didn't make any hypothesis about $y$, so the fact that $\Delta$ derives from $\Gamma$ and $\phi[y]$ actually means that you can derive $\Delta$ from $\Gamma$ and $\phi[y]$, for any value of the variable $y$. In other words, to derive $\Delta$ it is enough to suppose $\Gamma$ and the fact that there exists an $x$ such that $\phi[x/y]$, that is, under the hypotheses $\Gamma$ and $\exists x \phi[y/x]$ you can derive $\Delta$ (this is the conclusion of the rule $\exists\text{-}L$).

Taroccoesbrocco
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