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Having the inference rules

$$ \frac{\Gamma, A[x:t] \implies \Delta}{\Gamma, \forall x A \implies \Delta} \forall L $$

$$ \frac{\Gamma, A[x:y] \implies \Delta}{\Gamma, \exists x A \implies \Delta} \exists L$$

$$ \frac{\Gamma \implies \Delta, A[x:y]}{\Gamma\implies\Delta,\forall x A} $$

$$ \frac{\Gamma \implies \Delta, A[x:t]}{\Gamma\implies\Delta,\exists x A} $$

I want to prove $\forall x \neg P(x) \implies \neg \exists y P(y)$. I already tried

$$\dfrac{\dfrac{\dfrac{\dfrac{P(t)\implies P(t)}{\neg P(t), P(t)\implies}}{\neg P(t), \exists y P(y) \implies}}{\forall x \neg P(x), \exists y P(y) \implies}}{\forall x \neg P(x) \implies \neg \exists y P(y)}$$

but in the second step, I cannot use $t$, can I? How can I prove this?

Taroccoesbrocco
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1 Answers1

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Your second step (from $\lnot P(t), P (t) \implies$ to $\lnot P(t), \exists y P (y) \implies$) is wrong for two reasons, according to the restrictions to the inference rule $\exists L$ (see here and here):

  1. $t$ should be a variable, not a generic term;
  2. if you consider $t$ as a variable, it must not occur free anywhere in the sequent $\lnot P(t), \exists y P (y) \implies$, but it does occur free in $\lnot P(t)$.

It is easy to fix this problem: you have to swap the rule $\exists L$ and $\forall L$ in your derivation. Indeed, in your second step you have to apply the inference rule $\forall L$ (which does not have any restriction), in this way you get the sequent $\forall x \lnot P(x), P(y) \implies$, where no variable occurs free except the one you want to quantify with $\exists$ and then you can apply the inference rule $\exists L$, so as to get $\forall x \lnot P(x), \exists y P(y) \implies$ where no variable occurs free. So, a correct derivation in the sequent calculus is the following:

$$\dfrac{\dfrac{\dfrac{\dfrac{P(y)\implies P(y)}{\neg P(y), P(y)\implies}}{ \forall x \lnot P(x), P(y) \implies}\forall L}{\forall x \neg P(x), \exists y P(y) \implies}\exists L}{\forall x \neg P(x) \implies \neg \exists y P(y)}$$

Taroccoesbrocco
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