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I wanted to develop a tree-format natural deduction of the following tautology.

$$ \exists_{x \in S}(P(x)) \quad \implies \quad ( \forall_{y \in S}(P(y) \implies Q(y)) \implies \exists_{z \in S}(Q(z)) \; ) $$

Intuitively I can see how it is true, and I can also see how the proof should work in outline form, informally:

  • $\exists_{x \in S}(P(x))$ means there exists an $a$ such that $P(a)$
  • $\forall_{y \in S}(P(y) \implies Q(y)$ gives us $Q(a)$ from $P(a)$.
  • From $Q(a)$ we get $\exists_{z \in S}(Q(z)))$

The following is my tree-format deviation.

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Question: I am concerned my use of the constant symbol $a$ violates the rules for scoping constants used in sub-proofs. Is there an error in the above proof?


Discussion

I think the $a$ introduced in the assumption $[P(a)]^1$ is fine in the conclusion of the $\implies$-elim rule, $Q(a)$.

But I don't think that $a$ should be used in the line $P(a) \implies Q(a)$ because that is not strictly within the scope of the assumption $[P(a)]^1$.

Perhaps I should have $P(b) \implies Q(b)$?

But thinking further, $P(a) \implies Q(a)$ can't be used without the assumption of $a$ and $P(a)$, so maybe the structure of my tree is wrong?

For reference, this answer (link) gives a non-tree natural deduction proof.

My reference for "constant symbol" is this chapter (pdf).

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The constant is not introduced by the assumption $[P(a)]^1$ (which can be used multiple times in the proof, anyway); it is introduced by the $\exists$-elim rule at the bottom, and it is “in scope” anywhere in the tree above the right premise of the $\exists$-elim rule. Your proof is just fine.

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  • $\begingroup$ Thanks, this is helpful. I didnt' think of it this way - that a constant is introduced by an elimination rule and can be used anywhere in the relevant sub-tree. $\endgroup$ Commented Oct 12 at 11:30
  • $\begingroup$ (I have noticed I haven't finished the proof with two additional $\implies$-intro rules). $\endgroup$ Commented Oct 12 at 11:31

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