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Let $P$ be a formula with two variables. Given the premises:

  1. $\forall x \exists y P(x,y)$
  2. $\forall x \forall x' \forall y \forall y' (P(x, y) \land P(x',y') \to y = y')$

I would like to derive $\exists y \forall x P(x,y)$.

Informally, the idea is to introduce a function $(f(x) = y) \leftrightarrow P(x,y)$ using the premises and then show that $f$ is a constant function. However, as I am new to formal logic, I struggle to construct a formal proof using the rules of inference of a natural deduction system (e.g. the Fitch system). Therefore I would like to ask for a formal proof of this result.

My question is not that I don't understand the quantifier rules mentioned in What set of formal rules can we use to safely apply Universal/Existential Generalizations and Specifications?, but it's rather that I don't know how to formalize the idea I have presented above into a formal proof, as it requires introducing a function in the middle of the proof and derive further statements from it. And maybe the proof should not proceed in the way I proposed after all, in that case, what should be done to construct a proof of this statement?

FR09
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2 Answers2

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Hint

From 1st premise derive $P(a,b)$ for some $b$ and assume the negation of the conclusion $\lnot \exists y \forall x P(x,y)$, that is: $\forall y \exists x \lnot P(x,y)$.

From it, we have $\lnot P(c,b)$ for some $c$ and using again 1st premise: $P(c,d)$ for some $d$.

Now use 2nd premise with $P(a,b)$ and $P(c,d)$ to conclude with $b=d$.

From $\lnot P(c,b)$ we have $\lnot P(c,d)$ by laws of equality, and we have reached the desired contradiction.

Mauro ALLEGRANZA
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The idea is to use the formula $$\forall_x\left(\exists_yP(x,y)\Rightarrow \exists_y \forall_{x'}P(x',y)\right)$$ as the in-between step (I leave out the proof as an exercise).

Then, since $\exists_y \forall_{x'}P(x',y)$ does not have the variable $x$ free, we get $$\exists_x\exists_y P(x,y) \Rightarrow \exists_y\forall_{x'}P(x',y)$$ By assumption 1 (since in non-empty domain universal quantifier implies the existensial quantifier) we have $\exists_x\exists_y P(x,y)$ and by modus ponens $\exists_y\forall_{x'}P(x',y)$, which is the same as $\exists_y\forall_{x}P(x,y)$.

Kulisty
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