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Using natural deduction rules prove the statement:

$∀x∀y(P(x, y) → ¬P(y, x)),∀x∃yP(x, y)$ can be deduced to $¬∃v∀zP(z, v)$

I don't want to be given an answer, but a hint on how to start because I am stuck for a long time on this problem not knowing how to start efficiently.

user737163
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  • Well, I don't know predicate logic rules, but in English it's pretty clear. The first premise is that $P$ is anti-symmetric, the second is that everything (In your universe of discourse) is in the domain of $P$. From that they want you to conclude that there's nothing that could be a universal second element of $P$, which is clear because from statement 2, the reverse order pair would exist for one element, and statement 1 would say that can't happen. Can you convert that into your rules? – Alan Nov 12 '21 at 20:25

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Obviously here you can use proof by negation together with formal inference rules regarding quantifiers allowed in your system, and then a useful hint would be the predicate $P$ here can be interpreted as a strict partial order and thus $∃v∀zP(z,v)$ must be false when $z=v$, which is certainly an unavoidable special case for $∃v∀zP(z,v)$, while the given premise $∀x∃yP(x,y)$ can avoid such special case...

cinch
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  • why must the predicate be false for $z=v$? What if it handles the special case and is true? – user737163 Nov 14 '21 at 12:29
  • @user737163 pls note $∃v∀zP(z,v)$ is a universal condition stronger than $∀x∃yP(x,y)$, and order of mixed quantifiers matters. If you use natural number N with its usual < binary relation (strict partial order) as a model, then clearly when $z=v$, the assumed conclusion $∃v∀zP(z,v)$ must be false, thus a contradiction. And the previous 2 premises are used to formally prove this strict partial order property. Since you stated you only need hint, I think this is a critical one. Hope you now "see" it.. – cinch Nov 14 '21 at 19:46