I am trying to prove the validity of the following: $$(\exists x Px\to \forall y Qy) \to \forall z (Pz \to Qz)$$ but I am completely stumped. Can anyone offer any insight?
2026-04-08 06:08:59.1775628539
How can I prove that this sentence is valid?
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1
Using Natural Deduction :
1) $∃xPx→∀yQy$ --- premise [a]
2) $Pz$ --- premise [b]
3) $∃zPz$ --- from 2) by $∃$-intro
4) $∀yQy$ --- from 1) and 3) by $\to$-elim
5) $Qz$ --- from 4) by $∀$-elim
6) $Pz→Qz$ --- from 2) and 5) by $\to$-intro, discharging [b]
7) $∀z(Pz→Qz)$ --- from 6) by $∀$-intro.
Thus, from 1) and 7):
We can also easily prove the validity of the formula in a "semantical way".
For contradiction, assume that $∃xPx→∀yQy$ is true while $∀z(Pz→Qz)$ is false.
Thus, for some $a$: $Pa→Qa$ is false, i.e. $Pa$ is true and $Qa$ is false.
But if $∃xPx→∀yQy$ is true, then either $∃xPx$ is false or $∀yQy$ is true.
In the first case we have a contradiction with $Pa$ true and in the second case we have a contradiction with $Qa$ false.