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I'm trying to derive the following as a theorem in a FOL Hilbert system:

$$∀x∃yQ(xy) → ∃yQ(yy) $$

But I'm a bit confused as to how one should interpret the incidence of two variables in $Q$. Is there variable collision or are the two statements separated by the logical implication not related in that sense? Is this an instantiation?

Tankut Beygu
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ne.ko_92
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    (1) The conditional has operational precedence over quantifier scope binding. The statement is $(\forall x~(\exists y~Q(xy)))~\to~(\exists y~Q(yy))$ . (2) Collisions may be averted by alpha-substitution, $(\forall x~(\exists y~Q(xy)))~\to~(\exists z~Q(zz))$ (3) Before trying to prove it, determine whether it is valid. ... – Graham Kemp Mar 15 '22 at 02:34
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    Hint: Consider a domain of discourse that contains precisely two distinct objects: ${ 0,1}$. – Dan Christensen Mar 15 '22 at 02:44

2 Answers2

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Beside the points stated in the comments, if you are not somehow familiar with the formula, you ought to check its validity, whether there is any counter-instance to it, preferably, by a semantic tableaux method.

A counter-instance has to make $\forall x\exists yQxy$ true and $\exists yQyy$ false. Eventually, you will see that in a domain with one element $\{a\}$ the formula does not have a counter-instance, but in domains with two $\{a, b\}$ (or more) elements, it does (the tableau will not close): There exists an assignment of truth values which verifies $Qab$ and $Qba$, but falsifies $Qaa$ and $Qbb$. Hence, it is not a valid formula.

Tankut Beygu
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Just posted by a different user: Naming random variables.

The point is that while it is technically correct to recycle a variable among different scopes (in the given sentence, each quantifier's scope does not extend across the material conditional "$→$"), many dislike this practice for poor communication and being misleading.

Christian's answer here is careful about not recycling variable names, whereas mine isn't so much.

ryang
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