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Suppose $x, y, z > 1$ are integers, let:

$p(x,y)$ : $x$ is a factor of $y$

$q(x,y,z)$ : $z$ = $\text{GCD}(x,y)$

$r(x)$ : $x$ is prime.

Check if the following argument is valid or not.

$(\forall x \exists y)p(x,y) \implies r(x)$ ,

$(\forall x)(\forall y)(\forall z)(q(x,y,z))$ ,

$(\exists x)(\forall y)(p(x,y) \lor r(x))$

$\therefore (\forall y)(\exists z)(\exists x)q(x,y,z)$
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It is a valid argument since premise is false.

In premises $\forall x \forall y \forall z (q(x,y,z))$ is false, take counter example $x=2, y = 3, z=5$ in this premise is false.

Since premise is false the argument is valid, since it is not possible to have all the premises true and conclusion false.
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Quick revision:

an argument is valid when it always gives “true” as output irrespective of the input feed to it. Otherwise, it is invalid

Therefore, the only way to prove the given argument invalid is to somehow come up with T → F

 

Out of the given three premises,

(∀x)(∀y)(∀z)(q(x,y,z)) is false.

If we take the values of x,y, and z as 2,3 and 4 respectively, z != GCD(x,y), as GCD of (2,3) is 6, while z is 2.

Now, as per the truth table of implication, we know that the only way it can be invalid is the case T → F.

But we already got an F on the left hand side. Thus, we can say that it doesn’t matter whether F → T or F → F. As both of them will result in tautology. Hence, it is a valid argument.