Suppose that the discourse domain is $\{2,3,\ldots\}$ and
$p(x,y)$: $x$ is a factor of $y$
$q(x,y,z)$: $z = GCF(x,y)$
$r(x)$: $x$ is prime.
Check whether this argument is valid:
$$ \forall x\exists y\;(p(x,y) \implies r(x))\\ \forall x\forall
y\forall z\;(\neg q(x,y,z))\\\exists x\forall y\;(p(x,y) \vee
r(x)) $$$$\therefore\quad\forall y\exists z\exists x\;q(x,y,z).$$
I can't decide on my answer because:
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the first and second premises can be disproved and so they're false, and a valid argument can't have false premises
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the second premise and the conclusion contradict each other.
Best Answer
Every argument with inconsistent premises corresponds to a conditional with an unsatisfiable antecedent, that is, a conditional that is true regardless of interpretation, so is automatically valid. For example, this argument is valid: $$1=1\,\text{ and }\,1\ne1; \text{therefore }1=1.$$
The second premise of the above valid argument is false and contradicts its conclusion.
(A false premise makes an argument unsound, but doesn’t invalidate it.)
In my first sentence, “regardless of interpretation” refers to the fact that an argument's validity depends on its logical form. So, the above chunk of information (called an interpretation) is just a red herring.
This argument is invalid, but the red-herring interpretation above gives it a false premise so doesn't reveal this. On the other hand, the following counter-interpretation gives the argument no false premise and a false conclusion:
\begin{align}&\text{discourse domain}=\mathbb R\\P(x,y)\quad&{:\Leftrightarrow}\quad x+y=x+y\\ Q(x,y,z)\quad&{:\Leftrightarrow}\quad x+y+z\ne x+y+z\\ R(x)\quad&{:\Leftrightarrow}\quad x=x.\end{align}
Incorrect: in the following interpretation, the second premise and conclusion are both false: \begin{align}\text{discourse domain}=\mathbb Z\\Q(x,y,z)\quad{:\Leftrightarrow}\quad x=y=z=0.\end{align}
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Not necessarily: this argument, in real analysis, has false premises and conclusion, and is invalid, because in complex analysis its premise is true and its conclusion false: $$\exists x\,x^2{<}0; \text{therefore }1\ne1.$$