2

I can't decide whether the argument below is valid or not.

On the one hand, the argument's first two premises and conclusion are false, so it is valid.

On the other hand, the argument's second premise and the conclusion are inconsistent with each other, so it is invalid.

Let $x,y,z$ be integers greater than $1,$ and define

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

$q(x,y,z)$: $z = \operatorname{GCF}(x,y)$

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

$$ \forall x\exists y\;(p(x,y) \implies r(x))\\ \forall x\forall y\forall z\;(\neg q(x,y,z))\\\exists x\forall y\;(p(x,y) \vee r(x)) $$

$$\therefore\quad\forall y\exists x\exists z\;q(x,y,z).$$

ryang
  • 44,428
Yenus
  • 33
  • @Yenus is the following "argument" "valid"? $$0\ne0$$$$\therefore0=0$$ – Anne Bauval Apr 27 '23 at 08:23
  • The argument you gave is invalid but my point is the second premise says that for all x,y and z we can't find any number that satisfies q(x,y,z) but the conclusion says that for some z and x and for all y we can find a number that satisfies q(x,y,z). If one of them is true then the other can't be true. Other question I can see from this is if we're given false premises and we know the conclusion that these premises imply is also false then can the argument be considered valid? – Yenus Apr 27 '23 at 10:57
  • @Yenus Anne’s presented argument is valid in FOL and invalid in PL. On the other hand, the first argument that I gave below is valid in both PL and FOL, while the second one invalid in both PL and FOL. – ryang Sep 09 '24 at 12:07

1 Answers1

1

the argument's first two premises and conclusion are false, so it is valid.

No. The arguments ∃x (x^2 < 0); therefore, 1≠1 and ∀x∀y x=y; therefore, 1≠1, each of which also has a false premise and false conclusion, are invalid, because each has a true premise and false conclusion in complex analysis, and when the universe is restricted to $\{3\},$ respectively.

(The above tells us that a false premise also doesn't automatically validate an argument. However, since every conditional with an unsatisfiable antecedent is logically true, an inconsistent set of premises does invalidate an argument.)

the argument's second premise and the conclusion are inconsistent with each other, so it is invalid.

No. The argument 1=1 and 1≠1; therefore, 1=1 has a premise and conclusion in contradiction, and (as explained above) is valid.

Let $x,y,z$ be integers greater than $1,$ and define

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

$q(x,y,z)$: $z = \operatorname{GCF}(x,y)$

$r(x)$: $x$ is prime. $$ \forall x\exists y\;(P(x,y) \implies R(x))\\ \forall x\forall y\forall z\;(\neg Q(x,y,z))\\\exists x\forall y\;(P(x,y) \vee R(x))\\\therefore\quad\forall y\exists x\exists z\;Q(x,y,z).$$

The following countermodel shows that the given argument's logical form can have a false conclusion without any false premise; so, the given argument must be invalid:

\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}

Reply to comment

You can validly reach false conclusions from every false premise, for instance, $2=4,$ therefore $2+2=4+2,$ therefore $4/2=6/2,$ therefore $2=3.$

@David So is the argument you gave in your explanation valid? If so, it means that arguments with false premises which reach false conclusions are valid arguments, right?

No, while David's example argument is certainly valid under arithmetic axioms, it isn't in itself logically valid (see this by defining 2=4 and x+y=y and x/y=y, noting that + and / are arithmetical rather than logical operators).

An argument being (logically) valid just means that its corresponding conditional is valid, that is, true regardless of the context. After all, argument validity depends just on argument form, and is independent of the argument's specific content.

ryang
  • 44,428
  • Thanks a lot for the help. So in order to determine the invalidity of the argument you looked for values that make the premises true and the conclusion false.But why did you use the set of real numbers as your domain of discourse and can you clarify the interpretation you used to prove the invalidity? I'm having a hard time understanding it and additionally, do you know of any sites or books that explain this further? – Yenus Apr 27 '23 at 21:49
  • @Yenus The countermodel still works when we change its domain to $\mathbb C$ or ${2,3,\ldots}.$ :⇔ means 'denotes'. – ryang Sep 09 '24 at 13:20