# Confusion in determining an argument's validity?

I understand that an argument is basically an implication in the sense that: $$premise1 \land premise2 \land premise3.... \to conclusion$$

And an argument would be considered valid when such an implication is a tautology. What I don't understand is do we consider the actual truth value of the statements when validating an argument?

Lets say this is an argument:

Premise: 36 is divisible by 9

Conclusion: 36 is divisible by 3

Now this argument, when taken as $$p \to q$$ where $$p$$ is a premises and $$q$$ is a conclusion is invalid. Whereas the intuitive sense by understanding this argument says that the argument holds according to mathematical rules.

My question is when we formulate a truth table of the above $$p \to q$$, there is one condition when $$p = T$$, $$q = F$$ when the argument fails. But intuitively and in all practical senses, $$p = T$$, $$q = F$$ is not a possible combination and hence why we can ignore it saying the argument is valid?

Or an argument is in fact when we don't talk in concrete statements and rather abstract statement variables, which for any statements should work?

Lastly, what is the difference between an argument being valid and implication (of the form $$p \land q \land r \to z$$) being true?

## 1 Answer

An argument where the premises are $$P_1, \dots P_n$$ and the conclusion is $$C$$ is valid if and only if the formula $$(P_1 \land \dots \land P_n) \to C$$ is a tautology, i.e. the conclusion is true whenever all the premises are true.

If you formalize your argument as a formula $$p \to q$$ where $$p$$ stands for "36 is divisible by 9", and $$q$$ stands for "36 is divisible by 3", then your argument is not valid, from a logical point of view.

Why we have the feeling that your argument is actually valid? Because you are implicitly assuming more premises, i.e.:

"9 is divisible by 3"

"divisibility is a transitive relation".

The problem is: how can one express transitivity in logic (which is needed to make the argument valid)? You should move from propositional logic to first-order logic. There is no way to make your argument valid if it is formalized in propositional logic.

In first-order logic, let $$D(x,y)$$ be the relation "$$x$$ is divisible by $$y$$". Your argument becomes (where $$\forall$$ stands for "for all"):

premise 1: $$D(36,9)$$

premise 2: $$D(9,3)$$

premise 3 (transitivity): $$\forall x \forall y \forall z ((D(x,y) \land D(y,z)) \to D(x,z))$$.

conclusion: $$D(36,3)$$.

You can easily prove that this argument is logically valid, i.e. $$D(36,3)$$ holds whenever $$D(36,9)$$, $$D(9,3)$$ and $$\forall x \forall y \forall z ((D(x,y) \land D(y,z)) \to D(x,z))$$ hold. Indeed, as $$\forall x \forall y \forall z ((D(x,y) \land D(y,z)) \to D(x,z))$$ holds, in particular $$(D(36,9) \land D(9,3)) \to D(36,3)$$ holds (take $$x = 36$$, $$y = 9$$ and $$z = 3$$). Since both $$D(36,9)$$ and $$D(9,3)$$ hold, then $$D(36,9) \land D(9,3)$$ holds and hence $$D(36,3)$$ holds by modus ponens with $$(D(36,9) \land D(9,3)) \to D(36,3)$$ and $$D(36,9) \land D(9,3)$$.

Concerning your last question, there is a subtle but crucial difference between saying that an argument (where the premises are $$P_1, \dots, P_n$$ and the conclusion is $$C$$) is valid and saying that the formula $$(P_1 \land \dots \land P_n) \to C$$ is true. To simplify the explanation, I suppose the argument is formalized in propositional logic, but in first-order logic it is analogous.

Saying that an argument (where the premises are $$P_1, \dots, P_n$$ and the conclusion is $$C$$) is valid means that the formula $$(P_1 \land \dots \land P_n) \to C$$ is a tautology, i.e. $$C$$ is true whenever $$P_1, \dots P_n$$ are all true. In other words, it is impossible to find a situation (technically, a valuation) where $$P_1, \dots, P_n$$ are all true and $$C$$ is false: every interpretation that makes $$P_1, \dots, P_n$$ true, makes $$C$$ true as well. If you know truth tables, this means that you are saying something about all the rows of the truth table of the formula $$(P_1 \land \dots \land P_n) \to C$$.

Saying that a formula $$(P_1 \land \dots \land P_n) \to C$$ is true means that you are in a specific interpretation that makes $$(P_1 \land \dots \land P_n) \to C$$ true (i.e. you are in an interpretation that either makes one of the $$P_i$$'s false or makes $$C$$ true, according to the truth table of $$\land$$ and $$\to$$). So, you are talking about a specific interpretation, i.e. a specific row of the truth table of the formula $$(P_1 \land \dots \land P_n) \to C$$, you are not talking about all the rows of the truth table of the formula $$(P_1 \land \dots \land P_n) \to C$$.

• So in all essence, the argument I provided can NOT be proven valid through propositional/aristotelian logic right except if I explicitly add more premises (which by the way have been implicitly asserted). – Parker Queen Oct 7 '18 at 19:06
• Lastly, can you explain the difference between an argument being valid and the conditional of the form stated above true? – Parker Queen Oct 7 '18 at 19:07
• @ParkerQueen - Exactly. – Taroccoesbrocco Oct 7 '18 at 19:07
• @ParkerQueen - Concerning your last question, do you know what a valuation is (in propositional logic)? – Taroccoesbrocco Oct 7 '18 at 19:11
• I understand that propositional statements are either true or false and that implication is a statement and argument is a set of propositions etc.... But considering the 'Principal of Inference' (conditional) form of an argument, what stops me from saying an argument is basically a statement? – Parker Queen Oct 7 '18 at 19:19