Need help understanding valid arguments

Question

I don't understand the following parapgraph in "Discrete Mathematics and Its Applications (Rosen)": "The argument form with premises p1, p2,...,pn and conclusion q is valid, when (p1 AND p2 AND ... AND pn) --> q is a tautology."

This proposition is false when (p1 AND p2 AND ... AND pn) is true but q is false. So how can this ever be a tautology?

Answer 1

In propositional calculus we start with a set of connectives : $\lnot, \land, \ldots$ and a set $S$ of sentence symbols (or propositional letters) : $p_1, p_2, \ldots$.

Then we define inductively the set of well-formed formulas.

We introduce the notion of truth assignment :

a truth assignment $v$ for a set $S$ of sentence symbols is a function

$v : S \to \{ F, T \}$

assigning either $T$ or $F$ to each symbol in $S$.

[...]

Let $\overline S$ be the set of wffs that can be built up from $S$ by the conncetives. We define an extension $\overline v$ of $v$, that assigns the correct truth value to each wff in $S$ in the obviuos way :

$\overline v (\lnot \alpha)=T$ if $\overline v (\alpha)=F$,

etc.

Now consider a set $\Sigma$ of wffs (the set of premises) and another wff $\tau$ (the conclusion).

We say that $\Sigma$ (tautologically) implies $\tau$, written :

$\Sigma \vDash \tau$

iff every truth assignment for the sentence symbols in $\Sigma$ and $\tau$ that satisfies every member of $\Sigma$ also satisfies $\tau$.

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Note

If $\Sigma = \{ p_1,\ldots, p_n \}$ and $\tau$ is a proposition in the sentence symbols in $\Sigma$, then we can re-phrase the above definition in the terms of your textbook, saying that :

the argument form with premises $p_1, p_2,\ldots,p_n$ and conclusion $\tau$ is valid iff every truth assignment for the sentence symbols $p_1, p_2,\ldots,p_n$ that satisfies every $p_i$ also satisfies $\tau$.

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With the above definitions, it is easy to prove that :

$\alpha \vDash \beta$ iff $\vDash (\alpha \rightarrow \beta)$.

This theorem can be generalized to :

if $\Sigma = \{ \alpha_1,\ldots, \alpha_n \}$, then

$\Sigma \vDash \beta$ iff $\vDash (\alpha_1 \land \ldots \land \alpha_n) \rightarrow \beta$.

Answer 2

Specific examples might make things clearer. If your premises are simply the sentences of the form ‘$p$’ and ‘$q$’, and your conclusion is a sentence of the form ‘$r$’, then your argument would not be valid because the sentence ‘$p \wedge q \to r$’ is not a tautology.

However observe what happens if the above conclusion is ‘$p$’ rather than ‘$r$’. In this case we have a valid argument because the sentence ‘$p \wedge q \to p$’ is a tautology. The argument in this case being that if our premises in conjunction are true, then each premise is true individually.

For another example, suppose the premises are sentences of the form ‘$p$’ and ‘$p \to q$’ and the conclusion a sentence of the form ‘$q$’. In this case the sentence which is tested to determine validity is ‘$p \wedge ( p \to q ) \to q$’ which is a tautology, therefore the argument is a valid one. Specifically this is Modus Ponens.

On the other hand, if the premises have the form ‘$p \to q$’ and ‘$q$’, and the conclusion ‘$p$’, then this is another example of an argument which is not valid because the sentence ‘$( p \to q ) \wedge q \to p$’ is not a tautology. This is the formal fallacy called Affirming the Consequent.

The thing to bear in mind is that in your definition the sentence ‘$ p _1 \wedge \dots \wedge p _n \to q$’ is a schema in which premises and conclusion of specific forms are substituted. Only then do you judge the sentence as tautology or not.