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As in the title: Why do we know that a set $S$ of sentences implies the sentence $A$ if the set $S \cup \lnot A$ is inconsistent?

I "know" it's because if $S \cup \lnot A$ is inconsistent, then $S \cup A$ must be consistent. But why must $S \cup A$ be consistent if $S \cup \lnot A$ is inconsistent? Do we assume that $S$ is consistent, and notice that if we add $\lnot A$ to it, the union is inconsistent, and from there deduce that if instead of $\lnot A$ we would've added $A$, the set would've been consistent? If so, how can we deduce this? What makes it impossible for $S \cup \lnot A$ and $S \cup A$ both to be inconsistent or consistent?

I don't know if it matters, but I'm talking about basic propositional logic.

user265554
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2 Answers2

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I "know" it's because if $S\cup\{\neg A\}$ is inconsistent, then $S\cup\{A\}$ must be consistent.

No. That's not why it's the case, and indeed that statement is false (e.g. take $S$ itself to be something inconsistent). It is true if we assume $S$ is consistent, but we don't need to do this.

The point is that proof by contradiction is built into classical logic: in $S$, we can argue roughly as follows:

"Suppose $\neg A$. Then (since we're assuming that $S\cup\{\neg A\}$ is inconsistent) we can prove a contradiction. So our hypothesis $\neg A$ was false, hence $A$ is true."

Note that a consequence of proof by contradiction is that an inconsistent theory proves every sentence.

Noah Schweber
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  • So there's nothing to really "understand", since it's just a rule? – user265554 Oct 16 '17 at 15:51
  • @user265554 Well, you should understand proof by contradiction and the reason for accepting it as a logical principle. Think of the statement "$S$ proves $\varphi$" as meaning "If all the statements in $S$ are true, then so is $\varphi$." If $S$ is inconsistent, it can't be the case that all the statements in $S$ are true, so "If all the statements in $S$ are true, then so is $\varphi$" is vacuously true (in the same way that "all flying elephants are pink" is true). The motivation behind proof by contradiction is discussed in many places on this site, if you're curious. – Noah Schweber Oct 16 '17 at 15:58
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To say that $S \cup \{ \lnot A \}$ is inconsistent is to say that it is unsatisfiable, i.e. there is no valuation $v$ such that $v(\sigma)=$ t for every $\sigma \in S$ and $v(\lnot A)=$ t.

Consider now a valutaion $v$ such that $v(\sigma)=$ t for every $\sigma \in S$.

By the above we must have $v(\lnot A)=$ f, and thus $v(A)=$ t.

This implies that:

$S \vDash A$.

Mauro ALLEGRANZA
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