# Propositional Logic: Entailment

Given the sentences, S1, S2, if S1 |= S2 then all models that satisfy S1 also satisfy S2, how is the following statement correct?

A ∧ ¬A |= B

How can something and not something equate to true?

I am stuck trying to work out how true and false can still entail true, actually how can A and ¬A occur at the same time?

Thanks.

It's not easy to give a full explanation without knowing your axioms and rules, however, $$A \land \neg A$$ is a contradiction, and it is by exploiting that, that you can prove whatever you like.

First, let $$\top = A \lor \neg A$$.

We know that "true or whatever else" must be true. Since $$\top$$ is a tautology, then $$\top \lor B$$ must be true.

But since $$A \land \neg A = \neg (A \lor \neg A) = \neg \top$$ is assumed to be true we conclude the following:

• from $$\top \lor B$$ and $$\neg \top$$
• $$B$$

It all hinges on the fact that $$A \lor \neg A$$ is a tautology and that you assume that the negation of a tautology is true.

So to answer your question "actually how can $$A$$ and $$\neg A$$ occur at the same time?" It cannot, unless you have a contradiction. When you assume a contradiction, you can prove anything.

• So does this mean I can also prove its false? Feb 1 at 11:14
• What do you mean by it? You can also prove $\neg B$ by assuming a contradiction, yes. In fact, if you assume a contradiction, you can prove anything. Feb 1 at 13:38
• Could whoever down voted this answer, which actually makes sense to me explain why? It would be useful for my understanding. Feb 1 at 20:48

Think of it this way.

There are no elements of the empty set $$\emptyset$$. Moreover, for all sets $$A$$, $$\emptyset \subseteq A$$.

But what does subset actually mean? $$A \subseteq B$$ means that any element of $$A$$ must be an element of $$B$$. That is, $$\forall x \in A, x \in B$$.

It follows that for all sets $$A$$, $$\forall x \in \emptyset, x \in A$$. Every single element of $$\emptyset$$, every one of them, every none of them, is an element of every other set.

Assuming that $$A \wedge \neg A$$ is false in your logic (there are logics that admit controlled contradiction), then there are no models satisfy $$S_1$$. It follows that all models that satisfy $$S_1$$, every one of them, every none of them, also satisfy $$S_2$$.