# What restrictions lay on truth sets of predicates A(x) and B(x) if the given predicate is true

Suppose we are given 2 predicates $$A(x)$$ and $$B(x)$$ with domain $$M$$.

Suppose next we are given the following predicate $$\neg (A(x) \land B(x)) \land (\forall x(A(x) \rightarrow B(x)))$$ which we know is true, so $$\neg (A(x) \land B(x)) \land (\forall x(A(x) \rightarrow B(x))) = 1$$

The question is how does it restrict the truth sets of $$A(x)$$ and $$B(x)?$$

It is obvious that we have $$\neg (A(x) \land B(x)) = 1 \\ A(x) \land B(x) = 0\\ A(x) = 0 \lor B(x) = 0$$ So from that we get that either truth set for $$A(x)$$ is $$E_A \neq M$$ or truth set for $$B(x)$$ is $$E_B \neq M$$.

But knowing that $$\forall x(A(x) \rightarrow B(x)) = 1\\ \forall x(\neg A(x) \lor B(x)) = 1$$ I have no idea how to link it to useful information on truth sets of $$A(x)$$ and $$B(x)$$, any suggestions?

You are getting hung up on truth sets when there is only one variable in the problem. Focus on one element $$x$$ of $$M$$ and ask whether $$A(x)$$ and $$B(x)$$ can be true because all the elements are equivalent. You have found that both $$A(x)$$ and $$B(x)$$ cannot both be true but $$A(x) \implies B(x)$$. You should be able to derive that $$A(x)$$ is false and $$B(x)$$ can be anything. Check that in the original axiom and it works.
1 $$\neg (A(x) \land B(x)) \land (\forall x(A(x) \rightarrow B(x)))$$
2 $$\neg (A(x) \land B(x)) \equiv (\neg A(x)) \lor (\neg B(x))$$
$$(\forall x(A(x) \rightarrow B(x))) \equiv (\neg A(x)) \lor B(x)$$
3 $$(\neg A(x) \lor \neg B(x)) \land (\neg A(x) \lor B(x)) \equiv (\neg A(x))$$