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 $$(\neg A(x) \lor \neg B(x)) \land (\neg A(x) \lor B(x)) \equiv (\neg A(x))$$


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