0
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
-1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.