If $(A \lor C) \implies B$ and $(C \implies (\lnot A \land B))$ both hold, is $B$ true?

Question

Problem Statement

If the two implications $$D \equiv \left[\left(A \lor C\right) \implies B\right]$$ and $$E \equiv \left[C \implies \left(\lnot A \land B\right)\right]$$ both hold, is $B$ true?

My Attempt

Here is the truth table:

$$ \begin{array}{c|c|c|c|c|c} A & B & C & D & E & D \land E \\ \hline T & T & T & T & F & F\\ T & T & F & T & T & T\\ T & F & T & F & F & F\\ T & F & F & F & T & F\\ F & T & T & T & T & T\\ F & T & F & T & T & T\\ F & F & T & F & F & F\\ F & F & F & T & T & T\\ \end{array} $$

Conclusion

The last row is an instance of when $D$ and $E$ are both true, and yet $B$ is false. This occurs when all of $A$, $B$ and $C$ are false. Hence, the proposition does not hold, unless at least one of $A$, $B$ or $C$ is true.

Question

Is this conclusion valid? If so, is there a way to shorten the proof?

Answer 1

Your conclusion is fine. You can shorten the argument by stating it in a more informal fashion, but it’s essentially still just the truth table argument:

Suppose that $B$ is false. Since $D$ is true, $A\lor C$ must be false, which means that $A$ and $C$ must both be false. But then the falsity of $C$ ensures that $E$ is true, even though $B$ is false, so we cannot conclude from the truth of $D$ and $E$ that $B$ must be true.