Use a truth table to answer the following question.Which, if any, set of truth values assigned to the atomic sentences shows that the following argument is invalid? $\begin{array}{l}A\cup B\\ \frac{A}{\sim B}\end{array}$

A) A: T B: T
B) A: T B: F
C) A: F B: T
D) A: F B: F
E) None-the argument is valid.

The argument is valid because there is no row in the truth table where the premises are true and the conclusion is false. The argument structure is such that if $A$ is true, then $\sim B$ (not B) must also be true for the conclusion to follow. However, since the premises include $A\cup B$ (A or B), and the conclusion is $\sim B$ , the only way the argument could be invalid is if both $A$ and $B$ were false, which is not an option given the structure of the argument. Therefore, the argument is valid under all possible truth values for $A$ and $B$ .