p→Q RAS-Q→PAS

There are a few ways to approach this problem, but one common method is to use the rules of logical inference and truth tables. Here is one possible solution:

First, we can use the rule of modus ponens to infer that if p → Q and p is true, then Q must also be true. Similarly, if -Q → PAS and -Q is true, then PAS must also be true. So we have:

1. p → Q (premise)
2. -Q → PAS (premise)
3. p (assumption)
4. Q (modus ponens from 1 and 3)
5. -Q (assumption)
6. PAS (modus ponens from 2 and 5)

Now we can use the rule of conjunction to combine Q and PAS into a single statement. That is, if Q is true and PAS is true, then (Q ∧ PAS) must also be true. So we have:

7. Q ∧ PAS (conjunction from 4 and 6)

Finally, we can use the rule of modus tollens to infer that if RAS is false (i.e., not true), then p must also be false. That is, if p → Q and Q ∧ PAS and RAS is false, then p must be false. So we can write:

8. RAS → ¬p (premise)
9. ¬RAS (assumption)
10. ¬p (modus tollens from 8 and 9)

Therefore, the argument is valid and the conclusion is that p is false when RAS is not true. Note that we did not use the assumption of RAS in our proof, but it was necessary to include it as a premise in order to derive the conclusion.