Proof of P∧R from p→Q, R∧S, and Q⇒P∧S

There are a few ways to approach this problem, but one possible method is to use a proof by contradiction. Here's one way to do it:

1. Assume ~(P∧R)
2. From 1, it follows that ~P∨~R (De Morgan's law)
3. From 2 and p→Q, we can derive ~Q∨~R (modus ponens and contrapositive)
4. From 3 and R∧S, we can derive ~Q∨S (conjunction elimination)
5. From 4 and Q⇒P∧S, we can derive ~(P∧~S) (modus ponens and contrapositive)
6. From 5 and R∧S, we can derive ~(P∧~R) (conjunction elimination)
7. From 6 and ~(P∧R) (our assumption), we can derive a contradiction
8. Therefore, ~(~(P∧R)) (proof by contradiction)
9. From 8 (double negation) and ~(P∧R) (our assumption), we can conclude ~(~P∨~R)
10. From 9 (De Morgan's law), we can derive P∧R (negation elimination)

So we have shown that if p→Q, R∧S, and Q⇒P∧S, then P∧R must also be true.