Question

P ⇒ q, r ⇒ s, p ∨r q ∧ s is invalid

Answer 1

My explanation and answer are down here↓:

Step-by-step explanation:

p→q is false only when p is true and q is false.

∴p→(q∨r) is false when p is true and (q∨r) is false, and

q∨r false when both q,r are false.

Hence T,F,F.