Using the five axioms below prove: p→q A1: p→~y A2: ~r→q A3: p→~z A4: x→ q or z A5: r→x or y Do I have to take the contrapositive of some of the axioms to begin this proof?
Yes, that would be the simplest thing to do. The very first "axiom" gives you p-> ~y but there is no "~y-> " so you cannot continue directly. However, you do have "A5: r->x or y which has contrapositive ~(x or y)= (~x) and (~y)->~r and then both "A2: ~r-> q" and "A4: x-> q or z".
Am I on the right track with this? Conclusions Justifications 1. p Given 2. ~z or ~y All cases 3. ~z Case 1 4. ~x A4 5. ~r A5 6. q A2