Help with indirect logic proof please!

LCharette

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?

HallsofIvy

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".

LCharette

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