
#1
Mar2812, 09:40 AM

P: 9

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? 



#2
Mar2812, 10:01 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,879

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".




#3
Mar2812, 01:03 PM

P: 9

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 


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