Question

Suppose I want to prove the following statement by contradiction:

P⟶(Q∧Z)

If (Q∧Z) is false, then either: (i) Q is false and Z is true; (ii) Q is true and Z is false; (iii) Q and Z are false.

Do I need to consider all possible cases in which (Q∧Z) is false and arrive to a contradiction or it suffices to show a contradiction in only one possible of the three possible cases?

Thanks for your help!

Answer 1

any one will do it P -> Q ^ Z means P->Q and P->Z
so set P to true and find a contradiction anywhere you can

Answer 2

This means that the proof is complete if I assume (say) case (i) to be true and arrive to a contradiction?

Answer 3

P and ~Q disproves P->Q^Z
I am not sure what you mean by prove by contradiction
It has been quite a while since academic use of logic for me
been using the real stuff over 40 years now

Answer 4

if you need to prove P, assume ~P and go find a contradiction
and this will not do that

Answer 5

but if you want to prove not P and you know Q and or Z
then assume P and this WILL prove it

Answer 6

I want to show that P implies (Q and Z). Hence, I start by assuming that P is true and ~(Q and Z) and then find a contradiction. But ~(Q and Z) gives me three possibilities so I am not sure if all what I need is to show that in on of these possibilities I get a contradiction...

Answer 7

any one will do.... as usual we overthought it...
forget those last 3 posts of mine, in this case

Answer 8

great! thanks a lot!

Answer 9

DeMorgan is the one I couldn't remember. Ask him! He will tell you all about this.