How Many Cases Do I Need to Consider for Proof by Contradiction?

[cris(c)]

Homework Statement

Suppose I want to prove the following statement by contradiction:
P \longrightarrow (Q \land Z)

Homework Equations

If (Q \land 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.

The Attempt at a Solution

Do I need to consider all possible cases in which (Q \land 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!

 

[mtayab1994]

If you want to prove the first one is true all you have to do is prove it by only one of the cases, because QandZ is false only if one of those cases occurs, but depending on what your teacher wants you can do all three cases.

 

[cris(c)]

If you want to prove the first one is true all you have to do is prove it by only one of the cases, because QandZ is false only if one of those cases occurs, but depending on what your teacher wants you can do all three cases.

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

 

[mtayab1994]

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

Yes because Q and Z could only be true if Q and Z are both true or if Q and Z are both false, therefore all you have to do is prove it using one case. Hope this helps you.

 

[cris(c)]

Yes because Q and Z could only be true if Q and Z are both true or if Q and Z are both false, therefore all you have to do is prove it using one case. Hope this helps you.

Think I got it. Thanks a lot!

 

[mtayab1994]

No Problem.

 

[cris(c)]

No Problem.

Hi again,

It appears that your answer is not completely correct or I am truly messed up. Negating (Q and Z) means that either (Q and not Z) or (Z and not Q). Hence, to actually show that P implies (Q and Z), don't we need to show that both of the above cases aren't possible true? I mean, showing that P implies (Z and not Q) does not exclude the possibility that (Q and not Z) still is true, does it?

Thanks!