Proof by contradiction for statement of the form P->(Q and R)

[christoff]

Say I have a statement like this:
P implies (Q1 and Q2).

If I wanted to prove this by contradiction, I would assume P and not(Q1 and Q2)=[(not Q1) or (not Q2)] both hold, and try to find a contradiction.

My question is... Am I done if I find a contradiction while assuming P and [(not Q1) and (not Q2)] ? Is this sufficient? Or do I need to find a contradiction in both the statements:
P and (not Q1),
P and (not Q2)

?

 

[christoff]

Never mind. I figured it out.

 

[HallsofIvy]

No. The negation of "Q and R" is "not Q or not R".
The negation of "if P then (Q and R)" is "If (not Q or not R) then not P".

(For those who read this thread and wondered).

 

[jbriggs444]

The contrapositive of "if P then (Q and R)" is "if (not Q or not R) then not P"

But the contrapositive has the same truth value as the original.

The negation of "if P then (Q and R)" is "P and (not Q or not R)"

 

[CellCoree]

I would say that finding a contradiction while assuming P and [(not Q1) and (not Q2)] is sufficient to prove the statement P implies (Q1 and Q2) by contradiction. This is because the assumption of P and [(not Q1) and (not Q2)] already covers the possibility of both P and (not Q1) and P and (not Q2) being true, and finding a contradiction in this case would prove that both Q1 and Q2 must be true for P to hold. However, if you wanted to be thorough, you could also try to find a contradiction in both P and (not Q1) and P and (not Q2) separately, to cover all possibilities. Ultimately, it depends on the level of rigor and detail required for the specific situation.