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

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To prove a statement of the form P implies (Q1 and Q2) by contradiction, one assumes P and the negation of (Q1 and Q2), which is (not Q1 or not Q2), to find a contradiction. It is not sufficient to only find a contradiction with both not Q1 and not Q2; instead, the focus should be on the overall negation. The correct negation of "if P then (Q and R)" is "P and (not Q or not R)." The contrapositive of the statement maintains the same truth value as the original statement. Understanding these logical relationships is crucial for effective proofs.
christoff
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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)

?
 
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Never mind. I figured it out.
 
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).
 
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)"
 

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