Negation of Implication: Tautology or Contradiction?

[anonymity]

If the negation of an implication is a contradiction, the implication is a tautology.

Is this correct? Because if the negation is never true, then it must be a tautology...No?

For example, I am working on a problem that, after a whole bunch of other stuff, the negation of my statement is P $\wedge$ $\neg$P $\wedge\neg$Q..which is NEVER true. And because this was the negation of an implication (IE, the only time the implication is ever false), the implication is always true...

 

[micromass]

That is correct.

In this case, I suspect that the original implication is $P\rightarrow P\vee Q$? This is indeed a tautology!

 

[anonymity]

No it was a real mess of statements. What I posted was actually just part of the final product (but that contradiction held all of the power, so to speak). I do remember something along the lines of that in it though.

Thanks for confirming =]

 

[Jamma]

I've never quite understood what it means to be a tautology, and I suppose there cannot be an actual definition of it- any mathematical proof just uses a string of implications to provide a new implication. But I certainly wouldn't say that a,b,c,n integers, n greater than 2 means that a^n+b^n cannot be equal to c^n !.

 

[blue_raver22]

Your reasoning is correct. If the negation of an implication is always false, then the implication itself must be always true. This means that the implication is a tautology, which is a statement that is always true regardless of the truth values of its components. In other words, the implication is a logical truth and cannot be contradicted.