Using negation of an for all statment in a proof by contradiction.

[torquerotates]

So if I want to prove. A=>B for all x.

Does the following work?

Suppose for contradiction, B is not true for all x, that is, there exists at least one x such that B is not true. In particular, assume that B is true for x=c and B isn't true for all other x. If I arrive at a contradiction, then A must imply B.


So does it work if I pick a single value of x such that B is true and let B not be true for all other values? This is a little confusing because the negation simply specifies the case for at least one x such that B isn't true. There could be more than one x such that B isn't true.

 

[JThompson]

No. In proof by contradiction you choose one arbitrary case for which you are assuming B is not true.
You are showing that it it impossible for there to be even one case where A is true and B is not true. However, this case does have to be arbitrary, because if it isn't arbitrary then you can't generalize to all x which means you cannot generate a contradiction.

Edit: You don't have to assume that there are any cases for which the implication holds (and in fact you can't, because you haven't proved that any exist), as you did above. It is irrelevant to this type of proof, anyway.

 

[Stephen Tashi]

So if I want to prove. A=>B for all x.

For proof by contradiction, you assume: There is an x such that it is false that A=>B.

"It is false that A implies B" is equivalent to the statement "A is true and B is false". (This may be the difficult part to grasp. Think of it this way: the only way you can provide a counterexample to "A implies B" is to provide an example where A is true and B is false. An example where A was false would not disprove "A implies B" )

So what you assume is "There exists an x such that A is true and B is false".