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by laichzeit0
3 hours ago
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What’s assumed and not explicitly stated is the law of the excluded middle. That A is true or A is false and those are the only 2 possibilities. If you assume the law of the excluded middle then it’s impossible that “A or not-A” is false. So it’s true. But “A or not-A” is true is equivalent to “A and not-A” is false (just apply DeMorgan). So proof by contradiction is you assuming something B is true and it leading to a “A and not-A” (contradiction) so clearly B must be false. |
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If P is a proposition and it is demonstrated that "P implies not P", from this it can be concluded that P cannot be true and this conclusion is valid in any kind of logic, even if the law of the excluded middle is false.
Only in bivalent logic, where the law of the excluded middle is true, from the fact that a proposition is not true it can be concluded that it is false.
This is a separate thing, which has nothing to do with the technique of demonstration by a variant of reductio ad absurdum, where the goal is to prove the implication from P to not P.