Right. And it's even harder than finding the chain of implication; it's finding the a Z to start from. That's because you only know not-Z because Z is absurd. But a false proposition implies anything. So /*logically*/ you can start from any absurdity but to get anywhere you have to get to a Z that's implied by not-Y, which is implied by not-W, etc. Brent On 2/2/2017 5:56 PM, William R Somsky wrote:
If the 'reducto' proof is:
not-X => A => B => ... => Y => Z, but Z is clearly false, hence not-not-X, ie X
you should be able to invert that to be
not-Z (which is clearly true) => not-Y => ... => not-B => not-A => X
but good luck trying to find that from starting at the not-Z end.
On 2017-02-02 17:15, David Wilson wrote:
The reduction ad absurdum argument to prove X generally goes not-X => ... => false, a contradiction, therefore X is true. e.g. not-X => A, not-X => not-A, hence x => A and not-A => F, therefore X is true. I've always found such arguments, while effective, in a way unsatisfying, because by the time you find your contradiction, you have wander far from your premise. I was wondering if there might always be a way to turn such an argument around, so that a proof of not-X => false, can be flipped to prove the contrapositive, true => X. I'm obviously not talking about building the argument around not-X => false and then invoking the contrapositive, but rather inverting the entire argument into a direct argument where X appears only as the last statement proven, not as part of an earlier supposition.
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