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by jacb
1991 days ago
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That's not how arguments from contradiction work. You've assumed H (halting problem is decidable), prove B (beauty decidable), then say "but actually !H, therefore !B". All you've proven is that H => B. An actual argument from contradiction is "assume H, using H we can prove an obviously false statement, therefore not-H". And indeed, you have assumed something false (H), and shown a contradiction (various proofs for !H), so your original assumption was wrong (!H). A concrete example: say Beauty(R,D) is trivially decidable (say it returns (popcount(R)+popcount(D))%2. Your argument "proves" this property is undecidable, but it's definitely decidable. So some step in your argument is wrong. Another concrete example: Assume the halting problem is decidable. I would like a cookie. But the halting problem is not decidable. Therefore I would not like a cookie. |
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