Proof assistant demos at ICMS 2006
Exercise
Problem [B2 from IMO 1972]
f and g are real-valued functions defined on the real line.
For all x and y,
f(x + y) + f(x - y) = 2 f(x) g(y).
f is not identically zero and |f(x)| ≤ 1 for all x.
Prove that |g(x)| ≤ 1 for all x.
Solution
Let k be the least upper bound for |f(x)|.
Suppose |g(y)| > 1.
Then
2k ≥ |f(x + y)| + |f(x - y)| ≥ |f(x + y) + f(x - y)| = 2 |g(y)||f(x)|,
so |f(x)| ≤ k/|g(y)|.
In other words, k/|g(y)| is an upper bound for |f(x)| which is less than k.
Contradiction.
Demos
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Video of the demo
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Video of the demo
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Video of the demo
- Polished version with both the demo proof and Hales' proof: pdf doc (doc does not mean "Word" here, but "ProofPower") (formalization times: about 77 minutes and about 35 minutes; de Bruijn factors: about 2.4 and about 2.0)
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Video of the demo
(last modification 2006-09-21)