34 lines
1.5 KiB
Python
34 lines
1.5 KiB
Python
# Perform the proofs under the assumption that n is a nonnegative integer input constant
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input: n -> integer.
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assume: n >= 0.
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# p/1 is an auxiliary predicate
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output: q/1.
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# Verify that q computes the floor of the square root of n
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spec: exists N (forall X (q(X) <-> X = N) and N >= 0 and N * N <= n and (N + 1) * (N + 1) > n).
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# Multiplication with positive numbers preserves the order of integers
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axiom: forall N1, N2, N3 (N1 > N2 and N3 > 0 -> N1 * N3 > N2 * N3).
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# Induction principle instantiated for p.
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# This axiom is necessary because we use Vampire without higher-order reasoning
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axiom: forall N1 (p(N1) and forall N2 (N2 >= N1 and not p(N2) -> not p(N2 + 1)) -> forall N2 (N2 >= N1 -> p(N2))).
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#axiom: p(0) and forall N (N >= 0 and p(N) -> p(N + 1)) -> forall N p(N).
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lemma(forward): forall X (p(X) <-> exists N (X = N and N >= 0 and N * N <= n)).
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lemma(forward): forall X (q(X) <-> exists N2 (X = N2 and N2 >= 0 and N2 * N2 <= n and (N2 + 1) * (N2 + 1) > n)).
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lemma(forward): forall N1, N2 (N1 >= 0 and N2 >= 0 and N1 < N2 -> N1 * N1 < N2 * N2).
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lemma(forward): forall N (N >= 0 and p(N + 1) -> p(N)).
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lemma(forward): not p(n + 1).
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lemma(forward): forall N1, N2 (q(N1) and N2 > N1 -> not q(N2)).
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lemma(forward): forall N (N >= 0 and not p(N + 1) -> (N + 1) * (N + 1) > n).
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lemma(backward): forall N1, N2 (q(N1) and q(N2) -> N1 = N2).
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axiom: forall N1, N2 (p(N1) and not p(N1 + 1) and p(N2) and not p(N2 + 1) -> N1 = N2).
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lemma(backward): forall X1 (q(X1) -> p(X1) and exists X2 (exists N (X2 = N + 1 and N = X1) and not p(X2))).
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