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Example 2 as proven with Vladimir

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Patrick Lühne 2020-06-02 00:57:49 +02:00
parent bd108886dd
commit 789cb4f8f8
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GPG Key ID: 05F3611E97A70ABF
1 changed files with 51 additions and 4 deletions
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examples/example-2.lemmas
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@ -1,9 +1,9 @@
# Multiplication with positive numbers preserves the order of integers
axiom: forall N1, N2, N3 (N1 > N2 and N3 > 0 -> N1 * N3 > N2 * N3).
#axiom: forall N1, N2, N3 (N1 > N2 and N3 > 0 -> N1 * N3 > N2 * N3).
# Induction principle instantiated for p.
# This axiom is necessary because we use Vampire without higher-order reasoning
axiom: forall N1 (p(N1) and forall N2 (N2 >= N1 and not p(N2) -> not p(N2 + 1)) -> forall N2 (N2 >= N1 -> p(N2))).
#axiom: forall N1 (p(N1) and forall N2 (N2 >= N1 and not p(N2) -> not p(N2 + 1)) -> forall N2 (N2 >= N1 -> p(N2))).
#axiom: p(0) and forall N (N >= 0 and p(N) -> p(N + 1)) -> forall N p(N).
lemma(forward): forall X (p(X) <-> exists N (X = N and N >= 0 and N * N <= n)).
@ -14,7 +14,54 @@ lemma(forward): not p(n + 1).
lemma(forward): forall N1, N2 (q(N1) and N2 > N1 -> not q(N2)).
lemma(forward): forall N (N >= 0 and not p(N + 1) -> (N + 1) * (N + 1) > n).
lemma(backward): forall N1, N2 (q(N1) and q(N2) -> N1 = N2).
axiom: forall N1, N2 (p(N1) and not p(N1 + 1) and p(N2) and not p(N2 + 1) -> N1 = N2).
# Added completed definition of p/1:
# forall X1 (p(X1) <-> exists N1 (0 <= N1 and N1 <= n and X1 = N1)
# and exists X2 (exists N2, N3 (X2 = N2 * N3 and N2 = X1 and N3 = X1) and X2 <= n))
lemma(backward): forall N1, N2 (N1 >= 0 and N2 >= 0 and N1 * N1 <= N2 * N2 -> N1 <= N2).
lemma(backward): forall N (p(N) <-> 0 <= N and N <= n and N * N <= n).
lemma(backward): forall N (p(N) -> N * N <= n).
lemma(backward): forall N (not p(N) and N >= 0 -> N * N > n).
lemma(backward): forall N (N >= 0 -> N * N < (N + 1) * (N + 1)).
lemma(backward): forall N1, N2 (p(N1) and not p(N2) and N2 >= 0 -> N1 * N1 <= n and N2 * N2 > n).
lemma(backward): forall N1, N2 (p(N1) and not p(N2) and N2 >= 0 -> N1 * N1 < N2 * N2).
lemma(backward): forall N1, N2 (p(N1) and not p(N2) and N2 >= 0 -> N1 < N2).
lemma(backward): forall N1, N2 (p(N1) and not p(N1 + 1) and p(N2) and not p(N2 + 1) -> N1 = N2).
lemma(backward): exists N (forall X (q(X) <- X = N) and N >= 0 and N * N <= n and (N + 1) * (N + 1) > n).
lemma(backward): exists N (q(N) and N >= 0 and N * N <= n and (N + 1) * (N + 1) > n).
#axiom: forall N, M (q(N) <- p(N) and not p(M) and M > N).
#axiom: forall N1, N2 (p(N1) and not p(N1 + 1) and p(N2) and not p(N2 + 1) -> N1 = N2).
#lemma(backward): forall N1, N2 (q(N1) and q(N2) -> N1 = N2).
#lemma(backward): forall N (p(N) and N > 0 -> p(N - 1)).
#lemma(backward): forall N (q(N) <- p(N) and not p(N + 1)).
lemma(backward): forall X1 (q(X1) -> p(X1) and exists X2 (exists N (X2 = N + 1 and N = X1) and not p(X2))).
lemma(backward): forall X1 (q(X1) <- p(X1) and exists X2 (exists N (X2 = N + 1 and N = X1) and not p(X2))).