Fix induction axiom in example 2
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input: n -> integer.
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input: n -> integer.
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axiom: forall N1, N2, N3 (N1 > N2 and N3 > 0 -> N1 * N3 > N2 * N3).
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axiom: forall N1, N2, N3 (N1 > N2 and N3 > 0 -> N1 * N3 > N2 * N3).
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axiom: p(0) and forall N (N >= 0 and p(N) -> p(N + 1)).
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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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assume: n >= 0.
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assume: n >= 0.
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