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Proof step 8 from proof pot-mon (previous step)

m ^ (n - 1) < m ^ n1 ↔ ¬ n1 < n ⊦ m = 1, m = 0, m ^ n < m ^ n1 * m ↔ ¬ n1 < n

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 9:
n = m0 + 1 ∧ n ≠ 0, m ^ m0 < m ^ n1 ↔ ¬ n1 < n ⊦ m = 1, m = 0, m ^ n < m ^ n1 * m ↔ ¬ n1 < n

Rule argument

spec: nat inst: name: elim-pred-c seq: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1 subst: [n, m] → [n, m0] flags:

Additional information

Used simplifier rules: Pot-zero: ⊦ n ^ 0 = 1;%pot-zero: ⊦ m ≠ 0 → m ^ n ≠ 0;%zero-01: ⊦ ¬ m < 0; from specification nat-basic1%one: ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1); from specification nat-basic1%zero-06: ⊦ m ≠ 0 → (m * n = 0 ↔ n = 0); from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult%one-01: ⊦ m0 * m = 1 ↔ m0 = 1 ∧ m = 1; from specification nat-mult%n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: n, m0