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Proof step 7 from proof pot-mon (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
m ^ (n - 1) < m ^ n1 ↔ ¬ n1 < n ⊦ m ^ n < m ^ n1 * m ↔ ¬ n1 < n, m = 0, m = 1

Rule argument

Additional information

Used simplifier rules: Pot-zero: ⊦ n ^ 0 = 1;%pot-zero: ⊦ m ≠ 0 → m ^ n ≠ 0;%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m < n + 1 ↔ ¬ n < m; (s)%n0 ≤ n ⊦ n - n0 < m ↔ n < m + n0; (s)% ⊦ m0 * m = 1 ↔ m0 = 1 ∧ m = 1; (ws)% ⊦ ¬ m < 0; (s)%mult-c: ⊦ m * n = n * m; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult% ⊦ m ≠ 0 → (m * n = 0 ↔ n = 0); (s)% ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1); (s)% ⊦ m < 1 ↔ m = 0; (s)% ⊦ m ≤ n ↔ ¬ n < m; (ss)