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Proof step 9 from proof pot-mon (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
m ^ m0 < m ^ n1 ↔ m0 < n1 ⊦ m = 1, m = 0, m ^ (m0 + 1) < m ^ n1 * m ↔ m0 < n1

Rule argument

Additional information

Used simplifier rules: oz: ⊦ m + 1 ≠ 0; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2