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Proof step 12 from proof pot-mon (previous step)

m ^ m0 < m ^ n1 ↔ m0 < n1 ⊦ m ^ m0 * m < m ^ n1 * m ↔ m0 < n1, m = 1, m = 0

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: mult-c: ⊦ m * n = n * m; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult% ⊦ m0 ≠ 0 → (m0 * m < m0 * n ↔ m < n); (s)