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Proof step 3 from proof pot-mon (previous step)

∀ m, n. (m ^ n < m ^ n1 ↔ n < n1) ∨ m = 1 ∨ m = 0 ⊦ m ^ n < m ^ n1 +1 ↔ n < n1 +1, m = 1, m = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
∀ m, n. (m ^ n < m ^ n1 ↔ n < n1) ∨ m = 1 ∨ m = 0 ⊦ m = 1, m = 0, m ^ n < m ^ (n1 + 1) ↔ ¬ n1 < n

Rule argument

Additional information

Used simplifier rules: 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%succ: ⊦ m +1 = m + 1; from specification nat-basic2