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Proof step 2 from proof pot-mon (previous step)

⊦ m ^ n < m ^ n0 ↔ n < n0, m = 1, m = 0

User applied rule structural induction

And obtained the following new goals

Proof step 3:
∀ m, n. (m ^ n < m ^ n1 ↔ n < n1) ∨ m = 1 ∨ m = 0 ⊦ m ^ n < m ^ n1 +1 ↔ n < n1 +1, m = 1, m = 0

Rule argument

indvar: n0

Additional information

Used simplifier rules: Pot-zero: ⊦ n ^ 0 = 1;%pot-zero: ⊦ m ≠ 0 → m ^ n ≠ 0;%nat freely generated by 0, +1;;%one-01: ⊦ m < 1 ↔ m = 0; from specification nat-basic1%zero-01: ⊦ ¬ m < 0; from specification nat-basic1