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Proof step 4 from proof pot-mon (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ n ^ (m + 1) = n ^ m * n
Proof step 6:
∀ m, n. (m ^ n < m ^ n1 ↔ n < n1) ∨ m = 1 ∨ m = 0 ⊦ m ^ n < m ^ n1 * m ↔ ¬ n1 < n, m = 0, m = 1

Rule argument

spec: <toplevel> name: rec-01 seq: ⊦ n ^ (m + 1) = n ^ m * n subst: [m, n] → [n1, m] flags: <all paths>

Additional information

Used terms: n1, m