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Proof step 6 from proof pot-mon (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 7:
(m ^ (n - 1) < m ^ n1 ↔ n - 1 < n1) ∨ m = 1 ∨ m = 0 ⊦ m = 1, m = 0, m ^ n < m ^ n1 * m ↔ ¬ n1 < n

Rule argument

pos: left 1 terms: m, n - 1 abort: F discard: T computed: F

Additional information

Used terms: m, n - 1