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Proof step 14 from proof weakmon-lem (previous step)

ii(m1) = ii(m0), ii(m0) < ii(m1 + 1), m0 ≤ m1 + 1, ∀ m0, ii. m0 ≤ m1 ∧ (∀ m. ii(m) ≤ ii(m0)) ∧ (∀ n. ii(n) ≤ ii(n + 1)) → ii(m1) = ii(m0), ∀ m. ii(m) ≤ ii(m0), ∀ n. ii(n) ≤ ii(n + 1) ⊦ ii(m1 + 1) = 0, ii(m1 + 1) = ii(m0)

User applied rule all left

And closed the goal!

Rule argument

pos: left 5 terms: m1 + 1 abort: F discard: F computed: F

Additional information

Used simplifier rules: le-ls-simp: ⊦ m ≤ n → (n < m ↔ false); from specification nat Used terms: m1 + 1