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Proof step 11 from proof weakmon-lem (previous step)

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), m0 ≤ m1 → ii(m1) = ii(m0) ⊦ ii(m1 + 1) = ii(m0)

User applied rule case distinction

And obtained the following new goals

Proof step 12:
ii(m1) = ii(m0), 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) = ii(m0)

Rule argument

pos: left 5

Additional information

Used simplifier rules: lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat%les: ⊦ m < n → (n ≤ m + 1 ↔ n = m + 1); from specification nat%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2