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Proof step 28 from proof orderedls-all (previous step)

m1 + 1 < n, a < a0, m1 < # y, ordered<(a0 ' + y), n ≤ 1 + # y, ∀ m0, n0. m0 < n0 ∧ ¬ # y < n0 → (a0 ' + y)[m0] < (a0 ' + y)[n0] ⊦ y = [], (a0 ' + y)[m1] < (a0 ' + y)[n - 1]

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 29:
n = m + 1, m1 < # y, a < a0, m1 + 1 < n, ordered<(a0 ' + y), n ≤ 1 + # y, ∀ m0, n0. m0 < n0 ∧ ¬ # y < n0 → (a0 ' + y)[m0] < (a0 ' + y)[n0] ⊦ n = 0, y = [], (a0 ' + y)[m1] < (a0 ' + y)[m]

Rule argument

spec: nat name: elim-pred-c seq: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1 subst: [n, m] → [n, m1]

Additional information

Used simplifier rules: n0 < n ⊦ ¬ m + (* n) < (* n0); (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: n, m1