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Proof step 16 from proof orderedls-all (previous step)

n ≤ 1 + # y, ordered<(a0 ' + y), a < a0, (a0 ' + y)[n - 1] < a0, ∀ m0, n0. m0 < n0 ∧ ¬ # y < n0 → (a0 ' + y)[m0] < (a0 ' + y)[n0] ⊦ n = 0, a < (a0 ' + y)[n - 1]

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 17:
n = m + 1, n ≤ 1 + # y, ordered<(a0 ' + y), a < a0, (a0 ' + y)[m] < a0, ∀ m0, n0. m0 < n0 ∧ ¬ # y < n0 → (a0 ' + y)[m0] < (a0 ' + y)[n0] ⊦ n = 0, a < (a0 ' + y)[m]

Rule argument

spec: nat inst: name: elim-pred-c seq: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1 subst: [n, m] → [n, m0] flags:

Additional information

Used simplifier rules: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: n, m0