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Proof step 23 from proof orderedls-all (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 24:
m = m1 + 1, n ≤ 1 + # y, ordered<(a0 ' + y), a < a0, m < n, m ≤ # y, ∀ m0, n0. m0 < n0 ∧ ¬ # y < n0 → (a0 ' + y)[m0] < (a0 ' + y)[n0] ⊦ m = 0, (a0 ' + y)[m1] < (a ' + a0 ' + y)[n]

Rule argument

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

Additional information

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