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Proof step 6 from proof orderedls-all (previous step)

ordered<(x0) ↔ (∀ m, n. m < n ∧ n < # x0 → x0[m] < x0[n]) ⊦ a < x0.first ∧ ordered<(x0) ↔ (∀ m, n. m < n ∧ ¬ # x0 < n → (a ' + x0)[m] < (a ' + x0)[n]), x0 = []

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 7:
x0 = a0 + y, ordered<(x0) ↔ (∀ m, n. m < n ∧ n < # x0 → x0[m] < x0[n]) ⊦ a < a0 ∧ ordered<(x0) ↔ (∀ m, n. m < n ∧ ¬ # x0 < n → (a ' + x0)[m] < (a ' + x0)[n]), x0 = []

Rule argument

spec: list-data inst: name: elim-list-c seq: x ≠ [] ⊦ a = x.first ∧ y = x.rest ↔ x = a + y subst: [x, a, y] → [x0, a0, y] flags:

Additional information

Used simplifier rules: x ≠ [] ⊦ a = x.first ∧ y = x.rest ↔ x = a + y; (elim) Used terms: x0, a0, y