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Proof step 6 from proof sum-all-indices-le (previous step)

1 + # ints1 = # ints0, ints0.first + Σ ints0.rest < i + Σ ints1, ∀ n. ¬ # ints1 < n → (i ' + ints1)[n] ≤ ints0[n], ∀ ints. (∀ n. n < # ints1 → ints1[n] ≤ ints[n]) ∧ # ints1 = # ints → Σ ints1 ≤ Σ ints ⊦ ints0 = []

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Rule argument

spec: list-data inst: intlista name: elim-list-c seq: ints ≠ [] ⊦ i = ints.first ∧ ints0 = ints.rest ↔ ints = i + ints0 subst: [ints, i, ints0] → [ints0, i0, ints2] flags:

Additional information

Used simplifier rules: ints ≠ [] ⊦ i = ints.first ∧ ints0 = ints.rest ↔ ints = i + ints0; (elim) Used terms: ints0, i0, ints2