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Proof step 3 from proof sum-all-indices-le (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ ints ≠ [] → Σ ints = ints.first + Σ ints.rest
Proof step 5:
ints0 ≠ [], 1 + # ints1 = # ints0, ints0.first + Σ ints0.rest < i + Σ ints1, ∀ ints. (∀ n. n < # ints1 → ints1[n] ≤ ints[n]) ∧ # ints1 = # ints → Σ ints1 ≤ Σ ints, ∀ n. ¬ # ints1 < n → (i ' + ints1)[n] ≤ ints0[n] ⊦

Rule argument

spec: inst: name: sum-rec-01 seq: ⊦ ints ≠ [] → Σ ints = ints.first + Σ ints.rest subst: [ints] → [ints0] flags: rewrite, current

Additional information

Used simplifier rules: size-01: ⊦ # [] = 0; from specification list-data[intlista]%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2 Used terms: ints0