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Proof step 2 from proof sum-all-indices-le (previous step)

∀ ints0. (∀ n. 0 ≤ n ∧ n < # ints1 → ints1[n] ≤ ints0[n]) ∧ # ints1 = # ints0 → Σ ints1 ≤ Σ ints0, ∀ n. 0 ≤ n ∧ n < #(i + ints1) → (i + ints1)[n] ≤ ints0[n], #(i + ints1) = # ints0 ⊦ Σ(i + ints1) ≤ Σ ints0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
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] ⊦

Rule argument

Additional information

Used simplifier rules: sum-rec: ⊦ Σ(i ' + ints) = i + Σ ints;%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%lez: ⊦ 0 ≤ m; from specification nat%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs%a: ⊦ (ints + ints0) + ints1 = ints + ints0 + ints1; from specification list[intlista]%len-02: ⊦ # i ' = 0 +1; from specification list[intlista]%len: ⊦ #(ints + ints0) = # ints + # ints0; from specification list[intlista]%c: ⊦ i + ints = i ' + ints; from specification list[intlista]