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Proof step 1 from proof sum-all-indices-le (conclusion)

∀ n. 0 ≤ n ∧ n < # ints → ints[n] ≤ ints0[n], # ints = # ints0 ⊦ Σ ints ≤ Σ ints0

User applied rule structural induction

And obtained the following new goals

Proof step 2:
∀ 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

Rule argument

indvar: ints

Additional information

Used simplifier rules: sum-empty: ⊦ Σ [] = 0;%intlist freely generated by [], +;;%size-01: ⊦ # [] = 0; from specification list-data[intlista]%l: ⊦ # ints = 0 ↔ ints = []; from specification list-data[intlista]%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs%lez: ⊦ 0 ≤ m; from specification nat%zero-01: ⊦ ¬ m < 0; from specification nat-basic1%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs