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Proof step 5 from proof min-del-nlz-01 (previous step)

¬ ints0 ≠ [], min((i ' + ints0) -l i0) < 0, min((i ' + ints0) -l i0) < min(i ' + ints0), 0 < min(i ' + ints0), 0 < min(ints0) → (∀ i0. ¬ min(ints0 -l i0) < 0) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
min(i ' -l i0) < 0, min(i ' -l i0) < i, 0 < i ⊦

Rule argument

Additional information

Used simplifier rules: min-one: ⊦ min(i ') = i;%min-empty: ⊦ min([]) = 0;%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%transitivity: ⊦ i < i0 ∧ i0 < i1 → i < i1; (forward) from specification oelem[intlista]%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%a: ⊦ (ints + ints0) + ints1 = ints + ints0 + ints1; from specification list[intlista]%app: ⊦ ints + [] = ints; from specification list[intlista]