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Proof step 3 from proof product-negative (previous step)

¬(Π ints0 < 0 ∧ (∀ i. i ∈ ints0 → 0 ≤ i)), Π(i + ints0) < 0, ∀ i0. i0 ∈ i + ints0 → 0 ≤ i0 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
i * Π ints0 < 0, ∀ i0. i0 = i ∨ i0 ∈ ints0 → 0 ≤ i0 ⊦ Π ints0 < 0 ∧ (i0 ∈ ints0 → 0 ≤ i0)

Rule argument

Additional information

Used simplifier rules: product-app: ⊦ Π(ints + ints0) = Π ints * Π ints0;%product-one: ⊦ Π i ' = i;%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%in-17: ⊦ i ∈ i0 ' ↔ i = i0; from specification list[intlista]%in-append: ⊦ i ∈ ints + ints0 ↔ ¬ ¬(i ∈ ints ∨ i ∈ ints0); from specification list[intlista]%a: ⊦ (ints + ints0) + ints1 = ints + ints0 + ints1; from specification list[intlista]%c: ⊦ i + ints = i ' + ints; from specification list[intlista]