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Proof step 6 from proof product-negative (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
0 < Π ints0, i < 0, i < Π ints0, ∀ i0. i0 = i ∨ i0 ∈ ints0 → 0 ≤ i0 ⊦ i = 0

Rule argument

Additional information

Used simplifier rules: forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%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]%ls-le: ⊦ i < i0 → (¬ i0 < i ↔ true); from specification oelem[intlista]%lels-02: ⊦ i ≠ j → (i ≤ j ↔ i < j); from specification int-abs%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%zero-ls-03: ⊦ i ≤ 0 → (i * j < 0 ↔ ¬ ¬(0 < j ∧ i ≠ 0)); from specification int-mult%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs