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Proof step 22 from proof mod-by6 (previous step)

(6 * (i1 + 1) * (i1 + 1) + i1 * (i1 + 1) * (2 * i1 + 1)) % 6 = 0, 0 < i1, (i1 + 1) * (2 * i1 + 1) < 0 ⊦ (i1 * (i1 + 1) * (2 * i1 + 1)) % 6 = 0, i1 = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 23:
⊦ 0 < i → (i * j < 0 ↔ j < 0)

Rule argument

spec: int-mult name: zero-ls seq: ⊦ 0 < i → (i * j < 0 ↔ j < 0) subst: [j, i] → [2 * i1 + 1, i1 + 1] flags: <all paths>

Additional information

Used simplifier rules: ⊦ 0 < i ∧ 0 < j → 0 < i + j; (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2% ⊦ 0 < i ∧ abs(j0) < j → ¬ i * j + j0 < 0; (s)%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult Used terms: 2 * i1 + 1, i1 + 1