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Proof step 11 from proof mod-by6 (previous step)

(i0 * (i0 + 1) * (2 * i0 + 1)) % 6 = 0, (i0 + 1) * (2 * i0 + 1) + 1 < i0, 0 < i0, (i0 + 1) * (2 * i0 + 1) < 0, (i0 + 1) * (2 * i0 + 1) < i0 ⊦ (6 * (i0 + 1) * (i0 + 1) + i0 * (i0 + 1) * (2 * i0 + 1)) % 6 = 0, i0 = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 12:
⊦ 0 < i → (i * j < 0 ↔ j < 0)

Rule argument

spec: int-mult inst: name: zero-ls seq: ⊦ 0 < i → (i * j < 0 ↔ j < 0) subst: [j, i] → [2 * i0 + 1, i0 + 1] flags: rewrite

Additional information

Used simplifier rules: zero-02: ⊦ 0 < i ∧ 0 < j → 0 < i + j; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%modzero: ⊦ 0 < i ∧ abs(j0) < j → ¬ i * j + j0 < 0; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult Used terms: 2 * i0 + 1, i0 + 1