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Proof step 1 from proof square-even-01 (conclusion)

⊦ (i * (i + 1)) % 2 = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: int-div inst: name: square-even seq: ⊦ 0 ≤ i → (i * (i + 1)) % 2 = 0 subst: [i] → [i] flags: rewrite, current

Additional information

Used simplifier rules: even-mult: ⊦ (i * j) % 2 = 0 ↔ i % 2 = 0 ∨ j % 2 = 0;%modeven-neg-01: ⊦ i < 0 → ((i + 1) % 2 ≠ 0 ↔ (i + 1) % 2 = ~1 ∧ i % 2 = 0);%modeven-neg: ⊦ i < 0 ∧ i % 2 = 0 → (i + 1) % 2 = ~1;%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2 Used terms: i