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Proof step 2 from proof square-even (previous step)

0 ≤ i ⊦ (i * (i + 1)) % 2 = 0

User applied rule insert elim lemma

And obtained the following new goals

Rule argument

spec: int-div inst: name: div-elim-pospos seq: ⊦ 0 < j ∧ 0 ≤ i → ( i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ 0 ≤ j0 ∧ j0 < j ∧ 0 ≤ i0) subst: [i, j, i0, j0] → [i * (i + 1), 2, i0, j] flags: rewrite, current

Additional information

Used simplifier rules: lels-02: ⊦ i ≠ j → (i ≤ j ↔ i < j); from specification int-abs%two-ls: ⊦ 0 < j → (j < 2 ↔ j = 1); from specification int-basic1%zero-ge-02: ⊦ 0 ≤ i → (0 ≤ i * j ↔ ¬ ¬(i = 0 ∨ 0 ≤ j)); from specification int-mult%numle-02: ⊦ (* j) ≤ i0 + (* j0) ↔ j - j0 ≤ i0; from specification int-abs%numleledel-01: i0 ≠ 0, i0 ≤ i ⊦ (* i) ≤ j → (* i0) ≤ j; 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%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%square-even: ⊦ i * (i + 1) ≠ 2 * j + 1; from specification int-mult Used terms: i * (i + 1), 2, i0, j