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

0 ≤ i, 0 ≤ j ⊦ (i * (i + 1)) / 2 < (j * (j + 1)) / 2 ↔ i < j

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ i % 2 = 0 ∧ j % 2 = 0 ∧ 0 ≤ i ∧ 0 ≤ j → (i / 2 < j / 2 ↔ i < j)

Rule argument

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

Additional information

Used simplifier rules: square-even: ⊦ 0 ≤ i → (i * (i + 1)) % 2 = 0;%square-even: ⊦ 0 ≤ i → (i * (i + 1)) % 2 = 0;%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%square-ls-02: ⊦ 0 ≤ i ∧ 0 ≤ j → (i * (i + 1) < j * (j + 1) ↔ i < j); from specification int-mult%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 Used terms: j * (j + 1), i * (i + 1)