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Proof step 1 from proof square-le-04 (conclusion)

⊦ 0 ≤ i ∧ 0 ≤ j → (i * (i + 1) ≤ j * (j + 1) ↔ i ≤ j)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
0 ≤ i, 0 ≤ j ⊦ i * i + i ≤ j * j + j ↔ i ≤ j

Rule argument

Additional information

Used simplifier rules: dist-lsplit: ⊦ (i + i0) * j = i * j + i0 * j;%one: ⊦ 1 * i = i;%associativity: ⊦ i * j * j0 = (i * j) * j0;%commutativity: ⊦ i * j = j * i;%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2