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Proof step 9 from proof square-interval (previous step)

j * (j + 1) + 2 * j0 < (i + 1) * (i + 2), i * (i + 1) ≤ j * (j + 1) + 2 * j0, j0 ≤ j, 0 ≤ i, 0 ≤ j0 ⊦ i = j, j * (j + 1) + 2 * j0 < (j + 1) * (j + 2)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
(j * j + j) + 2 * j0 < 3 * i + i * i + 2, i * i < 2 * i + i * i + 2, 0 ≤ j, 2 * j + 2 ≤ 2 * j0, i * i + i ≤ (j * j + j) + 2 * j0, j0 ≤ j, 0 ≤ i, 0 ≤ j0 ⊦

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;%numsucc: ⊦ (* i0) * i + i = (i0 + 1) * i;%numls-04: ⊦ i + j < (* i1) * i + j0 ↔ j < (i1 - 1) * i + j0;%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs% ⊦ i < j ↔ ¬ j ≤ i; (ss)% ⊦ i + j < i + i0 ↔ j < i0; (s)%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2