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Proof step 14 from proof square-inject (previous step)

(i + j) * (i + j + 1) = (i1 + j0 - j) * 2 + j2, (i0 + j0) * (i0 + j0 + 1) = i1 * 2 + j1, j1 < 2, j2 < 2, 0 ≤ i1 + j0, j ≤ i1 + j0, 0 ≤ i1, 0 ≤ j1, 0 ≤ j0, 0 ≤ i0, 0 ≤ j, 0 ≤ i, 0 ≤ j2 ⊦ j1 = 1, i = i0 ∧ j = j0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
(i + j) * (i + j + 1) = (i1 + j0 - j) * 2 + j2, (i0 + j0) * (i0 + j0 + 1) = i1 * 2, j2 < 2, 0 ≤ i1 + j0, j ≤ i1 + j0, 0 ≤ i1, 0 ≤ j0, 0 ≤ i0, 0 ≤ j, 0 ≤ i, 0 ≤ j2 ⊦ i = i0 ∧ j = j0

Rule argument

Additional information

Used simplifier rules: add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2% ⊦ i + 0 = i; (s)% ⊦ i ≠ 1 ∧ 0 ≤ i → (i < 2 ↔ i = 0); (s)