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Proof step 11 from proof square-inject (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%add-zero-rev: ⊦ 0 + i = i; from specification int-basic2%sub-le-01: ⊦ i ≤ j - i0 ↔ i + i0 ≤ j; 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%sub-add-01: ⊦ i + (j - i0) = i + j - i0; from specification int-abs