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Proof step 7 from proof square-inject (previous step)

(i + j) * (i + j + 1) = i2 * 2 + j2 ∧ 0 ≤ j2 ∧ j2 < 2 ∧ 0 ≤ i2 ∧ 0 < 2 ∧ 0 ≤ (i + j) * (i + j + 1), i2 + j = i1 + j0, (i0 + j0) * (i0 + j0 + 1) = i1 * 2 + j1, 0 ≤ i1, 0 ≤ j1, 0 ≤ j0, 0 ≤ i0, 0 ≤ j, 0 ≤ i, j1 < 2 ⊦ i = i0 ∧ j = j0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
(i0 + j0) * (i0 + j0 + 1) = i1 * 2 + j1, i2 + j = i1 + j0, (i + j) * (i + j + 1) = i2 * 2 + j2, 0 ≤ i2, 0 ≤ j2, j1 < 2, 0 ≤ i, 0 ≤ j, 0 ≤ i0, 0 ≤ j0, 0 ≤ j1, 0 ≤ i1, j2 < 2 ⊦ 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%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%zero-ge-add-01: ⊦ 0 ≤ j ∧ 0 ≤ j0 ∧ 0 ≤ i0 → 0 ≤ i0 * j + j0; from specification int-mult%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs