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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, 0 ≤ j2, 0 ≤ i1 + j0, j ≤ i1 + j0, j2 < 2, 0 ≤ i1, 0 ≤ j0, 0 ≤ i0, 0 ≤ j, 0 ≤ i ⊦ i = i0 ∧ j = j0

User applied rule cut formula

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, 0 ≤ j2, 0 ≤ i1 + j0, j ≤ i1 + j0, j2 < 2, 0 ≤ i1, 0 ≤ j0, 0 ≤ i0, 0 ≤ j, 0 ≤ i ⊦ j2 = 1, i = i0 ∧ j = j0

Rule argument

fma: j2 = 1

Additional information

Used simplifier rules: square-even: ⊦ i * (i + 1) ≠ 2 * j + 1; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2 Used formulas: j2 = 1