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Proof step 8 from proof square-inject (previous step)

(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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 9:
⊦ i + (* j) = i0 + (* j0) ↔ i = i0 + (j0 - j)
Proof step 10:
(i0 + j0) * (i0 + j0 + 1) = i1 * 2 + j1, i2 = i1 + (j0 - j), (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

spec: int-abs inst: name: numadd seq: ⊦ i + (* j) = i0 + (* j0) ↔ i = i0 + (j0 - j) subst: [j0, i0, j, i] → [j0, i1, j, i2] flags: rewrite

Additional information

Used terms: j0, i1, j, i2