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Proof step 15 from proof div-div-pred-01 (previous step)

j0 + ((i1 * 4 + j1) + 3) * 2 = i0 + 6, j0 ≤ 0, i1 * 4 + j1 ≤ ~3, i1 * 4 + j1 < 0, abs(j0) < 2, abs(j1) < 4, i1 ≤ 0, j1 ≤ 0, i0 < ~7, i0 < 0 ⊦ i1 = i0 / 8

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 16:
⊦ (* j) = i0 + (* j0) ↔ j - j0 = i0
Proof step 17:
j0 + ((i1 * 4 + j1) + 3) * 2 - 6 = i0, j0 ≤ 0, i1 * 4 + j1 ≤ ~3, i1 * 4 + j1 < 0, abs(j0) < 2, abs(j1) < 4, i1 ≤ 0, j1 ≤ 0, i0 < ~7, i0 < 0 ⊦ i1 = i0 / 8

Rule argument

spec: int-abs inst: name: numadd-01 seq: ⊦ (* j) = i0 + (* j0) ↔ j - j0 = i0 subst: [j0, i0, j] → [6, i0, j0 + ((i1 * 4 + j1) + 3) * 2] flags: rewrite

Additional information

Used terms: 6, i0, j0 + ((i1 * 4 + j1) + 3) * 2