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Proof step 14 from proof div-div-pred-01 (previous step)

j0 + ((i1 * 4 + j1) + 3) * 2 = i0 + 6, i1 * 4 + j1 < 0, ((i1 * 4 + j1) + 3) * 2 + j0 < 0, i1 * 4 + j1 ≤ ~3, j0 ≤ 0, ((i1 * 4 + j1) + 3) * 2 + j0 < ~1, j1 ≤ 0, i1 ≤ 0, abs(j1) < 4, abs(j0) < 2 ⊦ i1 = i0 / 8

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
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

Rule argument

Additional information

Used simplifier rules: numlslsdel: j0 ≠ 0, j < j0 ⊦ i < (* j) → i < (* j0); from specification int-basic1%numls-03: ⊦ i + (* j) < (* j0) ↔ i < j0 - j; from specification int-abs%numlsf: ⊦ i < (* j) ∧ j < 0 → i < 0; (forward) from specification int-basic1%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2