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Proof step 39 from proof div-div-pred-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 40:
j0 + j1 * 2 ≤ ~8, i1 < 0, j0 + j1 * 2 < 0, 8 * i1 + j0 + j1 * 2 < 0, 8 * i1 + j0 + j1 * 2 < ~7, j1 ≤ 0, abs(j1) < 4, abs(j0) < 2, i1 * 4 + j1 < 0, i1 * 4 + j1 ≤ ~3, j0 ≤ 0 ⊦ (j0 + j1 * 2) / 8 = 0

Rule argument

Additional information

Used simplifier rules: zero-gels-03: ⊦ i < 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs%forward-le-04: ⊦ j ≤ i ∧ i ≤ 0 → j ≤ 0; (forward) from specification int-abs%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%numls-02: ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0; from specification int-abs%neg-13: ⊦ ~ i < j ↔ 0 < j + i; from specification int-abs%abs-08: ⊦ i < 0 → abs(i) = ~ i; from specification int-abs