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Proof step 12 from proof div-div-pred (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
abs(j1) < 2, j0 ≤ 0, abs(j0) < 2, 2 * (i1 * 2 + j1) + j0 < ~3, j1 ≤ 0, i1 ≤ 0, i1 * 2 + j1 < 0, 2 * (i1 * 2 + j1) + j0 < 0 ⊦ i1 = (2 * (i1 * 2 + j1) + j0) / 4

Rule argument

Additional information

Used simplifier rules: forward-le-04: ⊦ j ≤ i ∧ i ≤ 0 → j ≤ 0; (forward) from specification int-abs%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs