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

2 * (i1 * 2 + j1) + j0 = i0 * 4 + j2 ∧ j2 ≤ 0 ∧ abs(j2) < 4 ∧ i0 ≤ 0 ∧ 0 < 4, 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 = i0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
2 * (i1 * 2 + j1) + j0 = i0 * 4 + j2, abs(j2) < 4, abs(j1) < 2, j0 ≤ 0, abs(j0) < 2, i0 * 4 + j2 < ~3, j1 ≤ 0, i1 ≤ 0, i1 * 2 + j1 < 0, j2 ≤ 0, i0 ≤ 0, i0 * 4 + j2 < 0 ⊦ i1 = i0

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