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Proof step 25 from proof div-div-pred (previous step)

2 * j1 = i2 + 1, 2 * (i1 * 2 + j1) = (i0 * 4 + j2) + 1, abs(j2) < abs(i2), 4 ≤ abs(i2), abs(j2) < 4, abs(j1) < 2, i0 * 4 + j2 < ~3, j1 ≤ 0, i1 ≤ 0, i1 * 2 + j1 < 0, j2 ≤ 0, i0 ≤ 0, i0 * 4 + j2 < 0 ⊦ i1 = i0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 26:
2 * (i1 * 2 + j1) = (i0 * 4 + j2) + 1, 2 * j1 = i2 + 1, i0 * 4 + j2 < 0, i0 ≤ 0, j2 ≤ 0, i1 * 2 + j1 < 0, i1 ≤ 0, j1 ≤ 0, i0 * 4 + j2 < ~3, abs(j1) < 2, abs(j2) < 4, 4 ≤ abs(i2), abs(j2) < abs(i2) ⊦ i1 = i0, i2 = 0

Rule argument

Additional information

Used simplifier rules: abs-03: ⊦ 0 ≤ abs(i); from specification int-abs%abs-06: ⊦ 0 < abs(i) ↔ i ≠ 0; from specification int-abs%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%forward-le: ⊦ i ≤ j ∧ 0 < i → 0 < j; (forward) from specification int-abs%forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs