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Proof step 23 from proof div-div-pred (previous step)

j0 = ~1, 2 * (i1 * 2 + j1) + j0 = i0 * 4 + j2, abs(j2) < abs(2 * j1 + j0), 4 ≤ abs(2 * j1 + j0), i0 * 4 + j2 < 0, i0 ≤ 0, j2 ≤ 0, i1 * 2 + j1 < 0, i1 ≤ 0, j1 ≤ 0, i0 * 4 + j2 < ~3, abs(j0) < 2, j0 ≤ 0, abs(j1) < 2, abs(j2) < 4 ⊦ i1 = i0, 2 * j1 + j0 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) 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%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%neq-two: ⊦ 2 * i ≠ 1; from specification int-mult%sub-eq: ⊦ i - j = i0 ↔ i = i0 + j; 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%minus-add: i < 0 ⊦ (* i) + j = j - ~ i; from specification int-abs