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Proof step 26 from proof div-div-pred (previous step)

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

User applied rule cut formula

And obtained the following new goals

Rule argument

fma: j1 = ~1

Additional information

Used simplifier rules: nummult-ls: ⊦ 0 < i ∧ 0 ≤ j → ((* i) * i0 < (* j) ↔ i0 < (i + j - 1) / i);%minus-add: i < 0 ⊦ (* i) + j = j - ~ i; from specification int-abs%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%numadd-01: ⊦ (* j) = i0 + (* j0) ↔ j - j0 = i0; from specification int-abs%sub-ls: ⊦ i - j < i0 ↔ i < j + i0; from specification int-abs%lsone: ⊦ i < 1 ↔ i ≤ 0; from specification int-abs%zero-gels-04: ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs%lf-02: ⊦ i < j ∧ j = i0 → i < i0; (forward) from specification int-basic1%numlelsdel: j0 ≠ 0, j ≤ j0 ⊦ i < (* j) → i < (* j0); from specification int-abs Used formulas: j1 = ~1