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Proof step 18 from proof div-div-pred (previous step)

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

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: right 1

Additional information

Used simplifier rules: lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%lf-01: ⊦ i < j ∧ j < i0 → i +1 < i0; (forward) from specification int-basic1%trans: ⊦ i < j ∧ j < j0 → i < j0; (forward) from specification int-basic1%forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs%succ-01: ⊦ i +1 = i + 1; from specification int-basic2%numls-04: ⊦ i + j < (* i1) * i + j0 ↔ j < (i1 - 1) * i + j0; from specification int-mult%one: ⊦ 1 * i = i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%numls-12: ⊦ j < (* i1) * j + i ↔ 0 < (i1 - 1) * j + i; from specification int-mult%ls-02: ⊦ i < i + j ↔ 0 < j; from specification int-basic2%zero-gr: ⊦ 0 < i → (0 < i * j ↔ 0 < j); from specification int-mult%lenls: ⊦ i ≤ j → ¬ j < i; from specification int-abs