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Proof step 19 from proof div-div-pred (previous step)

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

User applied rule case distinction

And obtained the following new goals

Proof step 20:
2 * (i1 * 2 + j1) + j0 = i0 * 4 + j2, abs(j2) < 4, abs(j1) < 2, i0 * 4 + j2 < ~3, abs(j0) < 2, i1 * 2 + j1 < 0, i0 * 4 + j2 < 0, j0 ≤ 0, i1 ≤ 0, j1 ≤ 0, i0 ≤ 0, j2 ≤ 0 ⊦ i1 * 4 + 2 * j1 + j0 = i0 * 4 + j2, i1 = i0
Proof step 22:
2 * (i1 * 2 + j1) + j0 = i0 * 4 + j2, abs(j2) < 4, abs(j1) < 2, i0 * 4 + j2 < ~3, abs(j0) < 2, i1 * 2 + j1 < 0, i0 * 4 + j2 < 0, j0 ≤ 0, i1 ≤ 0, j1 ≤ 0, i0 ≤ 0, j2 ≤ 0 ⊦ i1 = i0, abs(2 * j1 + j0) < 4

Rule argument

pos: right 2

Additional information

Used simplifier rules: ⊦ i ≤ j ↔ ¬ j < i; (ss)%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs% ⊦ i < i + j ↔ 0 < j; (s)% ⊦ 0 < i → (0 < i * j ↔ 0 < j); (s)% ⊦ i ≤ j → ¬ j < i; (s)%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