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Proof step 12 from proof div-le-same-neg (previous step)

i0 * (i1 + 1) + j0 < 0, i0 < i1, j0 < i1, 0 < i1, abs(j0) ≤ i1, i0 ≤ 0, j0 ≤ 0 ⊦ 0 ≤ i1 * i0 + j0 ↔ i0 * (i1 + 1) + j0 = 0, i1 = 0

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 1

Additional information

Used simplifier rules: ⊦ i ≤ j ↔ ¬ j < i; (ss)%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%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%forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs%forward-lele: ⊦ j ≤ j0 ∧ j0 ≤ j1 → j ≤ j1; (forward) from specification int-abs%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs% ⊦ i + j < i + i0 ↔ j < i0; (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2% ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; (s)%ls-forward: ⊦ i * j < i * j0 → i ≠ 0; (forward) from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult% ⊦ i ≠ j → (i ≤ j ↔ i < j); (s)% ⊦ i < 0 ∧ j0 < j ∧ 0 < j → ¬ 0 ≤ i * j + j0; (s)