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Proof step 6 from proof div-minus-onediff-01 (previous step)

i + 1 = i0 + j, 0 < j, i < 0, ∃ j0. i = j0 * j ⊦ i0 / j = i / j, j = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
j0 * j + 1 = i0 + j, 0 < j, j0 < 0 ⊦ i0 / j = j0, j = 0

Rule argument

Additional information

Used simplifier rules: div-mult-03: j ≠ 0 ⊦ (i * j) / j = i;%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs% ⊦ 0 < i → (i * j < 0 ↔ j < 0); (s)%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult