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Proof step 5 from proof div-div (previous step)

j | i, i ≤ 0, j ≤ 0, ∃ i0. i * i0 = j ⊦ i = j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
i < 0, i * i0 | i, 0 ≤ i0 ⊦ i = 0, i0 = 1

Rule argument

Additional information

Used simplifier rules: ⊦ i < 0 ∧ 0 ≤ j → i < j; (s)%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs% ⊦ i < 0 → (i * j ≤ 0 ↔ 0 ≤ j); (s)% ⊦ i ≠ j → (i ≤ j ↔ i < j); (s)%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult% ⊦ i * j = i ↔ i = 0 ∨ j = 1; (ws)