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Proof step 17 from proof eq-one (previous step)

i ≤ 0 ⊦ ¬ i ≤ 0 ∧ ¬ 0 ≤ j, ¬ 0 ≤ i ∧ ¬ j ≤ 0, i * j = 1 ↔ abs(j) = 1 ∧ i = j, ¬ i ≤ 0 ∧ 1 < j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
i * j = 1, j ≤ 0, i < 0 ⊦ i = 0, ~1 = j ∧ i = ~1

Rule argument

Additional information

Used simplifier rules: zero-left: ⊦ 0 * i = 0;%associativity: ⊦ i * j * j0 = (i * j) * j0;%commutativity: ⊦ i * j = j * i;%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%numadd-01: ⊦ (* j) = i0 + (* j0) ↔ j - j0 = i0; from specification int-abs%negeq: ⊦ ~ i = j ↔ i + j = 0; from specification int-abs%abs-08: ⊦ i < 0 → abs(i) = ~ i; from specification int-abs%lels-02: ⊦ i ≠ j → (i ≤ j ↔ i < j); from specification int-abs%forward-le-04: ⊦ j ≤ i ∧ i ≤ 0 → j ≤ 0; (forward) from specification int-abs%anti: ⊦ i ≤ j → (j ≤ i ↔ i = j); from specification int-abs%zero-gels-04: ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs