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Proof step 20 from proof eq-one (previous step)

i * j = 1, j ≤ 0, i < 0 ⊦ 0 < 1 ∧ 0 < 1 ∧ 1 ≤ ~ i ∧ 1 < ~ j, i = 0, ~1 = j ∧ i = ~1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 21:
i * j = 1, j ≤ 0, i < 0, ~1 ≤ j ⊦ i = 0, ~1 = j ∧ i = ~1

Rule argument

Additional information

Used simplifier rules: forward-le-04: ⊦ j ≤ i ∧ i ≤ 0 → j ≤ 0; (forward) from specification int-abs%lels: ⊦ ¬ i < j ↔ j ≤ i; 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%numls-03: ⊦ i + (* j) < (* j0) ↔ i < j0 - j; from specification int-abs%neg-07: ⊦ i < ~ j ↔ i + j < 0; from specification int-abs%zero-gels-01: ⊦ i < 0 ∧ 0 ≤ j → i < j; from specification int-abs%neg: ⊦ 0 < ~ i ↔ i < 0; from specification int-abs%leone: ⊦ 1 ≤ i ↔ 0 < i; from specification int-abs