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Proof step 19 from proof eq-one (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 20:
i * j = 1, i < 0, j ≤ 0, ~1 ≤ j ⊦ i = 0, ~1 = j ∧ i = ~1

Rule argument

Additional information

Used simplifier rules: forward-lele: ⊦ j ≤ j0 ∧ j0 ≤ j1 → j ≤ j1; (forward) from specification int-abs% ⊦ i < j ↔ ¬ j ≤ i; (ss)%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2% ⊦ i + (* j) < (* j0) ↔ i < j0 - j; (s)% ⊦ i < ~ j ↔ i + j < 0; (s)% ⊦ i < 0 ∧ 0 ≤ j → i < j; (s)% ⊦ 0 < ~ i ↔ i < 0; (s)% ⊦ 1 ≤ i ↔ 0 < i; (s)