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Proof step 23 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 24:
i * j = 1, 0 ≤ j, 0 < i, j ≤ 1, 0 ≤ i ⊦ j = 1 ∧ i = 1

Rule argument

Additional information

Used simplifier rules: associativity: ⊦ i * j * j0 = (i * j) * j0;%commutativity: ⊦ i * j = j * i;%ls-01: ⊦ i < j → ¬ j < i; from specification int-basic1%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%anti: ⊦ i ≤ j → (j ≤ i ↔ i = j); from specification int-abs%lels-03: ⊦ i < j → (¬ ¬ j ≤ i ↔ false); from specification int-abs%abs-16: ⊦ 0 < j → abs(j) = j; from specification int-abs%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs