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Proof step 29 from proof eq-one (previous step)

1 * 1 < i * j ⊦ i * j = 1 ↔ abs(j) = 1 ∧ i = j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 30:
abs(j) = 1, 0 < j * j, 1 < j * j ⊦ j * j = 1

Rule argument

Additional information

Used simplifier rules: associativity: ⊦ i * j * j0 = (i * j) * j0;%commutativity: ⊦ i * j = j * i;%lf-02: ⊦ i < j ∧ j = i0 → i < i0; (forward) from specification int-basic1%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%forward-pos: ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward) from specification int-basic1