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Proof step 4 from proof eq-one (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
⊦ i * j = 1 ↔ abs(j) = 1 ∧ i = j, 0 < i ∧ 1 < j

Rule argument

Additional information

Used simplifier rules: ⊦ 1 ≤ i ↔ 0 < i; (s)