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Proof step 7 from proof eq-one (previous step)

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

User applied rule prop simplification

And obtained the following new goals

Proof step 8:
⊦ 0 < i ∧ j < 0, i < 0 ∧ 0 < j, i * j = 1 ↔ abs(j) = 1 ∧ i = j, 0 < i ∧ 1 < j

Rule argument

pos: right 1

Additional information