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Proof step 7 from proof grz (previous step)

0 < i ^ j, 0 < i, i ^ ~ j = 1 ∧ 0 < ~ j → abs(i) = 1 ⊦ i = 1, 0 ≤ j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
0 < i ^ j, 0 < i, j < 0 ⊦ i ^ ~ j = 1, i = 1, 0 ≤ j

Rule argument

Additional information

Used simplifier rules: ⊦ i < j ↔ ¬ j ≤ i; (ss)% ⊦ 0 < j → abs(j) = j; (s)% ⊦ 0 < ~ i ↔ i < 0; (s)