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Proof step 3 from proof grz (previous step)

0 < i ⊦ 0 < i ^ j ↔ i = 1 ∨ 0 ≤ j

User applied rule case distinction

And obtained the following new goals

Proof step 4:
0 < i, 0 < i ^ j ⊦ i = 1, 0 ≤ j

Rule argument

pos: right 1

Additional information

Used simplifier rules: one-01: ⊦ 1 ^ j = 1;%notneg-01: ⊦ 0 < i ∧ 0 ≤ j → ¬ i ^ j ≤ 0;% ⊦ i < j ↔ ¬ j ≤ i; (ss)