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Proof step 2 from proof grz (previous step)

0 < i ⊦ 0 < i ^ j ↔ i = 1 ∨ 0 ≤ j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
0 < i ⊦ 0 < i ^ j ↔ i = 1 ∨ 0 ≤ j

Rule argument

Additional information

Used simplifier rules: ⊦ 1 ≤ i ↔ 0 < i; (s)%numlsf-02: ⊦ (* i) < j → i + 1 ≤ j; (forward) from specification int-abs