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Proof step 6 from proof ls-pos (previous step)

1 < i ∧ 0 ≤ j → (i ^ j ≤ i ^ j0 ↔ j ≤ j0), i ^ j < i ^ j0, 0 < i, 1 < i, 0 ≤ j, j0 ≤ j ⊦

User applied rule case distinction

And obtained the following new goals

Proof step 7:
i ^ j ≤ i ^ j0 ↔ j ≤ j0, i ^ j < i ^ j0, 0 < i, 1 < i, 0 ≤ j, j0 ≤ j ⊦

Rule argument

pos: left 1

Additional information

Used simplifier rules: numlelsdel-01: i0 ≠ 0, i0 ≤ i ⊦ (* i) < j → (* i0) < j; from specification int-abs%zero-gels-04: ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs