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Proof step 7 from proof ls-pos (previous step)

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

User applied rule case distinction

And obtained the following new goals

Proof step 8:
i ^ j ≤ i ^ j0 ↔ j ≤ j0, i ^ j < i ^ j0, 0 < i, 1 < i, 0 ≤ j, j0 ≤ j ⊦

Rule argument

pos: left 6

Additional information

Used simplifier rules: ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; (s)