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Proof step 8 from proof ls-pos (previous step)

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 9:
1 < i, 0 < i, i ^ j < i ^ j0, j0 ≤ j, 0 ≤ j ⊦ j ≤ j0, i ^ j ≤ i ^ j0

Rule argument

pos: left 1

Additional information

Used simplifier rules: le-pos: ⊦ 1 < i ∧ 0 ≤ j → (i ^ j ≤ i ^ j0 ↔ j ≤ j0);% ⊦ i ≤ j → (j ≤ i ↔ i = j); (s)% ⊦ ¬ i < i; (s)