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Proof step 3 from proof ls-pos-03 (previous step)

0 < j, 0 ≤ i0, 0 ≤ i ⊦ i ^ j < i0 ^ j ↔ i < i0

User applied rule case distinction

And obtained the following new goals

Proof step 4:
0 < j, i ^ j < i0 ^ j, 0 ≤ i, 0 ≤ i0 ⊦ i < i0

Rule argument

pos: right 1

Additional information

Used simplifier rules: le-pos-01: ⊦ 0 ≤ i ∧ 0 ≤ i0 ∧ 0 < j → (i ^ j ≤ i0 ^ j ↔ i ≤ i0);% ⊦ i < j ↔ ¬ j ≤ i; (ss)% ⊦ i < j → (j ≤ i ↔ false); (as)