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Proof step 2 from proof ls-pos (previous step)

1 < i, 0 ≤ j ⊦ i ^ j < i ^ j0 ↔ j < j0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
0 < i, 1 < i, 0 ≤ j ⊦ i ^ j < i ^ j0 ↔ j < j0

Rule argument

Additional information

Used simplifier rules: forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%forward-pos: ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward) from specification int-basic1