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Proof step 5 from proof le-pos-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
0 < i, i0 < i, 0 < j, i ^ j ≤ i0 ^ j, 0 ≤ j, 0 ≤ i0, 0 ≤ i ⊦

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% ⊦ i < j → ¬ j < i; (s)% ⊦ i ≤ j ↔ ¬ j < i; (ss)% ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; (s)