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Proof step 10 from proof mod-ls-same-01 (previous step)

⊦ i % j < i ↔ 0 < i ∧ (abs(j) = 1 ∨ abs(j) ≤ i), j = 0, i < 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
0 ≤ i ⊦ i % j < i ↔ 0 < i ∧ (abs(j) = 1 ∨ abs(j) ≤ i), j = 0

Rule argument

Additional information

Used simplifier rules: forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs% ⊦ i < j ↔ ¬ j ≤ i; (ss)