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Proof step 5 from proof mod-zero-05 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
i < j, 0 < j, i < 0, 0 < j + i ⊦ i % j = i

Rule argument

Additional information

Used simplifier rules: forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs% ⊦ ~ i < j ↔ 0 < j + i; (s)% ⊦ i < 0 → abs(i) = ~ i; (s)% ⊦ i ≤ j ↔ ¬ j < i; (ss)% ⊦ ¬ abs(i) < i; (s)% ⊦ ¬ abs(i) < 0; (s)%forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs%forward-le-04: ⊦ j ≤ i ∧ i ≤ 0 → j ≤ 0; (forward) from specification int-abs% ⊦ i < j ↔ ¬ j ≤ i; (ss)