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Proof step 21 from proof modmod-mult-07 (previous step)

0 < j ⊦ (i % j * i0) % j = (i * i0) % j, i < 0, 0 ≤ i ∧ 0 ≤ i0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 22:
i0 < 0, 0 < j, 0 ≤ i ⊦ (i % j * i0) % j = (i * i0) % j

Rule argument

Additional information

Used simplifier rules: ⊦ i < 0 ∧ 0 ≤ j → i < j; (s)%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs% ⊦ i ≤ j ↔ ¬ j < i; (ss)% ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; (s)% ⊦ i < j ↔ ¬ j ≤ i; (ss)