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Proof step 5 from proof modmod-add-zero (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
0 ≠ j ⊦ i ≤ 0 ∧ i0 ≤ 0 → (i % j + i0) % j = (i + i0) % j
Proof step 7:
¬(0 ≠ j ∧ i0 ≤ 0 ∧ i ≤ 0), (i + i0 % j) % j = 0, 0 < j ⊦ (i + i0) % j = 0, 0 ≤ i0 ∧ 0 ≤ i

Rule argument

spec: int-div inst: name: modmod-add-02 seq: 0 ≠ j ⊦ i ≤ 0 ∧ i0 ≤ 0 → (i % j + i0) % j = (i + i0) % j subst: [i0, j, i] → [i, j, i0] flags: rewrite, current

Additional information

Used simplifier rules: add-c: ⊦ i + j = j + i; from specification int-basic2 Used terms: i, j, i0