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Proof step 3 from proof modmod-add-03 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
0 ≠ j ⊦ i ≤ 0 ∧ i0 ≤ 0 → (i % j + i0) % j = (i + i0) % j
Proof step 5:
i < 0, i0 < 0, i ≤ 0, i0 % j ≤ 0 ⊦ (i + i0 % j) % j = (i + i0) % j, 0 = j

Rule argument

spec: <toplevel> name: modmod-add-02 seq: 0 ≠ j ⊦ i ≤ 0 ∧ i0 ≤ 0 → (i % j + i0) % j = (i + i0) % j subst: [i0, j, i] → [i0 % j, j, i] flags: <all paths>

Additional information

Used simplifier rules: mod-grz-01: ⊦ i < 0 ∧ j ≠ 0 → ¬ 0 < i % j;% ⊦ i < 0 ∧ 0 ≤ j → i ≤ j; (s)% ⊦ i ≤ j ↔ ¬ j < i; (ss) Used terms: i0 % j, j, i