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Proof step 2 from proof mod-i2n (previous step)

0 ≤ i, 0 < j ⊦ i→n(i |%| j) = i→n(i) % i→n(j)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ 0 ≤ i % j → i |%| j = i % j
Proof step 4:
0 ≤ i % j, 0 ≤ i, 0 < j ⊦ i→n(i % j) = i→n(i) % i→n(j)

Rule argument

spec: int-mod inst: name: mod-pos-def seq: ⊦ 0 ≤ i % j → i |%| j = i % j subst: [j, i] → [j, i] flags: rewrite, current

Additional information

Used simplifier rules: mod-zero: ⊦ 0 < j ∧ 0 ≤ i → 0 ≤ i % j; from specification int-div Used terms: j, i