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Proof step 5 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 6:
⊦ 0 ≤ i ∧ 0 < j → i→n(i % j) = i→n(i) % i→n(j)

Rule argument

spec: intnat inst: name: i2n-mod seq: ⊦ 0 ≤ i ∧ 0 < j → i→n(i % j) = i→n(i) % i→n(j) subst: [j, i] → [j, i] flags: rewrite

Additional information

Used simplifier rules: zero-gels-04: ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs Used terms: j, i