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Proof step 10 from proof i2n-mod (previous step)

j0 < j, 0 < j, 0 ≤ j0, 0 ≤ i0, 0 ≤ j ⊦ i→n(j0) = i→n(j0) % i→n(j)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 11:
⊦ m < n → m % n = m
Proof step 12:
¬ i→n(j0) < i→n(j), j0 < j, 0 < j, 0 ≤ j0, 0 ≤ i0, 0 ≤ j ⊦ i→n(j0) = i→n(j0) % i→n(j)

Rule argument

spec: nat-div inst: name: mod-eq seq: ⊦ m < n → m % n = m subst: [n, m] → [i→n(j), i→n(j0)] flags: rewrite

Additional information

Used terms: i→n(j), i→n(j0)