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Proof step 11 from proof i2n-mod (previous step)

0 < j, j0 < j, 0 ≤ i0, 0 ≤ j0, 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 12:
⊦ m < n → m % n = m
Proof step 13:
j0 < j, 0 < j, 0 ≤ j, 0 ≤ j0, 0 ≤ i0 ⊦ i→n(j0) = i→n(j0) % i→n(j), i→n(j0) < i→n(j)

Rule argument

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

Additional information

Used terms: i→n(j), i→n(j0)