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Proof step 1 from proof i2n-mod (conclusion)
⊦ 0 ≤ i ∧ 0 < j → i→n(i % j) = i→n(i) % i→n(j)
System applied rule normalize
And obtained the following new goals
Proof step 2
:
0 < j, 0 ≤ i ⊦ i→n(i % j) = i→n(i) % i→n(j)
Rule argument
Additional information