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Proof step 1 from proof sub (conclusion)

j ≤ 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, j ≤ i ⊦ i→n(i - j) = i→n(i) - i→n(j)

Rule argument

Additional information