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Proof step 1 from proof add-n2i-num (conclusion)
⊦ 0 ≤ i → n→i(m) + (* i) = n→i(m + i→n(i))
Special step normalize
And obtained the following new goals
Proof step 2
:
⊦ 0 ≤ i → n→i(m) + i = n→i(m + i→n(i))