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Proof step 1 from proof i2n (conclusion)

⊦ i→n(i) * i→n(j) = i→n(i * j)

User applied rule cut formula

And obtained the following new goals

Proof step 2:
i < 0 ⊦ i→n(i) * i→n(j) = i→n(i * j)
Proof step 8:
⊦ i < 0, i→n(i) * i→n(j) = i→n(i * j)

Rule argument

fma: i < 0

Additional information

Used formulas: i < 0