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Proof step 2 from proof i2n (previous step)

i < 0 ⊦ i→n(i) * i→n(j) = i→n(i * j)

User applied rule cut formula

And obtained the following new goals

Proof step 3:
i < 0, j < 0 ⊦ i→n(i) * i→n(j) = i→n(i * j)
Proof step 5:
i < 0 ⊦ i→n(i) * i→n(j) = i→n(i * j), j < 0

Rule argument

fma: j < 0

Additional information

Used formulas: j < 0