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Proof step 9 from proof i2n (previous step)

0 ≤ i ⊦ i→n(i) * i→n(j) = i→n(i * j)

User applied rule cut formula

And obtained the following new goals

Proof step 10:
j < 0, 0 ≤ i ⊦ i→n(i) * i→n(j) = i→n(i * j)
Proof step 13:
0 ≤ i ⊦ i→n(i) * i→n(j) = i→n(i * j), j < 0

Rule argument

fma: j < 0

Additional information

Used formulas: j < 0