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Proof step 3 from proof nat (previous step)

i < 0 ⊦ i→n(i) / i→n(j) = i→n(i / j), j = 0

User applied rule cut formula

And obtained the following new goals

Proof step 4:
i < 0, j < 0 ⊦ i→n(i) / i→n(j) = i→n(i / j), j = 0
Proof step 6:
i < 0 ⊦ i→n(i) / i→n(j) = i→n(i / j), j = 0, j < 0

Rule argument

fma: j < 0

Additional information

Used formulas: j < 0