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Proof step 8 from proof nat (previous step)

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

User applied rule cut formula

And obtained the following new goals

Proof step 9:
j < 0 ⊦ i→n(i) / i→n(j) = i→n(i / j), j = 0, i < 0

Rule argument

fma: j < 0

Additional information

Used simplifier rules: natpos: 0 < j, 0 ≤ i ⊦ i→n(i) / i→n(j) = i→n(i / j);% ⊦ ¬ i < j → (j < i ↔ i ≠ j); (ss)% ⊦ i ≤ j ↔ ¬ j < i; (ss) Used formulas: j < 0