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Proof step 16 from proof div-eq-zero-01 (previous step)

1 < abs(j), i < 0, 0 < abs(j) + i ⊦ i / j = 0, 0 = j

User applied rule cut formula

And obtained the following new goals

Proof step 17:
i < j, 1 < abs(j), i < 0, 0 < abs(j) + i ⊦ i / j = 0, 0 = j
Proof step 19:
1 < abs(j), i < 0, 0 < abs(j) + i ⊦ i < j, i / j = 0, 0 = j

Rule argument

fma: i < j

Additional information

Used formulas: i < j