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Proof step 15 from proof mod-ls-same-01 (previous step)

0 < i ⊦ i % j < i ↔ abs(j) = 1 ∨ abs(j) ≤ i, i = 0, j = 1, j = 0

User applied rule cut formula

And obtained the following new goals

Proof step 16:
abs(j) = 1, 0 < i ⊦ i % j < i ↔ abs(j) = 1 ∨ abs(j) ≤ i, i = 0, j = 0, j = 1
Proof step 19:
0 < i ⊦ i % j < i ↔ abs(j) = 1 ∨ abs(j) ≤ i, abs(j) = 1, i = 0, j = 0, j = 1

Rule argument

fma: abs(j) = 1

Additional information

Used formulas: abs(j) = 1