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Proof step 9 from proof cong-equiv (previous step)

⊦ m = 0, m | n1 ↔ (n1 + n0) % m = n0 % m, n1 = 0

User applied rule case distinction

And obtained the following new goals

Proof step 10:
m | n1 ⊦ (n1 + n0) % m = n0 % m, m = 0, n1 = 0
Proof step 12:
(n1 + n0) % m = n0 % m ⊦ m | n1, m = 0, n1 = 0

Rule argument

pos: right 2

Additional information