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Proof step 32 from proof cong-equiv (previous step)

n = m0 * m + n0, n0 = (n1 + n) % m, n0 < m ⊦ m = 0, m | n1

User applied rule insert equation

And obtained the following new goals

Proof step 33:
n0 = (n1 + m0 * m + n0) % m, n0 < m ⊦ m = 0, m | n1

Rule argument

pos: left 1 flags: discard

Additional information