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Proof step 16 from proof cong-equiv (previous step)

n1 + m1 * m = m0 * m, n2 < m ⊦ n1 = 0, m | n1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 17:
⊦ m | n ↔ (∃ n0. m * n0 = n)
Proof step 18:
n1 + m1 * m = m0 * m, n2 < m ⊦ n1 = 0, ∃ n0. m * n0 = n1

Rule argument

spec: nat-prime inst: name: Divides seq: ⊦ m | n ↔ (∃ n0. m * n0 = n) subst: [n, m] → [n1, m] flags: rewrite, current

Additional information

Used terms: n1, m