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Proof step 5 from proof gcd-multdiv (previous step)

∃ n0. gcd(m, n) * n0 = m ⊦ gcd(m, n) | m * n0

User applied rule exists left

And obtained the following new goals

Proof step 6:
gcd(m, n) * n1 = m ⊦ gcd(m, n) | m * n0

Rule argument

pos: left 1

Additional information