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Proof step 9 from proof gcd-multdiv (
previous step
)
gcd(m, n) * n1 = m ⊦ ∃ n2. gcd(m, n) * n2 = (gcd(m, n) * n1) * n0
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 10
:
gcd(m, n) * n1 = m ⊦ ∃ n2. gcd(m, n) * n2 = m * n0
Rule argument
Additional information