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Proof step 7 from proof gcd-trans (
previous step
)
gcd(m, n0) | n0 ∧ n0 | n → gcd(m, n0) | n, n0 | n ⊦ gcd(m, n0) | gcd(m, n), gcd(m, n0) | n
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 8
:
n0 | n ⊦ gcd(m, n0) | gcd(m, n), gcd(m, n0) | n, gcd(m, n0) | n0
Rule argument
Additional information