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Proof step 1 from proof gcd-divides-both (conclusion)
⊦ j | gcd(i, i0) → j | i ∧ j | i0
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 2
:
j | gcd(i, i0) ⊦ j | i ∧ j | i0
Rule argument
Additional information