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Proof step 9 from proof gcd-one (
previous step
)
¬(∃ i0. 1 < i0 ∧ i0 | i ∧ i0 | j) ⊦ gcd(i, j) = 1
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 10
:
⊦ gcd(i, j) = 1, ∃ i0. 1 < i0 ∧ i0 | i ∧ i0 | j
Rule argument
Additional information