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Proof step 5 from proof gcd-rec-sub-01 (
previous step
)
¬(0 < i ∧ i < i0 + i) ⊦ gcd(i, i0 + i) = gcd(i, i0)
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 6
:
⊦ gcd(i, i0 + i) = gcd(i, i0), 0 < i ∧ i < i0 + i
Rule argument
Additional information