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Proof step 3 from proof num-gcd-pos (previous step)

1 < i, 0 < j, j < i, 0 < i ⊦ gcd(i, j) = gcd(j, i % j)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ 0 < j ∧ 0 ≤ i → gcd(i, j) = gcd(j, i % j)

Rule argument

spec: <toplevel> name: gcd-rec-pos seq: ⊦ 0 < j ∧ 0 ≤ i → gcd(i, j) = gcd(j, i % j) subst: [j, i] → [j, i] flags: <all paths>

Additional information

Used simplifier rules: ⊦ i ≤ j ↔ ¬ j < i; (ss)% ⊦ i < j → ¬ j < i; (s) Used terms: j, i