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Proof step 10 from proof gcd-rec-sub (previous step)

0 < i, 0 < i0, 1 < i0 + i, 0 ≤ i0 + i ⊦ gcd(i, (i0 + i) % i) = gcd(i, i0)

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: int-div name: mod-rec seq: 0 < j, j ≤ i ⊦ i % j = (i - j) % j subst: [j, i] → [i, i0 + i] flags: <all paths>

Additional information

Used simplifier rules: ⊦ i ≤ i + j ↔ 0 ≤ j; (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2% ⊦ i ≤ j ↔ ¬ j < i; (ss)% ⊦ i < j → ¬ j < i; (s) Used terms: i, i0 + i