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Proof step 56 from proof gcd-rec-sub-03 (previous step)

0 < i ⊦ gcd(i, i0 + i) = gcd(i0 + i, i0), i = 0, i < i0 + i

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 57:
⊦ i ≥ j ↔ ¬ i < j
Proof step 58:
0 < i, i ≥ i0 + i ⊦ gcd(i, i0 + i) = gcd(i0 + i, i0), i = 0

Rule argument

spec: int-abs name: ge-def seq: ⊦ i ≥ j ↔ ¬ i < j subst: [j, i] → [i0 + i, i] flags: <rotated> <all paths>

Additional information

Used terms: i0 + i, i