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Proof step 70 from proof gcd-rec-sub-01 (previous step)

0 < i, i0 < 0, abs(i0) < abs(i) ⊦ gcd(abs(i), abs(i) - abs(i0)) = gcd(abs(i), abs(i0)), i0 = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 71:
⊦ gcd(i, j) = gcd(j, i)
Proof step 72:
abs(i0) < abs(i), i0 < 0, 0 < i ⊦ gcd(abs(i), abs(i) - abs(i0)) = gcd(abs(i0), abs(i)), i0 = 0

Rule argument

spec: <toplevel> name: gcd-com seq: ⊦ gcd(i, j) = gcd(j, i) subst: [j, i] → [abs(i0), abs(i)] flags: <all paths>

Additional information

Used terms: abs(i0), abs(i)