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Proof step 29 from proof gcd-rec-sub-01 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 30:
⊦ i < j ↔ ¬(i = j ∨ j < i)
Proof step 31:
i < 0, 0 < i0, abs(i) = abs(i0) ∨ abs(i0) < abs(i) ⊦ gcd(i, i0 - abs(i)) = gcd(abs(i), abs(i0)), i = 0

Rule argument

spec: int-basic1 name: l seq: ⊦ i < j ↔ ¬(i = j ∨ j < i) subst: [j, i] → [abs(i0), abs(i)] flags: <all paths>

Additional information

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