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Proof step 42 from proof gcd-rec-sub-03 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 43:
⊦ i ≠ 0 → gcd(i, i) = abs(i)

Rule argument

spec: <toplevel> name: gcd-same seq: ⊦ i ≠ 0 → gcd(i, i) = abs(i) subst: [i] → [i] flags: <all paths>

Additional information

Used simplifier rules: gcd-zero: ⊦ gcd(0, 0) = 0;% ⊦ i < 0 → abs(i) = ~ i; (s)% ⊦ ~ i + i = 0; (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2 Used terms: i