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Proof step 6 from proof gcd-same (previous step)

gcd(abs(i), abs(i)) | abs(i) ⊦ gcd(abs(i), abs(i)) = abs(i), i = 0

User applied rule apply lemma

And obtained the following new goals

Rule argument

spec: <toplevel> name: gcd-max seq: ⊦ i0 | i ∧ i0 | j ∧ (i ≠ 0 ∨ j ≠ 0) → i0 ≤ gcd(i, j) subst: [i0, i, j] → [abs(i), abs(i), abs(i)]

Additional information

Used simplifier rules: divides-abs: ⊦ abs(i) | abs(j) ↔ i | j;%div-zero: ⊦ i | i;%gcd-nls-01: ⊦ i ≠ 0 → ¬ abs(i) < gcd(abs(i), j);% ⊦ abs(i) = 0 ↔ i = 0; (s)% ⊦ i ≠ j → (i ≤ j ↔ i < j); (s) Used terms: abs(i), abs(i), abs(i)