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Proof step 3 from proof gcd-mult-01 (previous step)

⊦ gcd(i * j, i) = abs(i)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ j ≠ 0 → gcd(i, j) = gcd(abs(j), abs(i) % abs(j))

Rule argument

spec: int-gcd inst: name: gcd-rec seq: ⊦ j ≠ 0 → gcd(i, j) = gcd(abs(j), abs(i) % abs(j)) subst: [j, i] → [i, i * j] flags: rewrite, current

Additional information

Used simplifier rules: gcd-zero: ⊦ gcd(0, 0) = 0;%gcd-zero-04: ⊦ gcd(i, 0) = abs(i);%abs-eqall: ⊦ abs(i * j) = abs(i) * abs(j); from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%abs-07: ⊦ abs(i) = 0 ↔ i = 0; from specification int-abs%mod-mult-04: j ≠ 0 ⊦ (i * j) % j = 0; from specification int-div%abs-01: ⊦ abs(abs(i)) = abs(i); from specification int-abs%zero-left: ⊦ 0 * i = 0; from specification int-mult Used terms: i, i * j