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Proof step 27 from proof gcd-mult-pos (previous step)

0 < i1, 0 < i, 0 ≤ j, j ≤ i, 0 ≤ i, ∀ j0, i0, i2. i→n(j0 * i0 + i2 * i0) < i→n(j * i1 + i * i1) → 0 < i0 ∧ 0 < i2 ∧ 0 ≤ j0 → gcd(i2 * i0, j0 * i0) = gcd(i2, j0) * i0 ⊦ gcd(i * i1, j * i1) = gcd(i, j) * i1

User applied rule insert rewrite lemma

And obtained the following new goals

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] → [j * i1, i * i1] flags: rewrite, current

Additional information

Used simplifier rules: gcd-zero-04: ⊦ gcd(i, 0) = abs(i);%zero-neq-01: ⊦ 0 < i → (i * j = 0 ↔ j = 0); from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%zero-left: ⊦ 0 * i = 0; from specification int-mult%abs-eqall: ⊦ abs(i * j) = abs(i) * abs(j); from specification int-mult%abs-16: ⊦ 0 < j → abs(j) = j; from specification int-abs%eq-pos: ⊦ 0 < i → (i * j = i * j0 ↔ j = j0); from specification int-mult Used terms: j * i1, i * i1