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Proof step 8 from proof exgcd-div (previous step)

gcd(i, j) * i0 = i ⊦ gcd(i, j) * (gcd(i, j) * i0) / gcd(i, j) = i

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 9:
j ≠ 0 ⊦ (j * i) / j = i

Rule argument

spec: int-div inst: name: div-mult seq: j ≠ 0 ⊦ (j * i) / j = i subst: [i, j] → [i0, gcd(i, j)] flags: rewrite

Additional information

Used simplifier rules: gcd-zero-zero: ⊦ gcd(i, j) = 0 ↔ i = 0 ∧ j = 0;%gcd-zero: ⊦ gcd(0, 0) = 0;%zero-left: ⊦ 0 * i = 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 Used terms: i0, gcd(i, j)