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Proof step 2 from proof divides-mult-02 (
previous step
)
i | j, i0 | j0 ⊦ i * i0 | j * j0
User applied rule insert rewrite lemma
And obtained the following new goals
Proof step 3
:
⊦ i | j ↔ (∃ i0. i * i0 = j)
Proof step 4
:
i | j, ∃ i1. i0 * i1 = j0 ⊦ i * i0 | j * j0
Rule argument
spec: int-gcd inst: name: divides seq: ⊦ i | j ↔ (∃ i0. i * i0 = j) subst: [j, i] → [j0, i0] flags: rewrite, current
Additional information
Used terms: j0, i0