There is a path from a certificate to each of its predecessors.
This should be intuitive, since predecessors is based on the DAG edges, which define the paths.
We use the path-to-previous theorem to prove this, unsurprisingly.
Theorem:
(defthm path-to-predecessor (implies (and (certificate-setp dag) (certificate-set-unequivocalp dag) (in cert1 dag) (in cert (predecessors cert1 dag))) (equal (path-to-author+round cert1 (certificate->author cert) (certificate->round cert) dag) cert)))