Notation lexfg := (e_lexfg _) (only parsing).

Theorem lexfg_cons_s : forall (a : terms) (s s' : sub_explicits), e_lexfg _ (e_P1 ws) (e_P2 ws) s s' -> e_lexfg _ (e_P1 ws) (e_P2 ws) (cons a s) (cons a s').