Hint Resolve lexfg_cons_t.

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').