Hint Resolve lexfg_env_s.

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