Hint Resolve multipar_context_independent.

Lemma consistent_mutual:
(forall S a A,   Typing S a A -> True) /\
(forall S phi,   PropWff S phi -> True) /\
(forall S D p1 p2, Iso S D p1 p2 -> Good S D -> (forall A1 B1 T1 A2 B2 T2, p1 = Eq A1 B1 T1 -> p2 = Eq A2 B2 T2-> (joins S D A1 A2 /\ joins S D B1 B2 /\ joins S D T1 T2))) /\
(forall S D A B T,   DefEq S D A B T -> Good S D -> joins S D A B) /\
(forall S,       Ctx S -> True).