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%% This example demonstrates when schematic polymorphism fails to prove
%% properties about polymorphic specifications

Define eq : A -> A -> prop by
  eq X X.

Type p   A -> o.

Theorem eq_non_schm[A,B] : forall (X:A) (Y:B),
  eq (p X) (p Y) -> false.
intros. %% The following case analysis fails for the following reason. %% Whether (p X) is equal to (p Y) depends on whether X and Y has %% identical types, i.e., whether A and B can be made equal. Since A %% and B are treated as "black boxes", examination of their identity %% is not allowed. % case H1. skip.
%% The following is a theorem that does not have %% a schematic proof. Theorem true_thm_with_no_schematic_proof [A,B]: forall (X:A) (F:A -> B), exists (Y:B), (eq (p X) (p Y)) \/ (eq (p X) (p Y) -> false).
%% This theorem has a proof as follows: %% Case 1: A = B. %% We choose Y to be X and show the right branch holds. %% Case 2: A =/= B. %% We choose Y to be (F X) and then try to prove the right branch. %% Since (p X) and (p Y) has different predicate heads, %% the right branch is trivially true. %% %% However, since the above proof inspects the types for A and B, %% it is not a schematic proof. In fact, the above theorem, also provable, %% does not have a schematic proof. skip.