I'm try to run some simple examples on apply ... with ... tactic from Pierce's "Software Foundations".
It seems that examples from book doesn't work for me:
Theorem trans_eq: forall (X: Type) (n m o: Type),
n = m -> m = o -> n = o.
Proof.
intros X n m o eq1 eq2. rewrite -> eq1. rewrite -> eq2. reflexivity.
Qed.
Example trans_eq_example' : forall (a b c d e f : nat),
[a;b] = [c;d] ->
[c;d] = [e;f] ->
[a;b] = [e;f].
Proof.
intros a b c d e f eq1 eq2.
(* If we simply tell Coq apply trans_eq at this point,
it can tell (by matching the goal against the
conclusion of the lemma) that it should instantiate X
with [nat], n with [a,b], and o with [e,f].
However, the matching process doesn't determine an
instantiation for m: we have to supply one explicitly
by adding with (m:=[c,d]) to the invocation of
apply. *)
apply trans_eq with (m:=[c;d]). apply eq1. apply eq2. Qed.
trans_eq_example' failed with error:
trans_eq_example' < apply trans_eq with (m:=[c;d]).
Toplevel input, characters 6-30:
> apply trans_eq with (m:=[c;d]).
> ^^^^^^^^^^^^^^^^^^^^^^^^
Error: Impossible to unify "?1707 = ?1709" with "[a; b] = [e; f]".
Additional information about Coq version:
coqtop -v
The Coq Proof Assistant, version 8.4pl4 (July 2014)
compiled on Jul 27 2014 23:12:44 with OCaml 4.01.0
How can I fix this error?
The issue is not the apply but a typo in your previous code. The definition of trans_eq should be:
Theorem trans_eq: forall (X:Type) (n m o: X), n = m -> m = o -> n = o.
Note that the type of n m o should be X, not Type.