On Thursday, 12 October 2017 20:43:01 UTC+2, Dimitris Tsementzis wrote:
>
>
> But there are two distinct TT-model homomorphisms from C_TT to C_TT*, one 
> which sends p(t0) to pq(t0) and one which sends p(t0) to qp(t0) (where 
> p(t0) is regarded as an element of Tm_{C_TT} (empty, B(B(T0))), i.e. of the 
> set of terms of B(B(T0)) in the empty context as they are interpreted in 
> the term model C_TT). 
>

There seems to be a gap in the proof here.  In a term model we quotient out 
by judgemental equality (correct me if this is wrong), so this step does 
not give a contradiction: rather, we conclude |- qp(t0) = pq(t0) : B(B(T0))

I would expect this equality to hold in the examples you have in mind.  If 
T = Type_0, B(T) = Type_1, and p(t) = t -> t, then the equality is q(t -> 
t) = q(t) -> q(t).  This holds if TT* includes the rule 

G |- A : Type_0           G |- B : Type_0
-------------------------------------------------------
G |- q(A -> B) = q(A) -> q(B) : Type_1

which I would expect you need in order for the category of TT-models to be 
isomorphic to the category of TT*-models.  So I suggest you also check the 
proof of this isomorphism in more detail.

--
Robin