Joint Embedding Property I want to show that any complete theory has JEP,
And that JEP does not imply comleteness.
I have trouble showing it, and I think I'm missing sometiong here.  
And another question:
If $T$ is model complete - why compeleteness is equivalent to JEP?
 A: Let T be complete, $A,B \models T$. Note that JEP is equivalent to satisfiability of
$$\Gamma := T \cup Diag(A) \cup Diag(B)$$
Let $\Delta$ be a finite subset of $Diag(B)$, $\phi$ the conjunction of formulas from $\Delta$, $c_1 \dots c_n$ the new constants occuring in $\phi$. Replace these with unused variables $v_1 \dots v_n$ to get $\phi'$. By completeness:
$$B\models \exists \overline{v}\phi' \Rightarrow A\models \exists \overline{v}\phi'$$ 
So by interpreting $c_1 \dots c_n$ by suitable witnesses $A\models \Delta$. By compactness $\Gamma$ is satisfiable.
Let T be model complete and $ A,B \models T$, let $C \models T$ s.t. A and B embed into C.
By modelcompleteness these embeddings are elementary, so for a sentence $\phi$
$$A \models \phi \Leftrightarrow C \models \phi \Leftrightarrow B \models \phi$$
Hence T is complete.
Also the empty theory over the the language containing no non-logical symbols has JEP, since  two models both embed into their union, but the sentence $\exists x \forall y : y=x$ is not decided.
A: To see that completeness implies joint embedding property, use compactness. For the other direction, consider the empty theory (in any language).
For the other question, this is just unfolding the definition.
