First Order Logic - coming up with an example I'm learning First order logic using a university textbook and it has the following question:

Give an example of languages $L'$ and $L$, $L \subseteq L'$, as well
  as formulas $A$ and $B$ in language $L'$ such that for every set of
  formulas $T$ in $L$:
If $T \vdash A$ then $T \vdash B$
But there exists a set of formulas $T'$ in $L$ such that $T' \cup \{A\} \nvdash B$

Ok I have been staring at this question for over 2 hours with no major breakthroughs, just trying to run different examples in my head. I just can't seem to find an example that if you can infer A and B from T then you can't infer B from T and A.
Any hints or help is appreciated!
Thank you
 A: 
I just can't seem to find an example that if you can infer A and B from T then you can't infer B from T and A.

That's because there can't be such an example. If $T \vdash A \land B$, then $T \vdash B$, and therefore $T, A \vdash B$, in any standard logic. 
As to the displayed problem you reproduce(?), should the last $T$ be $T'$? What's the relation between $T$ and $T'$? If there are no constraints, just take $T'$ to be the empty set!
A: Assuming you are working in first-order logic with equality built in, take $L$ to have no non-logical symbols. Take $L'$ to be $L$ extended with two constant symbols $c$ and $d$. Finally, take $A$ to be $c = d$ and $B$ to be $\forall x\cdot\forall y \cdot x = y$. Then if $T \subseteq L$ and $T \vdash A$, then $T \vdash B$, but with $T' = \emptyset$, $T' \cup \{A\} \not\vdash B$.
If you don't want to use equality, take $L$ to have a single unary predicate symbol $p$ and no other non-logical symbols and take $L'$ to extend $L$ by adding a constant symbol $c$. Then take $A$ to be $p(c)$ and $B$ to be $\forall x \cdot p(x)$.
