Reverse of Deduction Theorem Why is it "easy to see" that if $S \vdash (A\to B)$ then $S \cup\{A\} \vdash B$?
Let $\langle\varphi_1,\varphi_2,\ldots,\varphi_n\rangle$ be a proof of $(A\to B)$ from $S$: $\varphi_n$ is $(A\to B)$, and each $\varphi_k$ with $k&lt;n$ either is in $S$, is an axiom, or follows from earlier statements in the list by one of your inference rules. Then $\langle\varphi_1,\ldots,\varphi_n=(A\to B),A,B\rangle$ is a proof of $B$ from $S\cup\{A\}$: the presence of $A$ in the list is justified by the fact that $A\in S\cup\{A\}$, and $B$ follows from $(A\to B)$ and $A$ by modus ponens, which is surely one of your rules of inference.
It is not so easy to see!

The only axiom we have is the original single axiom DDpDqrDDtDttDDsqDDpsDps, and our only rule of inference other than uniform substitution is 

{D$\alpha$D$\beta$$\gamma$, $\alpha$} $\vdash$ $\gamma$.

This "if S⊢(A→B), then S∪{A}⊢B" involves $\rightarrow$.  I'll rewrite that as:
"if S$\vdash$Cab, then S ∪ {a} $\vdash$b."  Thus, using the definition 

Cpq := DpDqq, we want to show that:

"if S$\vdash$DaDbb, then S ∪ {a} $\vdash$b."

Suppose that we've proved DaDbb.  This means that in the rule above we can substitute $\alpha$ with "a", $\beta$ with "b", and $\gamma$ with "b" yielding:

{DaDbb, a} $\vdash$ b.  Now from that since we have $\vdash$DaDbb, and "a" consists of a hypothesis, $\vdash$b.