Deduction Theorem + Modus Ponens + What = Implicational Propositional Calculus? Implicational propositional calculus is a system of propositional calculus in which implication is the only logical connective, and all other connectives are defined with respect implication and a single false statement.  Consider the system of implicational propositional calculus with the following two rules of inference: the Deduction Theorem, which states that if by assuming P you can conclude Q then P implies Q, and Modus Ponens, which states that if P and P implies Q then Q.   
I'm guessing that this is not a complete system for the implicational propositional calculus, so my question is, what else do we need to add to make it complete?  Is Peirce's Law all we need?
Any help would be greatly appreciated.
Thank You in Advance.
 A: In order to verifiy if Peirce's law is sufficient, when added to Deduction Theorem and modus ponens, we can try to verify if the (complete) axiom system for propositional logic of Elliott Mendelson, Introduction to Mathematical Logic (4th ed - 1997) [page 35] can be derived under these assumptions.
(A1)  $\mathcal{B} \rightarrow ( \mathcal{C} \rightarrow \mathcal{B})$
We have that :
$B$ 1 --- assumption 
$C$ 2 --- assumption 
$B$ 3 --- assumption 
$C \rightarrow B$ 4 --- deduction theorem from 2 and 3
$B \rightarrow (C \rightarrow B)$ 5 --- deduction theorem from 1 and 4.
(A2)  $(\mathcal{B} \rightarrow ( \mathcal{C} \rightarrow \mathcal{D})) \rightarrow ((\mathcal{B} \rightarrow \mathcal{C}) \rightarrow (\mathcal{B} \rightarrow \mathcal{D}))$
We have that :
$B \rightarrow (C \rightarrow D)$ 1 --- assumption 
$B \rightarrow C$ 2 --- assumption 
$B$ 3 --- assumption 
$C$ 4 --- modus ponens from 2 and 3
$C \rightarrow D$ 5 --- modus ponens from 1 and 3
$D$ 6 --- modus ponens from 4 and 5
$B \rightarrow D$ 7 --- deduction theorem from 3 and 6
$(B \rightarrow C) \rightarrow (B \rightarrow D)$ 8 --- deduction theorem from 2 and 7
$(B \rightarrow (C \rightarrow D)) \rightarrow ((B \rightarrow C) \rightarrow (B \rightarrow D))$ 9 --- deduction theorem from 1 and 8.
We still have to derive 
(A3)  $(\lnot \mathcal{C} \rightarrow \lnot \mathcal{B}) \rightarrow ((\lnot \mathcal{C} \rightarrow \mathcal{B}) \rightarrow \mathcal{C})$.
But, according to Implicational propositional calculus 

the axiom system formed by (A1), (A2) and Peirce's law with the rule of inference modus ponens is semantically complete with respect to the usual two-valued semantics of classical propositional logic. 

Note. Peirce's law is necessary, due to the fact that $\nvdash_{(A1)(A2)}(Peirce)$. The addition of Peirce’s law is sufficient, due to some results of Tarski and Bernays.
In order to have "full" propositional calculus, we have to add the falsum symbol ($\bot$) and an additional axiom : Ex Falso Quodlibet ($\bot \rightarrow \mathcal{A}$).
A: Not sure what exactly you mean , which book are you studying?
Positive implicational logic can mean many different things:
Classcal implicational logic (classical logic where implication is the only connective)
axioms (one of many, many possibilities)


*

*$ P \to ( Q \to P) $ 

*$ (P \to ( Q \to R)) \to (( P \to Q) \to ( P \to R )) $

*$ ((P \to Q) \to P) \to  P $


The above together with axioms for other connectives:
And:


*

*$ P \to ( Q \to (P \land Q )) $ (and introduction)

*$ (P \land Q ) \to P  $ (and elimination)

*$ (P \land Q ) \to Q  $ (and elimination)


Or: 


*

*$ P \to (P \lor Q ) $ (or introduction)

*$ P \to (Q \lor P ) $ (or introduction)

*$ (P \to R) \to (( Q \to R) \to ((P \lor Q ) \to R)) $ (or elimination)


Or maybe something else alltogether, different authors have so there own names for the different systems.
A: You don't need to add more axioms actually here.  Instead of adding Peirce's law CCCpqpp or some other formula as an axiom, you could just add the rule
CC$\alpha$$\beta$$\alpha$ $\vdash$ $\alpha$.
A: As Mauro's answer indicates you could join 


*

*(⊥→A) [or C0a]


and Peirce's law


*(((A → B) → A) → A) [CCCabaa]


However, as Wajsberg's second paper on Metalogic of 1939 indicates (which you can find in the volume called "Polish Logic") you could just add the following formula as an axiom:
(((A → ⊥) →  ⊥) → A) [CCCp00p] 
