Proof for $ \vdash Q \rightarrow (P \rightarrow Q) $ in symbolic logic. I need help about proving the following sequent:
$$ \vdash Q \rightarrow (P \rightarrow Q) $$
I tried to prove it by trying to introduce a conditional proof, but somehow can't get the suppositions correctly in the beginning. I would appreciate some help. 
Additionally, here is another similar sequent I've supposedly proved, but not completely sure: 
$$ \vdash Q \rightarrow (P \rightarrow P) $$

Explanations for the proof:
CP: Conditional Proof Introduction A: Assumption
Leftmost Column: Score-board for keeping track of suppositions.
I'd appreciate any help and suggestions. 
A: Try the following:


*

*$Q$ (Assumption)

*--- $P$ (Assumption)

*--- --- $Q$ (Restating 1)

*--- $P \rightarrow Q$ (Conditional proof discharging assumption 2)

*$Q \rightarrow (P \rightarrow Q)$ (Conditional proof discharging assumption 1)


The tricky bit here is understanding what role line 3 plays: we need to restate the first assumption to make it the conclusion of the second inner proof which starts by assuming $P$.
A: The conditional proofs are rather similar.  For the second you have.


*

*Suppose $Q$.  Suppose $P$.  Reiterate $P$.   Discharge the assumptions.


[Well, the reiteration is redundant, but shows how they are similar.]
$$\def\fitch#1#2{\quad\begin{array}{|l}#1\\\hline#2\end{array}}\begin{array}{l}\begin{array}{l}\text{The above}\\\fitch{}{\fitch{~~1.~Q}{\fitch{~~2.~P}{~~3.~P}\\~~4.~P\to P}\\~~5.~Q\to(P\to P)}\end{array}&&\begin{array}{l}\text{What you had (also valid)}\\\fitch{}{\fitch{~~1.~Q}{\fitch{~~2.~P}{~}\\~~3.~P\to P}\\~~4.~Q\to(P\to P)}\end{array}\end{array}$$
Just do the same for the first proof.


*

*Suppose $Q$.  Suppose $P$.  Reiterate $Q$.   Discharge the assumptions.



tl;dr Basically what you need is the rule of reiteration.  If something was derivable under a subset of the current assumptions, you may inference it. 
In your system, anything derived (or assumed) under assumptions (1) or (2) may be inferenced to hold under (1)(2). 
A: Here are both proofs providing results similar to the other answers but in a Fitch-style proof checker:

Line 3 using reiteration (R) repeating line 1.

Note that line 2 is both the antecedent and consequent of the conditional represented as a subproof with only one line 2-2.

Kevin Klement's JavaScript/PHP Fitch-style natural deduction proof editor and checker http://proofs.openlogicproject.org/
