Proving $q\Rightarrow r \models (p\land q) \Rightarrow (p \land r)$ using only natural deduction. I'm trying to prove
$$q\Rightarrow r \models (p\land q) \Rightarrow (p \land r)$$
using only the natural deduction rules in this handout.
Any hints? I am not allowed to do transformational stuff, such as converting everything to CNF or DeMorgan's, unless they are proven. (I'm sure that will make things easier!). 
 A: $(1)\; q\rightarrow r \quad\text{premise}$


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*$(2)\; p\land q\quad\text{Assumption}$


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*$(3)\; p \quad\text{Conjunction Elimination}, (2)$

*$(4)\; q\quad \text{Conjunction Elimination}, (2)$

*$(5)\; r \quad \text{Modus Ponens}, (1), (4)$

*$(6)\; p\land r\quad \text{Conjunction Introduction}, (3),(5)$
$(7) \;(p \land q) \rightarrow (p\land r) \quad\text{Conditional Introduction}, (2 - 6)$
Remark:
Note that when the proposition you are asked to prove is an implication, that's almost always a sign-post that, after listing the premise(s), you'll want to begin with a sub-proof, led off by the assumption of the antecedent, with the aim of arriving at the consequent.
A: The main question here has already been answered, but in the comments you asked for natural deduction proofs of DeMorgan's laws.  I've used the rules cited in the question with two exceptions.  I'm using Fitch-style disjunction elimination ($\lor E$) and biconditional introduction ($\leftrightarrow I$) rules.
The $\leftrightarrow I$ is a trivial difference;  I cite the subproofs that start with $\phi$ and $\psi$ and end with $\psi$ and $\phi$, respectively, instead of deriving $\phi \to \psi$ and $\psi\to\phi$ individually and citing them. 
The $\lor E$ is more significant.  Disjunctive syllogism ($\lnot P, P \lor Q / Q$) is perfectly valid rule, but it's not typically taken as the elimination rule for disjunction in natural deduction systems.  In the tradition of Gentzen, many natural deduction systems have an introduction and an elimination rule for each connective.  For disjunction, the introduction rule is actually pair of rules which state that from either disjunct you may infer the disjunction:
$$ 
\begin{array}{c} \phi \\ \hline \phi \lor \psi \end{array}\lor I_L \qquad 
\begin{array}{c} \psi \\ \hline \phi \lor \psi \end{array}\lor I_R \qquad 
$$
Disjunction elimination is a bit more complicated.  It says that if $\rho$ is derivable from both $\phi$ and from $\psi$ and $\phi\lor\psi$ holds, then so does $\rho$ (and the assumptions are discharged). It captures proof by case reasoning. 
$$
\begin{array}{ccc}
 & [\phi] & [\psi] \\
 & \vdots & \vdots \\
\phi\lor\psi & \rho & \rho \\
\hline
& \rho
\end{array}\lor E
$$
Here is $(\lnot p \land \lnot q) \leftrightarrow \lnot(p \lor q)$:


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*$\lnot(p \lor q)$ Assume.


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*$p$ Assume.



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*$p \lor q$ by $\lor I$ from 2.



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*$\mathbf{false}$ by $\lnot E$ from 1 and 3.



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*$\lnot p$ by $\lnot I$ from 2–4.


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*$q$ Assume.



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*$p \lor q$ by $\lor I$ from 6.



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*$\mathbf{false}$ by $\lnot E$ from 1 and 7.



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*$\lnot q$ by $\lnot I$ by 6–8.


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*$\lnot p \land \lnot q$ by $\land I$ from 5 and 9.


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*$\lnot p \land \lnot q$ Assume.


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*$p \lor q$ Assume.



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*$p$ Assume.




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*$\lnot p$ by $\land E$ from 11.




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*$\mathbf{false}$ by $\lnot E$ from 13 and 14.




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*$q$ Assume.




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*$\lnot q$ by $\land E$ from 11.




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*$\mathbf{false}$ by $\lnot E$ from 16 and 17.




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*$\mathbf{false}$ by $\lor E$ from 12, 13–15, and 16–18.



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*$\lnot(p \lor q)$ by $\lnot I$ from 12–19.


*$\lnot(p \lor q) \leftrightarrow (\lnot p \land \lnot q)$ by $\leftrightarrow I$ from 1–10 and 11–20. 

A: Assume $p \land q$.
By simplification $\land \mathcal E_1$: $p$.
By simplification $\land \mathcal E_2$: $q$.
By the premise, $q \implies r$.
By modus ponens $\implies \mathcal E$: $r$.
By conjunction $\land \mathcal I$: $p \land r$.
By implication $\implies \mathcal I$: $p \land q \implies p \land r$.
