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!). 

$(1)\; q\rightarrow r \quad\text{premise}$


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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.
