The question is like this
The original Question:
By defining appropriate propositions, show that the following inference is valid. You will lose points if you do not give reasons for every statement you infer.
• If the prime interest rate goes up, then it is sufficient that unemployment goes down for prices to rise.
• However, unemployment goes down only if the prime interest rate goes up.
• Therefore, if prices do not go up, then unemployment does not go down.
-if P then Q for R (for example)
If the prime interest rate goes up, then it is sufficient that unemployment goes down for prices to rise
-However, Q if P
-Therefore, if $\neg R$ then $\neg Q$
So far, I have deduced
first statement $$p\implies(q\implies r)$$
second statement
$$p\implies q$$
and so it modus ponens can be used to deduce
$$q\implies r$$ $$q$$ $$\therefore r$$
which proves that
$$r\implies q$$
inturn proving the hypothesis
$$\neg r\implies \neg q$$
is this correct or have I gone horribly wrong at some point?
You don't have the symbolization correct. Yes, the statement
Does indeed translate to $P \rightarrow (Q \rightarrow R)$, assuming you are using:
$P$: Prime interest rate goes up
$Q$: Unemployment goes down
$R$: Prices rise
However, with this, the statement
translates to $Q \rightarrow P$, rather than $P \rightarrow Q$
You seem to be unfamiliar with the difference between '$P$ if $Q$' and '$P$ only if $Q$', so here is example.
We know that someone can be a bachelor only if they are unmarried. However, not everyone who is unmarried is a bachelor. Being unmarried is therefore not a sufficient condition for being a bachelor, but it is one of several necessary conditions for being a bachelor, for one also has to be male, and of adult age. So, 'bachelor only if unmarried' does not translate into $Unmarried \rightarrow Bachelor$. However, it does translate into $Bachelor \rightarrow Unmarried$, exactly because being unmarried is a necessary condition for being a bachelor!
In general, '$P$ only if $Q$' translates to $P \rightarrow Q$, which is the converse of '$P$ if $Q$'
With the corrected symbolization, you should be able to prove the desired result.