Natural deduction (Logic) proof help I'm very new to natural deduction and have been stuck trying to prove this argument all day:
$A\to ¬B,$ 
$¬B\to ¬C,$ 
Therefore, $C\to ¬A$.
I've been told I need to use modus tollens on the first premise, but I am not sure what to use it on/what to assume, as I thought using modus tollens on premise 1 would yield $B$, which I don't need, instead of $\lnot A.$ 
Thank you.
 A: Assuming the availability of Modus Tollens rule, we have :
1) $A \to \lnot B$ --- premise
2) $\lnot B \to \lnot C$ --- premise
3) $C$ --- assumed [a]
4) $\lnot C$ --- assumed [b]
5) $\lnot \lnot C$ --- from 3) and 4) by $\lnot$elim followed by $\lnot$-intro (also called Double Negation introduction), discharging [b]
6) $\lnot \lnot B$ --- from 5) and 2) by MT
7) $\lnot A$ --- from 6) and 1) by MT

8) $C \to \lnot A$ --- from 3) and 7) by $\to$-intro, discharging [a].


Note : without MT the proof is quite simple :
1) $A \to \lnot B$ --- premise
2) $\lnot B \to \lnot C$ --- premise
3) $C$ --- assumed [a]
4) $A$ --- assumed [b]
5) $\lnot B$ --- from 4) and 1) by $\to$-elim
6) $\lnot C$ --- from 5) and 2) by $\to$-elim
7) $\lnot A$ --- from 4) with 3) and 6) by $\lnot$-elim followed by $\lnot$-intro, discharging [b]

8) $C \to \lnot A$ --- from 3) and 7) by $\to$-intro, discharging [a].

A: If you have Hypothetical Syllogism in your tool box, the proof becomes far more direct.

Hypothetical Syllogism (HS) is a valid argument form in which two premises are conditionals, and the rule is that, whenever instances of "$\displaystyle P\to Q$", and "$\displaystyle Q \to  R$", appear on lines of a proof, "$\displaystyle P\to R$" can be placed on a subsequent line. 

In your argument, we then have
$(1) \;\;A\to \lnot B,\qquad$ (premise)
$(2) \;\;\lnot B \to \lnot C,\,\quad$ (premise)
$(3) \;\;A\to \lnot C,\qquad$ HS, $(1), (2)$
$(4) \;\;C \to \lnot A,\qquad$ contraposive of $(3).$

Note that instead of relying on the equivalence of an implication $p\to q$ with its contrapositive, $\lnot q \to \lnot p$, we can proceed from $(3)$ to 
$(4) \;\;\quad C \quad$ assumption
$(5) \;\;\quad \lnot\lnot C\quad$ double negation on $(4)$
$(6) \;\;\quad  \lnot A\quad$ ($3$, $5$, modus tollens)
$(7) \;\; C\to \lnot A \quad$ ($4-6$)
