Natural deduction using Modus Tollens

Question

I was thinking about how to proof this:

$(a \implies b) \vdash (\lnot a \implies c) \implies (\lnot c \implies b)$

I want to solve it using Modus Tollens. Has anyone got an hint on this?

My ideas so far:

$1: a \implies b$ premises

$2: \lnot a \implies c$ assume

$ ... $

$\lnot c \implies b$

$(\lnot a \implies c) \implies (\lnot c \implies b)$ implication introduction

Any ideas/hints?

Answer 1

\begin{array}{l|l:l} 1. & a\to b & \mathsf P1 \\ \quad 2. & \neg a \to c & \textsf{Assume}\\ \qquad 3. & \neg c & \textsf{Assume} \\ \qquad 4. & a & \textsf{Why?} \\ \qquad 5. & b & \textsf{How?} \\ \hline \quad 6. & \neg c \to b & 3,5 , \to\textsf{Introduction}\\ \hline 7. & (\neg a \to c)\to(\neg c \to b) & 2, 6 \to\textsf{Introduction} \end{array}