TITLE: If F→G is a consequence of F, then so is ¬G→¬F. A direct proof?
QUESTION [5 upvotes]: Homework question (introduction to logic):
"If $F \to G$ is a consequence of $\mathcal F$, then so is $\lnot G \to \lnot F$. We refer to this rule as $\to$-contrapositive. Verify this rule by giving formal proof."

My attempt at proving it (by contradiction):
$H: \lnot G \to \lnot F$
$(1)\quad\mathcal F\vdash F \to G \quad$ {assumption}
$(2)\quad\mathcal F \cup \{ \lnot H \} \vdash \lnot ( \lnot G \to \lnot F) \quad$ {assumption for contradiction}
$(3)\quad\mathcal F \cup \{ \lnot H \} \vdash \lnot(\lnot F \lor G)\quad$ {from $\vdash \lnot (\phi \to \psi) \equiv \psi \lor \lnot \phi$, and line: 2}
$(4)\quad\mathcal F \cup \{ \lnot H \} \vdash F \land \lnot G\quad \quad${from deMorgan's Law and line: 3}
$(5)\quad\mathcal F \cup \{ \lnot H \} \vdash F \quad \quad$ {$\land$ - elimination, line: 4}
$(6)\quad\mathcal F \cup \{ \lnot H \} \vdash \lnot G \quad \quad$ {$\land$ - elimination, line: 4}
$(7)\quad\mathcal F \cup \{ \lnot H \} \vdash G \quad \quad$ {$\to$ - elimination, lines: 1, 5}
$(8)\quad\mathcal F \cup \{ \lnot H \} \vdash G \land \lnot G\quad \quad$ {$\land$ - introduction, lines: 6, 7}
$\quad \quad\mathcal F \vdash H  .\square$
question 1: Can you verify if it's OK/ give hints what's wrong if it's not?question 2: Is there a direct proof?
[edited]

REPLY [2 votes]: A direct proof has the outline:
$H: \lnot G \to \lnot F$
$(1)\quad\mathcal F\vdash F \to G \quad$ {assumption}
$(2)\quad\mathcal F \vdash \lnot G$ {assumption for direct proof}
...
$(n)\quad\mathcal F \vdash \lnot F$ {reasons...}
The steps are very simlar to your proof by contradiction. I leave the details to be filled in by the reader.