Propositional Logic - Formal Proofs using natural deduction

Question

I have a question I have come across in an old exam paper which I am trying to work through.

It states that a formal proof must be given using the rules of natural deduction

Now generally what I do is I work backwards to see how I could derive the conclusion then I start working forward.

Two problems I am having:

1) For C → D ∨ E this does not look well formed to me as it is lacking parentheses so it seems ambiguous to me. I have chosen to rewrite this as follows (C → D) ∨ E.

2) Working backward initially I use conditional introduction in my sub-proof before my conclusion.

See below:

I have chosen to reiterate B line 7 as I want to be able to derive by elsewhere in my proof.

I am not sure how to proceed next. I thought of perhaps using contradiction elimination by finally deriving a contradiction and ultimately asserting C → D.

Some advice on how to proceed would be greatly appreciated.

Thanks

Answer 1

The usual convention for the omission of parentheses is that :

1) the negation symbol applies to as little as possible

2) $\land$ and $\lor$ apply to as little as possible, given the above convention.

Thus, $C \to D \lor E$ must be :

$C \to (D \lor E)$.

Answer 2

As @Mauro ALLEGRANZA points out, the convention is that $\vee$ and $\wedge$ bind more tightly than $\to$, so the formula is $C \to (D \vee E)$. Here's a sketch of how the deduction proceeds:

1. $\neg A \to B$
2. $C \to D \vee E$
3. $D \to \neg C$
4. $A \to \neg E$
5. $C \to \neg D \quad$ from 3.
6. $C \to E \quad$ from 2. and 5.
7. $E \to \neg A \quad$ from 4.
8. $C \to \neg A \quad$ from 6. and 7.
9. $C \to B \quad$ from 8. and 1.