# Thread: First Order Logic problem with conditionals

1. ## First Order Logic problem with conditionals

I am having trouble solving the following problem:

Prove that:
Code:
A -> (C /\ D)
follows from the premises
Code:
(A V D) -> C
(A /\ C) -> D
My first question is:
Are there any rules regarding what can you do when you have something of the form
Code:
(A V B) -> C?
Like, is it possible to decompose it in something like
Code:
(A -> C) V (B -> C)
? It seems like you can't. I've tried
Code:
-(A V B) V C
But then I don't know what to do with that, too.

Any guidance here would be appreciated

2. I'm used to the natural deduction rules, myself (elimination and introduction rules for each symbol). You want to assume $A$, and show that $C\land D$ is the result.

$(A\vee D)\to C$ assumption
$(A\land C)\to D$ assumption
$A$ assumption
$A\vee D\quad\vee$ intro
$C\quad\to$ elimination

Can you take it from here?

3. Ah. I think I did it, but it is kinda ugly

$A$
$A \land A$
$A \land (A \vee D)$
$A \land C$
(A /\C) /\ A
$D \land A$
$D \land (A \vee D)$
$D \land C$

Is it correct? How can I make this into something more pleasant and readable?

Thanks

4. Yeah, that works. You have a few redundant steps there. I would do it this way, picking up from where I left off:

1. $(A\vee D)\to C$ assumption
2. $(A\land C)\to D$ assumption
3. $A$ assumption
4. $A\vee D\quad\vee$ intro: 3
5. $C\quad\to$ elim: 1, 4
6. $A\land C\quad\land$ intro: 3, 5
7. $D\quad\to$ elim: 2, 6
8. $C\land D\quad\land$ intro: 5, 7
Therefore,
9. $A\to(C\land D)\quad\to$ intro: 3, 8.

5. Originally Posted by devouredelysium
How can I make this into something more pleasant and readable?
LaTeX Tutorial located on the LaTeX Help subforum.

Edit: Oh, I guess you can use LaTeX already. You can use \color{red} among some other color choices, to avoid having to drop out of LaTeX like that. For example:

$y=\color{red}x^2\color{black}+2x+1$

(I realise you may or may not have been referring to formatting.)

6. Originally Posted by Ackbeet
Yeah, that works. You have a few redundant steps there. I would do it this way, picking up from where I left off:

$(A\vee D)\to C$ assumption
$(A\land C)\to D$ assumption
$A$ assumption
$A\vee D\quad\vee$ intro
$C\quad\to$ elim
$A\land C\quad\land$ intro
$D\quad\to$ elim
$C\land D\quad\land$ intro
Therefore,
$A\to(C\land D)\quad\to$ intro.
I don't get how can you do step 3.
How can you Intro C without having it stated anywhere else?
Thanks

7. I've edited my last version of the proof to include line numbers, and references to line numbers. Does that answer your question?

8. Originally Posted by Ackbeet
I've edited my last version of the proof to include line numbers, and references to line numbers. Does that answer your question?
Yes! That was perfect!

Thanks