1. Mar 15, 2010

### nietzsche

Hi PF.

This is a homework question, but it's for symbolic logic. Figured that it didn't really make sense to post it with the math and science questions, although I'm sure people on there would know how to solve it.

I'm having trouble constructing a derivation for the following:

For all x, (Bx -> Ex).
There exists some y such that (Ay -> ~Ey).
Therefore, there exists some z such that (Az ^ ~Bz).

(Don't know how to do the symbols...)

I changed the second premise into (~Aa v ~Ea), but I don't know what to do with that. I can see that if I have Aa to begin with, then Aa -> ~Ea -> ~Bx.

But what if I have ~Ea to begin with? That's what I'm having trouble with.

Or maybe I'm looking at it the wrong way?

Hope someone on here can help me. Thanks in advance.

2. Mar 16, 2010

### Staff: Mentor

I don't think the Philosophy forum is right for this either, so I moved it to Calculus & Beyond. Hopefully somebody here can help. If you can post more info, that will help us all.

3. Mar 17, 2010

### Redbelly98

Staff Emeritus
Note to all:
C -> D means "if C, then D" or "C implies D"
~ means "not", i.e. logical negation
v means "or"
^ means "and"

Okay, so Aa → ~Bx. Maybe you can do something with that, but I don't see how we can get to the required (Aa ^ ~Ba) from there.

Perhaps you should make a truth table, to at least convince yourself that (Aa ^ ~Ba) is true and not an error in the book.

4. Mar 18, 2010

### kote

You can click the far right button on the format bar when you are typing in a message to get the latex reference. $$\exists$$ (\exists), $$\forall$$ (\forall), etc, are all in there under logic.

If you have ~Ea to begin with, then from your first premise you can get to ~Ba, which you are looking for in your conclusion. Does that help any? I'm not sure what else you were thinking here. I'm sure the assignment was due by now anyways though :tongue:.

BTW I just found this on the philosophy forum. Symbolic logic is required for majors .

5. Mar 18, 2010

### JSuarez

The argument, as it is, is not correct: from two conditionals, you cannot infer Aa (so you cannot infer its conjunction with anything else).

6. Mar 21, 2010

### Redbelly98

Staff Emeritus
Moderator's note:

We are here to provide help so that the OP, nietzsche, can solve the problem. We are not here to provide the actual solution.

Lacking further input from the OP, I urge people to refrain from providing more help at this point.