Is This Predicate Logic Proof Correct?

AI Thread Summary
The discussion revolves around the correctness of a predicate logic proof involving existential and universal quantifiers. The original poster presents their work on two problems, seeking validation and assistance with logical formatting in TeX. Key points include the use of conditional proof and reductio ad absurdum for hypothesizing contradictions. Participants emphasize the importance of understanding quantifier elimination and introduction rules, suggesting that the original poster review these concepts for clarity. The thread highlights the collaborative effort to refine logical proofs and improve understanding of formal logic techniques.
lazycritic
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Hey,

If anyone here likes working on these problems, I'm working on a couple more. I finished the first two. These go:

1. Vx Vy (Fxy ---> ~Fyx) |- -ExFxx
(now my work - maybe wrong
2. Vy (Fay ---> ~Fya) 1,VE
3. (Faa ---> ~Faa) 2,VE
4. | Faa H(for -->E)
5. | ~Faa 3,4, CON
6. Faa
7. ~~Faa 3,6, MT
(I find this hard to articulate in the logical format, but basically, if ~Faa is proved wrong by Faa in line six(~(~Faa)), then MT says that ~Faa can be infered...)
8. ~Faa
9. -ExFxx 8, EI(can I do this?)

Actually ended up working it out as I posted it. Is it right, though? Btw, does anyone know the tex way to do this?

Test: VxEy(Mxy -> Fxy)

Didn't work.
 
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I think I got the next problem, a theorom, but if anyone can look over it, I'd appreciate it.

4) |- ~Ex(Fx & ~Fx)

1. | ~[~Ex(Fx & ~Fx)] H(for con) - that how would I say I'm hypothesizing for a contradition?

2. | Ex(Fx & ~Fx) 1, DN
3. | Fa & ~Fa 2, E Elim.(tex would be nice here)
4. ~Ex(Fx & ~Fx) 1-3 Con?
 
Bump. Anyone mind checking these?
 
\forall \exists \neg \vee \wedge \rightarrow \leftrightarrow \vdash
Click the graphic to see the code; Follow the link for more.
lazycritic said:
1. Vx Vy (Fxy ---> ~Fyx) |- -ExFxx
(now my work - maybe wrong
2. Vy (Fay ---> ~Fya) 1,VE
3. (Faa ---> ~Faa) 2,VE
4. | Faa H(for -->E)
5. | ~Faa 3,4, CON
From 4 and 5 you can infer Faa -> ~Faa (which you already have). Conditional proof gives you a conditional- a conditional which says that your hypothesis implies whatever you can derive while assuming it.
Remember I don't know much about quantifier elimination and introduction. Maybe you could post your rules for these.
If you can get
3. Faa -> ~Faa
Then
4. ~Faa v ~Faa [3, Implication]
5. ~Faa [4, Tautology]
Of course, those are what my rules are called, you may use different names or different rules. But if you can see that (P -> Q) <=> (~P v Q) and (~P v ~P) <=> ~P, you can probably find the rules you need.

Patience, grasshopper :-p
 
lazycritic said:
how would I say I'm hypothesizing for a contradition?
That seems fine. There are two rules that use provisional hypotheses: Conditional proof and Reductio ad absurdum. You can find both rules here. I highly recommend you read those rules!
4) |- ~Ex(Fx & ~Fx)

1. | ~[~Ex(Fx & ~Fx)] H(for con) - that

2. | Ex(Fx & ~Fx) 1, DN
3. | Fa & ~Fa 2, E Elim.(tex would be nice here)
4. ~Ex(Fx & ~Fx) 1-3 Con?
Not by any rules I know. However, you can do:
1. | Fa & ~Fa [Reductio]
2. | Fa [1, Simplification]
3. | ~Fa [1, Simplification]
Now what can you infer from that?
 

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