Predicate Logic Universal and Existential quantifiers

  • Thread starter Mishada17
  • Start date
  • #1
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Hi, I'm taking an intro logic class and though I'm comfortable with most propositional logic, predicate logic is confusing me. I joined the forum to ask this particular question that I've been stuck on for a while. Any help would be appreciated - I'm having trouble finding information on the web since my teacher uses a different format that most of what I've seen elsewhere.

Homework Statement


∃x F(x) |− ~(∀x ~F(x))


Homework Equations



Lecture information may be found on these two pages:
http://people.cis.ksu.edu/~schmidt/301s11/Lectures/5natdedS.html
http://people.cis.ksu.edu/~schmidt/301s11/Lectures/6quantS.html

The Attempt at a Solution


1. ∃x F(x) premise
-----------------
2. a ∀x ~F(x) assumption - I think this is the right assumption because I want to ~it, right?
3. ~ F(a) ∀e 2
4. ∃a ~F(a) ∃i 3
-----------------
5. ∃x ~F(x) ∃e 1 2-4

... beyond here I have no idea. I think I need to find an _|_ but so far the closest I've come is ∃x~F(x) and ∃x F(x) and those apparently don't contradict because ~e won't work on them. :'(
 

Answers and Replies

  • #2
242
0
premise one says something is F, call it a, so F(a) is true.

your assumption (the correct one) says nothing is F, so, in particular, a is not F, so ~F(a) is true.

now its just a matter of putting it in notation your teacher will like.
 
  • #3
3
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I'm sorry, I still don't understand how to apply it.
 
  • #4
242
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well, you are doing a proof by reductio ad absurdum. so you need to show that your assumption leads to a contradiction. the statement 'F(a) and ~F(a)' is a contradiction, so if that follows directly from your assumption (and premises), and you believe your premises to be true, then the only conclusion is that your assumption is false, which is what you want to show.

hope this helps
 
  • #5
3
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It did, thank you. You were ten times more helpful than my teacher, who just said, "look at example 3..." Well, I figured it out and example 3 was no help, sir!

Here is my completed proof:
1. ∃x F(x) premise
----------------------
2. a F(a) assumption
---------------------------
3. ∀x ~F(x) assumption
4. ~ F(a) ∀e 3
5. _|_ ~e 2,4
---------------------------
6. ~(∀x ~F(x)) ~i 3-5
----------------------
7. ~(∀x ~F(x)) ∃e 1,2-6
 
  • #6
242
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cheers
 

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