Help Me Logic

blah

Ok, so I have a logic assignment, unfortunately logic is my kryptonite. I've looked thoroughly through the text book, my notes and some internet sites, all with little luck. The question relates to more complicated truth trees which are just going straight over my head. Could someone help break the process of doing these down?

Example question:

(∀x)((Bx & Dx) ⊃ Lx), ~(∀x)(Lx ⊃ Mx) ∴ (∃x)(Mx & ~Dx)

Ark

What's the question exactly? Do you have to draw the truth trees for those statements or something?

Also, is ∴ the same as the turnstile symbol?

blah

Ah sorry, yeah I need to use tree method to check the arguments for validity and then set a counterexample if they are invalid.

∴ means 'therefore' as I understand it. Which I think is the same as the turnstile symbol.

Vrtra Theory

Link:

http://www.wadsworth.com/philosophy_d/templates/student_resources/0534584829_hurley/trees/section4.htm

I'm not sure whether this is in sync with your textbook, but this is the way I remember doing it. Basically, you turn the predicate statement into a simple logical statement by assuming there is only one member in the universe. You then negate the conclusion and draw a truth tree: if there is any open path, you've found the set of variables that prove the original statement is invalid. If all paths are closed, you expand the universe to include two members and try again. You can't really prove validity in this way, you can only fail to prove invalidity.

I think the page explains it better than I can, but here's how I would do your example. (I'll skip the single member step, because it's relatively easy.)

Problem:

(∀x)((Bx & Dx) ⊃ Lx), ~(∀x)(Lx ⊃ Mx) ∴ (∃x)(Mx & ~Dx)

In the form of an argument:

(∀x)((Bx & Dx) ⊃ Lx)
~(∀x)(Lx ⊃ Mx)
(∃x)(Mx & ~Dx)

Now, assume the universe consists of a and b. (Essentially, copy every predicate for each member of the universe, using & for universal predicates and | for existential predicates.)

((Ba & Da) ⊃ La) & ((Bb & Db) ⊃ Lb)
~((La ⊃ Ma) & (Lb ⊃ Mb))
(Ma & ~Da) | (Mb & ~Db)

This is now the top of our truth tree. Convert all the "p⊃q" pairs into "~p|q" pairs:

(~(Ba & Da) | La) & (~(Bb & Db) | Lb)
~((~La | Ma) & (~Lb | Mb))
(Ma & ~Da) | (Mb & ~Db)

Simplify:

(~Ba | ~Da | La) & (~Bb | ~Db | Lb)
(La & ~Ma) | (Lb & ~Mb)
(Ma & ~Da) | (Mb & ~Db)

And negate the conclusion:

(~Ba | ~Da | La) & (~Bb | ~Db | Lb)
(La & ~Ma) | (Lb & ~Mb)
(~Ma | Da) & (~Mb | Db)

You can now treat this like a regular truth tree, evaluating it from the top down. I won't even try to draw branches, but immediately under this top section will be three branches "~Ba", "~Da", and "La". Under each of those branches will be three branches "~Bb", "~Db", and "Lb". Under all nine of those branches will be the two branches "La & ~Ma" and "Lb & ~Mb". Etc., etc., etc.

The goal is to trim down (ignore) any branches that will clearly be closed, and focus on branches that look like they will lead to an open path. If you find one, following it from bottom to top will tell you the values of "a" and "b" in your two-member universe that prove the original argument is invalid.


Hope that helps some. It's late so I don't guarantee the above is mistake-free .