How can I prove this?(Simple Logic Problems)?

Question

These are the two problems I cannot answer:

1.) All citizens who are not traitors are present. All officials are citizens. Some officials are not present. Therefore there are traitors.
Let C(x) = All citizens.
T(x) = All traitors.
P(x) = All who are present.
O(x) = All officials.

2.) A book is interesting only if it is well written. A book is well written only if it is interesting. Therefore, any book is both interesting and well written if it is either interesting or well written.
Let B(x) = All books.
I (x) = All things interesting
W (x) = All things well written.

The instruction is to construct a formal proof of validity.

I've tried all ways and all identities, but I just can't come up with the conclusion!

I'll give the best answer to the person who can answer just either of the two. (of course, both would be better!) ^_^

Thanks.. :)

Answer 1

Here's a proof of the first one... I'm assuming you want a deductive proof as opposed to one using truth trees. I'm also assuming this is for a formal logic class; if it's for a math class much of what I'm saying might not make sense to you.

I'm not sure about the operators you use so here are the ones I'm using:

Conditional (if-then): >

Negation (not): ~

Conjunction (and): &

Disjunction (or): v

Universal Quantifier (for all x): Vx

Existential Quantifier (there is an x): 3x

_E means I am applying the rule for elimination of an operator

_I means I am applying the rule for the introduction of an operator

I'm also not using parentheses in the predicates because I think it would look way too confusing with all the parentheses, so I'm letting:

Cx = x is a citizen, Tx = x is a traitor, etc...



Here are the proofs:

Problem 1:

P1. Vx (~Tx > Px)

P2. Vx (Ox > Cx)

P3. 3x (Ox & ~Px)

4........Oa & ~Pa.................. (Assumption; derived from line 3)

5.......~Ta > Pa.................. (derived from line 1, VE)

6....................~Ta............... (Assumption; looking for contradiction)

7.....................Pa................. (Derived from line 5-6)

8...................~Pa................ (From line 4, &E)

9........Ta.............................. (Lines 6-8, ~E by contradiction)

10.3xTx............................... (Line 9, 3E)

And Line 10 is what you wanted!

Did you transcribe the third premise correctly? If you used a conditional I could see why you had trouble proving this.



I'll come back for problem 2.

Ok here's problem 2. Part of the trick is symbolizing everything correctly, which is easy to do wrong here.



P1: Vx (Bx > (Ix > Wx))

P2: Vx (Bx > (Wx > Ix))

3: Ba > (Ia > Wa) ...................... (P1; VE)

4. Ba > (Wa > Ia) ...................... (P2; VE)

5...........Ba.................. ............... (Assumption for >I)

6...........Ia > Wa......... ................ (3, 5, >E)

7...........Wa > Ia........... .............. (4, 6, >E)

8...................(Ia v Wa) ............... (Assumption for >I)

9............ ............Ia.. ...................... (from 8; for vE)

10........ .............Wa. ..................... (6, 9, >E)

11............. ........Wa............................. (from 8; for vE)

12......... .............Ia................................ (7, 11, >E)

13......... .............Wa & Ia.... ..............(10, 12, &I)

14..... ...(Ia v Wa) > (Wa & Ia). ........ (8-13, >I)

15.Ba > ((Ia v Wa) > (Wa & Ia)) ......(5-14, >I)



And that monstrosity in lines 15 is what you were looking to prove. QED. If you have questions send me an e-mail.

Answer 2

1.)

O(x) is a part of C(x)

All elements of C(x) that are not also part of P(x) are part of T(x).

Thus, all elements of O(x) that are not part of P(x) are part of T(x).

This could only be false if any element of O(x) is not part of C(x).



2.)

An element of B(x) is part of I(x) only if it is part of W(x)

An element of B(x) is part of W(x) only if it is part of I(x)

Therefor, any element of B(x) is either part of both I(x) and W(x) or part of neither.

The conclusion is indeed if an element of B(x) is found a part of I(x) it must also be a part of W(x) because of the first rule. And if an element of B(x) is found part of W(x) it must also be part of I(x) because of the second rule.

Answer 3

#2) There's a set B(x) containing all the books...

I(x) = W(x) and these are the subsets of B(x)