Logic Proof

Philosophy Forums Philosophy
1

For my logic final we were given the following argument in which we needed to show the proof for using only the rules of inference and replacement. I’ve worked on this proof for days and I can’t get it and it is driving me nuts. I was wondering if anyone could figure it out for me.

Here’s the argument that we were given:

  1. XZ
  2. YZ // (XY)Z.

And as a reminder, here are the rules of inference:

Modus Ponens (MP)
pq
p____
q

Modus Tollens (MT)
pq
~q____
~p

Pure Hypothetical Syllogism (HS)
pq
qr__
pr

Disjunctive Syllogism (DS)
pq
~p___
q

Constructive Dilemma (CD)
(pq)*(rs)
pr___
qs

Conjunction (Conj
p
q__
p*q

Simplification (Simp)
p*q
p

Addition (Add)
p___
pq

And here are the rules of replacement:

De Morgan’s Rule (DM
~(pq) :: (~q~q)
~(pq) :: (~p~q)

Commutativity (Com)
(pq) :: (qp)
(pq) :: (qq)

Associativity (Assoc)
[(pq)r] :: [p(qr)]
[p*(qr)] :: [(pq)*r]

Distribution (Dist)
[p*(qr)] :: [(pq)(pr)]
[p(qr)] :: [(pq)(pr)]

Double Negation (DN)
p :: ~~p

Transposition (Trans)
(pq) :: (~q~p)

Material Implication (Impl)
(pq) :: (~p~q)

Material Equivalence (Equiv)
(pq) :: [(pq)(qp)]
(pq) :: [(pq)(~p*~q)]

Exportation (Exp)
[(p*q)r] :: [p(qr)]

Tautology (Taut)
p :: (pp)
p :: (p*p)

Of course I found a link to this as I was near done writing out all of the stoopid code that would’ve saved me a whole lot of time #-o
http://www.mathpath.org/proof/proof.inference.htm

2

X–>Z
Y–>Z

(XvY)–>Z

Start a subderivation
Assume (XvY)
Then use disjunction elimination
When you assume X you get Z and when you assume Y you get Z so disjunction elimination allows you to derive Z
Then end your subderivation
and you have (XvY)–>Z through conditional introduction

3

I think I just figured it out, but I’d like to run it by ya’ll see to see if I am in fact correct.

  1. XZ
  2. YZ //(XY)Z
  3. ~XZ 1 Impl
  4. ~YZ 2 Impl
  5. Z~X 3 Com
  6. Z~Y 4 Com
  7. (Z~X)*(Z~Y) 5,6 Conj
  8. Z(~X*~Y) 7 Dist
  9. Z~(XY) 8 DM
  10. ~(XY)Z 9 Com
  11. (XY)Z 10 Impl
4

1X–>Z
2Y–>Z

3|XvY…A
4||X…A
5||Z…1,4–>E
6||Y…A
7||Z…2,6–>E
8|Z…3-7vE
9(XvY)–>Z…3-8–>I

Your way works fine, but is overly complicated.

5

I’m just glad I finally figured it out. I’ve been working on it for a week just because I wanted to figure out the damn proof. I figured out earlier today that I had copied one of the rules of replacement incorrectly. Funny that it took me so long to get it right. #-o

6

Cheers for that.

7
  1. X->Z
  2. Y->Z
  1. X v Y A
  2. Z v Z 1,2,3 CD
  3. Z 4 T
  1. (X v Y) → Z 3-5 CP
8

The problem with step 3 however is that because you don’t have X alone you aren’t able to use Addition on it to get X v Y. The rules of inference can only be used on the main operator.

EDIT
Of course, I just realized I may be incorrect in assuming that ‘A’ is for addition. Even if it’s for Associativity it would be an incorrect use…

9

In Nothingness’ post, 3 is just an auxilary premise which is discharged in 6, if I’m not mistaken.

10

It’s a provisional assumption.

I should have used PA.

11

Ahhh. For that proof (in regards to my logic test) we were only able to use the rules of inference and replacement that I had listed.