Solutions to proofs in predicate logic

1. Oranges are sweet. Oranges are citrus fruits. So oranges are sweet citrus fruits.

 

1. (x) (Ox > Sx)

2. (x) (Ox > (Cx . Fx))                     / (x) [(Ox > (Sx . (Cx . Fx))]

 

 

2. Since tomatoes are vegetables, the tomatoes in the garden are vegetables.

1. (x) (Tx >Vx)                     / (x) ((Tx . Gx) > Vx)

 

 

3. Apples and pears grow on trees, so pears grow on trees.

1. (x) ((Ax v Px) > Gx)                  / (x)(Px > G)

 

 

4. Socrates is a man, and all men are mortal, so Socrates is mortal.

1. Ms

2. (x)(Mx > Ox)                       / Os

 

 

5. Carrots are vegetables and peaches are fruit. There are carrots and peaches in this garden. So there are vegetables and fruit in this garden.

1. (x)(Cx > Vx) . (x)(Px > Fx)

2. (∃x)(Cx . Gx) . (∃x)(Px . Gx)                     / (∃x)(Vx. Gx) . (∃x)(Fx . Gx)

 

 

6. Since there are no legumes in the garden, and beans and peas are legumes, there are no beans or peas in the garden.

1.~(∃x)(Lx . Gx) or (x)(Gx> ~Lx) (these are equivalent, as section 10.3 shows)

2. (x)((Bx v Px) > Lx)                   / (x) (Bx v Px) > ~Gx))

 

 

7. Only philosophers are logicians, and anyone who’s a philosopher is a thinker. Frege and Russell are logicians, so they’re also thinkers.

1. (x)(Lx > Px)

2. (x)(Px > Tx)

3. Lf . Lr                       / Tf . Tr

 

 

8. If Wittgenstein invented truth tables, then there was no truth-table logic in the 19th century. If Quine refuted the dogma of the analytic/ synthetic distinction, then not all of Leibniz’s ideas are correct. Wittgenstein did invent them, and Quine has refuted the distinction, so not all of Leibniz’s idea are correct.

1. Iw >(x) (Lx > ~Tx)

2. Rq > ~(x) (Dx > Cx)

3. Iw . Rq                    / ~(x)(Dx > Cx)

 

 

9. Some philosophers are idealists. Some scientists are materialists. If anyone is a philosopher, the scientists are skeptics if they are materialists. Therefore some scientists are skeptics.

1. (∃x)(Px . Ix)

2. (∃x)(Sx . Mx)

3. (∃x)Px > (x)(Sx > (Mx > Kx))                  / (∃x)(Sx . Kx)

 

10. If the artichokes in the kitchen are ripe, the guests will be surprised. If they are flavorful, the guests will be pleased. They are ripe and they are also flavorful, so the guests will be surprised as well as pleased.

1. (x)[(Ax . Kx) > Rx] > (x)(Gx > Sx)

2. (x)[(Ax . Kx) > Fx)] > (x)(Gx >Px)

3. (x)[(Ax . Kx) > (Rx . Fx)]                                     / (x)(Gx > (Sx . Px))

4. (Ax . Kx) > (Rx . Fx)                          UI 3

5. ~(Ax . Kx) v (Rx . Fx)                     IMP 4

6. [~(Ax . Kx) v Rx] . [~(Ax . Kx) v Fx]                  DIST 5 treating ~(Ax . Kx) as the “p”

7. ~(Ax . Kx) v Rx                                 SM 6

8. ~(Ax . Kx) v Fx                                    CM, SM 6

9. (Ax . Kx) > Rx                                       IMP 7

10. (Ax . Kx) > Fx                                       IMP 8

11.(x) ((Ax . Kx) > Rx)                                 UG9

12. (x) (Gx > Sx)                                       MP 1, 11

13. (x) ((Ax . Kx) > Fx)                                UG 10

14. (x) (Gx > Px)                                   13, 2 MP

15. Gx > Sx                                                UI 12

16. Gx > Px                                               UI 14

–17. Gx                                                      ACP

–18. Sx                                                   MP 17, 15

–19. Px                                                   MP 17, 16

–20. Sx . Px                                               CN 18, 19

21. Gx > (Sx . Px)                                       CP 17-20

22. (x) [Gx > (Sx . Px)]                                   UG 21

 

Just as IMP, DIST, IMP in 5, 6 and 8/ 9 generated conditional statements that were needed for MP, that idea could have been used in place of the CP after line 16 (IMP, CN, DIST, IMP).

And just as CP was used at 17-20, it could have been used (but it would be needed twice, very short both times) after line 4, to generate (Ax . Kx) > Rx (the first one), and then (Ax. Kx) > Fx (the second one).

Do not even imagine that I would put this proof on the exam.

One Response to Solutions to proofs in predicate logic

  1. Would love to perpetually get updated great blog ! .

Leave a Reply

Your email address will not be published. Required fields are marked *