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CS-645 Assignment 5 Proofs and Logic Exercises, Assignments of Computer Science

The solutions to assignment 5 of a logic and proofs course, focusing on proving various formulas using predicate logic. Topics include quantifiers, implications, and connectives. The document also includes exercises on validity and invalidity of formulas with free occurrences of x in c.

Typology: Assignments

2009/2010

Uploaded on 02/24/2010

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CS-645
Assignment 5
Jeff Rose
1. Assume there is no free occurrence of y in A(x), no free occurrence of x in B(y), and
no free occurrence of x in e. Prove each of the following formulas:
(a) y (x (A(x) B(y)) (x A(x) B(y)))
1. y (x (A(x) B(y)) P
2. x A(x)P
3. A(C) 2, EI
I am lost here because I don't know what to do when one of the quantifiers surrounds the
implication to be proved.
(b) x ((x = e) A(x)) ≡ A(e)
1. x ((x = e) A(x)) P
2. (D = e) A(D)1, EI
3. D = e2, Simp.
4. A(D) 2, Simp.
5. A(e) 3, 4, EE
6. x ((x = e) A(x)) A(e)1, 5, CP
7. A(e) P
8. ¬∃x ((x = e) A(x)) P for IP
9. x ¬((x = e) A(x)) 8, (7.4), E
10. ¬((e = e) A(e)) 9, UI
11. ¬(e = e) ¬A(e)10, E
12. ¬A(e)11, E
13. 7, 12, Conj.
14. x ((x = e) A(x)) 8, 13, IP
15. A(e) x ((x = e) A(x)) 7, 14, CP
16. (x ((x = e) A(x)) A(e)) (A(e) x ((x = e) A(x))) 6, 15, Conj.
QED 16, E
pf3

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Assignment 5 Jeff Rose

  1. Assume there is no free occurrence of y in A(x), no free occurrence of x in B(y), and no free occurrence of x in e. Prove each of the following formulas: (a) ∀ y (∀ x ( A ( x ) → B ( y )) → (∃ x A ( x ) → B ( y )))
  2. (^) ∀ y (∀ x ( A ( x ) → B ( y )) P
  3. x A ( x ) P
  4. A ( C ) 2, EI I am lost here because I don't know what to do when one of the quantifiers surrounds the implication to be proved. (b) ∃ x (( x = e ) ∧ A ( x )) ≡ A ( e )
  5. (^) ∃ x (( x = e ) ∧ A ( x )) P
  6. (^) ( D = e ) ∧ A ( D ) 1, EI
  7. D = e 2, Simp.
  8. A ( D ) 2, Simp.
  9. A ( e ) 3, 4, EE
  10. (^) ∃ x (( x = e ) ∧ A ( x )) → A ( e ) 1, 5, CP
  11. A ( e ) P
  12. (^) ¬∃ x (( x = e ) ∧ A ( x )) P for IP
  13. (^) ∀ x ¬(( x = e ) ∧ A ( x )) 8, (7.4), E
  14. ¬(( e = e ) ∧ A ( e )) 9, UI
  15. (^) ¬( e = e ) ∨ ¬ A ( e ) 10, E
  16. (^) ¬ A ( e ) 11, E
  17. (^) ⊥ 7, 12, Conj.
  18. x (( x = e ) ∧ A ( x )) 8, 13, IP
  19. (^) A ( e ) → ∃ x (( x = e ) ∧ A ( x )) 7, 14, CP
  20. (^) (∃ x (( x = e ) ∧ A ( x )) → A ( e )) ∧ ( A ( e ) → ∃ x (( x = e ) ∧ A ( x ))) 6, 15, Conj. QED 16, E
  1. In this exercise, predicates with numbers above 7.8 cannot be used without proof. For each formula, first show that the formula is invalid and then show that the formula is valid if there are no free occurrences of x in C. Be sure to indicate in your proofs which steps rely on the fact that there are no free occurrences of x in C. I have looked at these over the last two days and still have no idea under which circumstance(s) they are invalid with free occurrences of x in C. (a) ∀ x CC
  2. (^) ∀ x C P
  3. C 1, UI
  4. (^) ∀ x CC 1, 2, CP
  5. C P
  6. x C 4, UG (rely)
  7. (^) C → ∀ x C 4, 5, CP
  8. (^) (∀ x CC ) ∧ ( C → ∀ x C ) 3, 6, Conj. QED 7, E (b) ∀ x ( CA ( x )) → C ∨ ∀ x A ( x )
  9. (^) ∀ x ( CA ( x )) P
  10. (^) CA (Everything) 1, UI
  11. (^) C ∨ ∀ x A ( x ) 2,? * QED 1, 3, CP
  • I know step 3 is wrong, but I have no idea how to say that every possible x value can be plugged in there. (c) C ∨ ∀ x A ( x ) → ∀ x ( CA ( x ))
  1. C ∨ ∀ x A ( x ) P
  2. C P
  3. (^) ∀ x C 2, UG (rely)
  4. C → ∀ x C 2, 3 CP
  5. (^) ∀ x A ( x ) → ∀ x A ( x ) T
  6. (^) ∀ x C ∨ ∀ x A ( x ) 1, 4, 5, CD
  7. (^) ∀ x A ( x ) ∨ ∀ x B ( x ) → ∀ x ( A ( x ) ∨ B ( x )) T, (7.2c)
  8. x ( CA ( x )) 6, 7, MP QED 1, 8, CP