Exam 2 solutions

1. Letter used for flagging must be new to the proof, must not appear in premises or conclusion or outside the subproof in which it is flagged.

2.
(x) universal quantifier
F predicate letter
x individual variable
Fx propositional function
(x)Fx universal sentence

3. A singular sentence is any sentence containing a name. An existential sentence is a sentence that asserts that a propositional function is true of at least one thing. Some singular sentences are not existential. (e.g., John does not exist.)

4. truth functional compound

5. yes. no flagging restrictions apply.

6. Notation:

Ax = x gets an A.
Lx = x is lazy.
Bx = x is beer guzzling.
Px = x is a person.
Yx = x watches Baywatch 5 days a week.

Translation:

1. (x)(Px & Ax -> ~(Lx & Bx))
2. (Ex)((Px&Ax)&Yx)
3. (Ex)((Px&Yx) & ~Lx)

It was also acceptable to have the second conjunction in 1 replaced by a disjunction:

1. (x)(Px & Ax -> ~(Lx v Bx))

Though this is not the better of the two translations, this translation is the only one on which teh proof is valid. See note on quesiton 10.

7.

1. (y)(By -> Cy)
2. (x)(Ax -> Bx)
3. -> flag a FSUG
4. | Aa -> Ba 2 UI
5. | Ba -> Ca 1 UI
6. |- Aa -> Ca 4, 5 HS
7. (x)(Ax -> Cx) 3-6 UG
8. (x)(~Ax v Cx) 7 CE

8.

1. Fc
2. (x)(~Gx -> ~Fx)
3. ~Gc -> ~Fc 2 UI
4. Gc 1, 3 MT
5. (Ex)(Gx) 4, EG

9.

01. (y){~[(Dy v Sy) -> Ay ] -> ~(Ry v Sy) }
02. |-> flag a FSUG
03. | |-> Ra & Da ACP
04. | | Ra 3 SIMP
05. | | Da 3 SIMP
06. | | Ra v Da 4, ADD
07. | | Da v Sa 5, ADD
08. | | ~[(Da v Sa) -> Aa ] -> ~(Ra v Sa) 1 UI
09. | | (Da v Sa) -> Aa 6, 8 MT
10. | | Aa 7, 9 MT
11. | |- Ra & Da -> Aa 3-10 CP
12. |- (x)( Rx & Dx -> Ax) 2-11 UG

10. NOTE: Due to instructor error, the better of the two translations on question 6 is actually invalid as a proof. Therefore, all students were given full credit for this proof. What follows is a proof for the less accurate, but valid, of the two possible translations for 6.

1. (x)((Px &Ax) -> ~(Lx v Bx))
2. Ex((Px & Ax) & Yx)
3. (Pa&Aa)&Ya 3, EI flag a
4. (Pa&Aa) -> ~(La v Ba) 1 UI
5. Pa&Aa 3, SIMP
6. ~(La v Ba) 4, 5 MP
7. ~La & ~Ba 6 DEM
8. ~La 7 SIMP
9. (Pa&Aa)&~La 5, 8 CONJ
10. Ex((Px&Ax)&~Lx) 9, EG

11.

01. |-> (z)[Az -> ~(Gz v Mz)] ACP
02. | |-> flag a FSUG
03. | | | -> (Aa & Ma) AIP
04. | | | Aa -> ~(Ga v Ma) 1 UI
05. | | | Aa 3 SIMP
06. | | | ~(Ga v Ma) 4, 5 mp
07. | | | ~Ga & ~Ma 6 DEM
08. | | | ~Ma 7 SIMP
09. | | | Ma 3 SIMP
10. | | | ~Ma & Ma 8, 9 CONJ
11. | | |- ~(Aa & Ma) 3-10 IP
12. | |- (z)~(Az & Mz) 1-12 CP
13. |- (z)[Az -> ~(Gz v Mz)] -> (z)~(Az & Mz) 1-12 CP

12.

1. (x)Jx -> (Ex)~Sx
2. ~(x)Sx -> (Ex)~Hx
3. ~((Ex)~Hx) -> (x)Sx 2 CONT
4. (x)Hx -> (x)Sx 3 QN
5. (x)Hx -> ~(Ex)~Sx 4 QN
6. ~ (Ex)~Sx -> ~(x)Jx 1 CONT
7. ~ (Ex)~Sx -> (Ex)~Jx 6 QN
8. (x)Hx -> (Ex)~Jx 5, 7 HS

13.

01. |-> ~((Ez)Dz v (Ez)~Gz) ACP
02. | ~(Ez)Dz & ~(Ez)~Gz 1 DEM
03. | ~(Ez)Dz 2 CONJ
04. | (z)~Dz 3 QN
05. | |-> flag a FSUG
06. | | ~Da 4 UI
07. | | ~Da V Pa 6 ADD
08. | | Da -> Pa 7 CE
09. | |-(x)(Dx -> Px) 5-8 UG
10. |- ~((Ez)Dz v (Ez)~Gz) -> (x)(Dx -> Px) 1-9 CP