Practice Exam #4 Solutions

 

Part I Translation (4 points each, 20 points total)

 1.Everyone who is taller than Ignat is shorter than Tony.

Txy: x is taller than y
 Sxy: x is shorter than y

i: Ignat
a: tony 

(x)(Txi ⊃ Sxa)

2.There is someone who likes everyone.

Lxy: x likes y

( ∃x)(y)Lxy 

3. No hard-drinking Texans knit.

Hx: x is hard-drinking
Kx: x knits
Tx: x is a Texan.

(x)((Tx & Hx) ⊃ ~Kx) 

4. No film critic loves every movie.

 Fx: x is a film critic

Mx: x is a movie

Lxy: x loves y

(x)(y)((Fx & My) ⊃ ~Lxy)  or (x)(Fx ⊃ (y)(My ⊃ ~Lxy))  Note: You can prove that these two are logically equivalent!

5. Some dark chocolate is neither sweet nor high in carbohydrates.

Dx: x is dark
Cx: x is chocolate
Sx: x is sweet
Hx: x is high in carbohydrates

(∃x)((Dx &Cx) & ~(Sx v Hx))

6. Ignat likes everyone who likes him.

(x)(Lxi ⊃ Lix)

Lxy: x likes y

i: Ignat

Part II Translation with Identity (5 points each, 10 points total)

1. There are exactly two fish.

Fx: x is a fish.

(∃x)(∃y)(((Fx & Fy) & x ≠y) &  (z)((Fz ⊃((z = x) v (z = y)))

2. Bertha is the tallest philosopher.

b: Bertha

Txy: x is taller than y

Px: x is a philosopher

(x)((Px & x ≠ b)  ⊃ Tbx) & Pb

 
 Part III Testing for Validity Using the Model Universe Method (10 points total)

 

1. (∃x)(Sx ⊃ Tx), (x)Tx, // (∃x)Sx

domain: {a}

(Sa ⊃ Ta), Ta, // Sa

Invalid when

Sa: F
Ta: T

Part IV Testing for Validity Using the Truth-Tree Method (10 points each, 20 points total)

 

1.  ∃x)(Ax & ~Bx), (x)(Bx ⊃ ~Cx) // (x)(Cx ⊃ ~Ax)  invalid; see:  tree

 

3. (x)(y)(Rxy ⊃ ~Txy), (∃x)( y)(Txy & ~ Uxy) // (∃x)(∃y)(Rxy ⊃ Uxy) valid; see: tree

 

Part V Proofs (15 points each; 30 points total.)

 

1. (x)(~Ax ⊃ Kx), (∃x)~Kx // (∃x)(Ax v ~Bx)

 

1. (x)(~Ax ⊃ Kx) premise    
2.  (∃x)~Kx premise    
3. ~Ka E.I., flag a 2  
4. (~Aa ⊃ Ka) U.I. 1  
5. ~~Aa M.T. 3,4  
6. Aa D.N. 5  
7. Aa v ~Ba Disj. 6  
8. (∃x)(Ax v ~Bx) E.G. 7  


 

  2. // (x)Fxa  ⊃ (x)(∃y)Fxy

1. (x)Fxa Assump., CP SP1  
2. flag b flagging asump. for UG SP1 SP2
3. Fba UI, 1 SP1 SP2
4. (∃y)Fby E.G., 3 SP1 SP2
5. (x)(∃y)Fxy U.G. 2-4 SP1  
6. (x)Fxa ⊃ (x)(∃y)Fxy CP 1-5    

 

Part VI Essay (10 points)

 

Can you determine whether (∃x)(y)~Txy is logically true? Explain.

 

Answer: To determine whether this wff is logically true, we do a tree for its negation. (provide details). Show that the attempt to instantiate the wff obtained after applying the truth-tree rules for quantifier negation leeds to an undecidable tree. Therefore, we cannot determine whether the original wff is logically true.

 

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