Question 1

All cheetahs are faster than some tigers. Everything is striped if, and only if, it is a tiger. So, if some things are cheetahs, then some things have stripes. (Cx: x is a cheetah; Tx: x is a tiger; Sx: x has stripes; Fxy: x is faster than y)
-Which of the following is the best translation into F of this argument?
A) 1. (∀x)[Cx ⊃ (∃y)(Ty • Fyx)] 2. (∀x)[(Sx ⊃ Tx) • (Tx ⊃ Sx)] / (∃x)Cx ⊃ (∃x)Sx
B) 1. (∀x)(∃y)[Cx • (Ty • Fxy)] 2. (∀x)Sx ≡ (∀x)Tx / (∃x)Cx ⊃ (∃x)Sx
C) 1. (∀x)[Cx ⊃ (∃y)(Ty • Fxy)] 2. (∀y)(Sy ≡ Ty) / (∃x)Cx ⊃ (∃x)Sx
D) 1. (∀x)[Cx ⊃ (∃y)(Ty • Fyx)] 2. (∀x)(Sx ≡ Tx) / (∃x)(Cx ⊃ Sx)
E) 1. (∀x)[(∀y)(Fxy • Ty) ⊃ Cx] 2. (∀x)(Sx ⊃ Tx) • (∀x)(Tx ⊃ Sx) / (∃x)(Cx ⊃ Sx)

Accepted Answer

The correct translation is C because it accurately represents the argument's premises and conclusion. The first premise states that for all cheetahs, there exists some tiger that the cheetah is faster than, which is correctly translated as "∀x[Cx ⊃ (∃y)(Ty • Fxy)]". The second premise states that something is striped if and only if it is a tiger, accurately translated as "∀y(Sy ≡ Ty)". The conclusion that if some things are cheetahs, then some things have stripes, is correctly represented as "∃x Cx ⊃ ∃x Sx".

Question 2

All cheetahs are faster than some tigers. Everything is striped if, and only if, it is a tiger. So, if some things are cheetahs, then some things have stripes. (Cx: x is a cheetah; Tx: x is a tiger; Sx: x has stripes; Fxy: x is faster than y)
-Which of the following propositions is an immediate (one-step) consequence in F of the given premises?
A) Sx ≡ Ty
B) Sx ≡ (∀x)Tx
C) Ca ⊃ (∃y)(Ty • Fya)
D) Sx ⊃ Tx
E) Cx ⊃ (∃y)(Ty • Fxy)

Accepted Answer

- Premise 1: All cheetahs are faster than some tigers. ( ∀x (Cx → ∃y(Ty ∧ Fxy)) )
- Premise 2: Everything is striped if, and only if, it is a tiger. ( ∀x (Sx ↔ Tx) )
- Conclusion: If some things are cheetahs, then some things have stripes. ( ∃x(Cx) → ∃y(Ty ∧ Fxy ∧ Sy))

In other words, if there are at least some cheetahs (which implies that there are at least some things that are not tigers), then there must also be at least some tigers that are faster than those cheetahs AND have stripes. This follows from the fact that all tigers (and only tigers) are striped, as well as from the fact that all cheetahs are faster than some tigers.

Question 3

All cheetahs are faster than some tigers. Everything is striped if, and only if, it is a tiger. So, if some things are cheetahs, then some things have stripes. (Cx: x is a cheetah; Tx: x is a tiger; Sx: x has stripes; Fxy: x is faster than y)
-Which of the following claims can also be derived from the premises of this argument?
A) (∀x)∼Sx ⊃ (∀x)∼Cx
B) (∀x)(∃y)[Fxy ⊃ (Cy • Tx)]
C) (∃x)(Cx • Tx)
D) ~(∃x)(Cx • Tx)
E) (∀x)(∃y)[Fxy ⊃ (Cx • Ty)]

Accepted Answer

The answer of All cheetahs are faster than some tigers....

Question 4

All philosophers respect each other. Some philosopher doesn't study some philosopher. Anything which respects something without studying it is open-minded, if ignorant. So something is open-minded and ignorant. (Ix: x is ignorant; Ox: x is open-minded; Px: x is a philosoper; Rxy: x respects y; Sxy: x studies y)
-Which of the following is the best translation into F of this argument?
A) 1. (∀x)(∀y)[(Px • Py) ⊃ (Rxx • Ryy)] 2. (∃x)(∃y)[(Px • Py) • ~Sxy]
3) (∀x)(∀y)[(Rxy • ~Sxy) ⊃ (Ox $ \lor $ Ix)] / (∃x)(Ox • Ix)
B) 1. (∀x)(∀y)[(Px • Py) ⊃ (Rxy ≡ Ryx)] 2. (∃x)(∃y)[(Px • Py) • ~Sxy]
3) (∀x)(∀y)[(Rxy • ~Sxy) ⊃ (Ox $ \lor $ Ix)] / (∃x)(Ox • Ix)
C) 1. (∀x)(∀y)[(Px • Py) ⊃ (Rxy • Ryx)] 2. (∃x)(∃y)[(Px • Py) • ~Sxy]
3) (∀x)(∀y)[(Rxy • ~Sxy) ⊃ (Ox $ \lor $ Ix)] / (∃x)(Ox • Ix)
D) 1. (∀x)(∀y)[(Px • Py) ⊃ (Rxy • Ryx)] 2. (∃x)[Px • (∃y)(Py • ∼Sxy)]
3) (∀x)(∀y)[(Rxy • ∼Sxy) ⊃ (Ox • Ix)] / (∃x)(Ox • Ix)
E) 1. (∀x)(∀y)[(Px • Py) ⊃ (Rxx • Ryy)] 2. (∃x)(∃y)[(Px • Py) • ~Sxy]
3) (∀x)(∀y)[(Rxy • ~Sxy) ⊃ (Ox $ \lor $ Ix)] / (∃x)(Ox $ \lor $ Ix)

Accepted Answer

- Premise 1: (∀x)(∀y)[(Px • Py) ⊃ (Rxy • Ryx)] - this translates as "For all philosophers x and y, if x respects y then y respects x"
- Premise 2: (∃x)[Px • (∃y)(Py • ∼Sxy)] - this translates as "There exists a philosopher x who studies some philosopher y"
- Premise 3: (∀x)(∀y)[(Rxy • ∼Sxy) ⊃ (Ox • Ix)] - this translates as "For all philosophers x and y, if x respects y without studying them, then x is open-minded and ignorant"
- Conclusion: (∃x)(Ox • Ix) - this translates as "There exists a philosopher who is both open-minded and ignorant"

Question 5

All philosophers respect each other. Some philosopher doesn't study some philosopher. Anything which respects something without studying it is open-minded, if ignorant. So something is open-minded and ignorant. (Ix: x is ignorant; Ox: x is open-minded; Px: x is a philosoper; Rxy: x respects y; Sxy: x studies y)
-Which of the following propositions is an immediate (one-step) consequence in F of the
Given premises?
A) Ox • Ix
B) (∀y)[(Px • Py) ⊃ (Ryy • Ryy)]
C) (∃y)[(Px • Py) ~ Sxy]
D) Pa • (∃y)(Py • ∼Say)
E) (∀y)[(Rxy • ~Sxy) ⊃ (Ox $ \lor $ Ix)]

Accepted Answer

The premises state that all philosophers respect each other and some philosopher doesn't study some philosopher. This directly leads to the conclusion that there is a philosopher who respects but does not study another philosopher, which is captured by choice D.

Question 6

All philosophers respect each other. Some philosopher doesn't study some philosopher. Anything which respects something without studying it is open-minded, if ignorant. So something is open-minded and ignorant. (Ix: x is ignorant; Ox: x is open-minded; Px: x is a philosoper; Rxy: x respects y; Sxy: x studies y)
-Which of the following claims can also be derived from the premises of this argument?
A) Rab • ∼Sab
B) ~Rab • ~Sab
C) ~Rba • ~Sab
D) Rba • Sab
E) Rba • Sbb

Accepted Answer

Given the premises, it logically follows that there is at least one philosopher who respects another without studying them. This is captured by "Rab • ~Sab," which states that a philosopher (a) respects another philosopher (b) but does not study them. This aligns with the premises that all philosophers respect each other and that some philosopher doesn't study some philosopher, leading to the conclusion that something (or someone in this context) is open-minded and ignorant.

Question 7

determine whether the given argument is valid or invalid. If it is invalid, select a counterexample.
-1. (∀x)(Ax ⊃ Dex)
2) (∃x)(Bx • Dxe)
3) (∀x)(Bx ⊃ Ax) / (∃x)[Ax • (Dxe • Dex)]
A) Valid
B) Invalid. Counterexample in a domain of one member, in which: Ae: True Be: False Dee: False
C) Invalid. Counterexample in a domain of two members, in which:
Aa: True Ba: True Daa: True Dea: False
Ae: True Be: True Dae: False Dee: False
D) Invalid. Counterexample in a domain of two members, in which:
Aa: True Ba: False Daa: True Dea: True
Ae: False Be: False Dae: False Dee: True
E) Invalid. Counterexample in a domain of two members, in which: Aa: False Ba: True Daa: False Dea: False
Ae: True Be: True Dae: True Dee: False

Accepted Answer

The argument is valid because the premises logically lead to the conclusion. Given that for all x, if Ax then Dex (1), and there exists an x such that Bx and Dxe (2), and for all x, if Bx then Ax (3), it logically follows that there exists an x such that Ax and (Dxe and Dex), satisfying the conclusion. The premises ensure that any x that is Bx is also Ax, and since there exists at least one Bx that is Dxe, and all Ax are Dex, the conclusion necessarily follows.

Question 8

determine whether the given argument is valid or invalid. If it is invalid, select a counterexample.
-1. (?x)(?y)(Pxy • ?Pyx)
2) (?x)[(?y)Pxy ? (?y)Qxy] / (?x)(?y)(Qxy • ?Pyx)
A) Valid
B) Invalid. Counterexample in a domain of three members, in which:
$\begin{array}{lll}
\text { Paa: True } & \text { Pab: True } & \text { Pac: False } \
\text { Pba: False } & \text { Pbb: True } & \text { Pbc: True } \
\text { Pca: True } & \text { Pcb: False } & \text { Pcc: True } \
\text { Qaa: False } & \text { Qab: False } & \text { Qac: True } \
\text { Qba: True } & \text { Qbb:True } & \text { Qbc: True } \
\text { Qca: True } & \text { Qcb:True } & \text { Qcc: True }
\end{array}$

C) Invalid. Counterexample in a domain of three members, in which:
$\begin{array}{lll}
\text { Paa: True } & \text { Pab: True } & \text { Pac: False } \
\text { Pba: False } & \text { Fbb: False } & \text { Pbc: False } \
\text { Pca: False } & \text { Pcb: False } & \text { Pcc: True } \
\text { Qaa: False } & \text { Qab: False } & \text { Qac: True } \
\text { Qba: True } & \text { Qbb:True } & \text { Qbc: False } \
\text { Qca: True } & \text { Qcb: True } & \text { Qcc: True }
\end{array}$

D) Invalid. Counterexample in a domain of three members, in which:
$\begin{array}{lll}
\text { Paa: False } & \text { Pab: Ture } & \text { Pac: False } \
\text { Pba: False } & \text { Pbb: True } & \text { Pbc: True } \
\text { Pca: True } & \text { Pcb: True } & \text { Pcc: True } \
\text { Qaa: False } & \text { Qab:True } & \text { Qac: True } \
\text { Qba: False } & \text { Qbb:True } & \text { Qbc: True } \
\text { Qca: True } & \text { Qcb:True } & \text { Qcc: False }
\end{array}$

E) Invalid. Counterexample in a domain of three members, in which:
$\begin{array}{lll}
\text { Paa: True } & \text { Pab: True } & \text { Pac: False } \
\text { Pla: False } & \text { Pbb: False } & \text { Ploc: True } \
\text { Pca: True } & \text { Pcb: False } & \text { Pcc: True } \
\text { Qaa: False } & \text { Qab: False } & \text { Qac: True } \
\text { Qba: False } & \text { Qbb: False } & \text { Qbc: False } \
\text { Qca: True } & \text { Qcb: True } & \text { Qcc: False }
\end{array}$

Accepted Answer

The argument is invalid because the premises do not logically guarantee the conclusion. The counterexample provided shows a scenario where the premises can be true, but the conclusion does not necessarily follow, demonstrating the argument's invalidity.

Question 9

determine whether the given argument is valid or invalid. If it is invalid, select a counterexample.
-1. (?x)[Tx ? (?y)(Sy • Wxy)]
2) (?x)(Sx • Vx)
3) (?x)(Tx • Rx) / (?x)[(Rx • Vx) • (?y)Wxy]
A) Valid
B) Invalid. Counterexample in a domain of one member, in which: Ta: True Sa: False Waa: True
Ra: False Va: True
C) Invalid. Counterexample in a domain of two members, in which:
$\begin{array}{lll}
\text { Ta: False } & \text { Sa: False } & \text { Waa: True } \
\text { Tb: True } & \text { Sb: True } & \text { Wab: False } \
\text { Ra: True } & \text { Va: False } & \text { Wba: False } \
\text { Rb: False } & \text { Vb: False } & \text { Wbb:True }
\end{array}$

D) Invalid. Counterexample in a domain of two members, in which:
$\begin{array}{lll}
\text { Ta: False } & \text { Sa: True } & \text { Waa: True } \
\text { Tb: True } & \text { Sb: True } & \text { Wab: True } \
\text { Ra: False } & \text { Va: False } & \text { Wba: False } \
\text { Rb: False } & \text { Vb:True } & \text { Wbb: False }
\end{array}$

E) Invalid. Counterexample in a domain of two members, in which:
$\begin{array}{llll}
\text { Ta: True } & \text { Sa: True } & \text { Waa: True } & \
 \text { Tb: False } & \text { Sb: True } & \text { Wab: True } \
 \text { Ra: True } & \text { Va: False } & \text { Wba: True } \
\text { Rb: False } & \text { Vb:True } & \text { Wbb: True }
\end{array}$

Accepted Answer

The answer of determine whether the given argument is valid...

Question 10

determine whether the given argument is valid or invalid. If it is invalid, select a counterexample.
-1. (?x)[Lx ? (?y)(My • Nxy)]
2) (?x)[Lx • (?y)Nxy] / (?x)[Mx • (?y)Nyx]
A) Valid
B) Invalid. Counterexample in a domain of one member, in which:
La: True Ma: True Naa: False
C) Invalid. Counterexample in a domain of two members, in which:
$\begin{array}{llll}
\text { La: False } & \text { Ma: True } & \text { Naa: True } & \text { Nba: True } \
\text { Lb:True } & \text { Mb: True } & \text { Nab: True } & \text { Nbb: False }
\end{array}$

D) Invalid. Counterexample in a domain of two members, in which:
$\begin{array}{llll}
\text { La: True } & \text { Ma: False } & \text { Naa: True } & \text { Nba: False } \
\text { Lb: False } & \text { Mb: True } & \text { Nab: True } & \text { Nbb: False }
\end{array}$

E) Invalid. Counterexample in a domain of two members, in which: La: True Ma: False Naa: True Nba: True
Lb: False Mb: True Nab: False Nbb: False

Accepted Answer

The answer of determine whether the given argument is valid...

Question 11

select the best translation into predicate logic, using the following translation key: 
d: Diego
s: Sean
Dx: x is on the Dean's list 
Px: x is a philosophy major 
Sx: x is a student
Vx: x is a valedictorian
-There is exactly one philosophy major on the Dean's list.
A) (∃x){(Px • Dx) • (∀y)[(Py • Dy) ⊃ y=x]}
B) (Pd • Ps) ⊃ [(Dd $ \lor $ Ds) • ~(Dd • Ds)]
C) (∀x)(∀y){[(Px • Dx) • (Py • Dy)] ⊃ x=y}
D) (∃x){(Px • Dx) ⊃ (∀y)[(Py • Dy) ⊃ x=y]}
E) (Pd • Dd) $ \lor $ (Ps • Ds)

Accepted Answer

This statement reads "there exists exactly one x such that x is a philosophy major on the Dean's list, and for all y, if y is a philosophy major on the Dean's list, then y is that same x." This can be translated as (∃x){(Px • Dx) • (∀y)[(Py • Dy) ⊃ y=x]} in predicate logic. The other options do not accurately capture the meaning of the English sentence or contain unnecessary complexities.

Question 12

select the best translation into predicate logic, using the following translation key: 
d: Diego
s: Sean
Dx: x is on the Dean's list 
Px: x is a philosophy major 
Sx: x is a student
Vx: x is a valedictorian
-There are exactly two philosophy majors on the Dean's list.
A) Pd • Ps • Dd • Ds
B) (∃x)(∀y)[(Px • Py • Dx • Dy) ⊃ x=y]
C) (∀x)(∀y)(∀z)[(Px • Py • Pz • Dx • Dy • Dz) ⊃ (x =y $ \lor $ x=z $ \lor $ y=z)]
D) (∃x)(∃y){Px • Dx • Py • Dy • x≠y • (∀z)[(Pz • Dz) ⊃ (z=x $ \lor $ z=y)]}
E) ~(∃x)(∃y)(∃z)(Px • Py • Pz • Dx • Dy • Dz • x≠y • x≠z • y≠z)

Accepted Answer

Option D translates to "There exist two distinct individuals x and y such that x and y are philosophy majors on the Dean's list, and for any individual z, if z is both a philosophy major and on the Dean's list, then z must be either x or y." This accurately represents the statement "there are exactly two philosophy majors on the Dean's list." None of the other options capture the idea of exactly two individuals meeting the criteria.

Question 13

select the best translation into predicate logic, using the following translation key: 
d: Diego
s: Sean
Dx: x is on the Dean's list 
Px: x is a philosophy major 
Sx: x is a student
Vx: x is a valedictorian
-At least four philosophy majors are on the Dean's list.
A) (∃x)(∃y)(∃z)(∃w)[Px • Dx • Py • Dy • Pz • Dz • Pw • Dw •
(x=y $ \lor $ x=z $ \lor $ x=w $ \lor $ y=z $ \lor $ y=w $ \lor $ z=w)]
B) (∀x)(∀y)(∀z)(∀w)[(Px • Py • Pz • Pw • Dx • Dy • Dz • Dw) ⊃
(x=y $ \lor $ x=z $ \lor $ x=w $ \lor $ y=z $ \lor $ y=w $ \lor $ z=w)]
C) (∃x)(∃y)(∃z)(∃w)(Px • Dx • Py • Dy • Pz • Dz • Pw • Dw • x≠y • x≠z • x≠w • y≠z • y≠w • z≠w)
D) (∀x)(∀y)(∀z)(∀w)[(Px • Py • Pz • Pw • Dx • Dy • Dz • Dw) ⊃
(x≠y • x≠z • x≠w • y≠z • y≠w • z≠w)]
E) (∃x)(∃y)(∃z)(∃w)[(Px • Dx • Py • Dy • Pz • Dz • Pw • Dw) ⊃
(x≠y • x≠z • x≠w • y≠z • y≠w • z≠w)]

Accepted Answer

Option C correctly translates the statement "At least four philosophy majors are on the Dean's list" into predicate logic by asserting the existence of four distinct individuals (x, y, z, w) who are all philosophy majors and on the Dean's list, with the inequalities ensuring they are distinct individuals.

Question 14

select the best translation into predicate logic, using the following translation key: 
d: Diego
s: Sean
Dx: x is on the Dean's list 
Px: x is a philosophy major 
Sx: x is a student
Vx: x is a valedictorian
-All philosophy majors except Sean are on the Dean's list.
A) (∃x)(Px • Dx • x≠s)
B) Ps • ∼Ds • (∀x)[(Px • x≠s) ⊃ Dx]
C) (∀x)[Dx ⊃ (Px ⊃ x≠s)]
D) (∀x)(Px ⊃ Dx) • Ps • ~Ds
E) (∀x)[Px ⊃ (Dx ⊃ x≠s)]

Accepted Answer

The statement "All philosophy majors except Sean are on the Dean's list" can be translated as "For all x, if x is a philosophy major and x is not Sean, then x is on the Dean's list." Using the provided translation key, this can be written as: (∀x)[(Px ∧ x≠s) → Dx]. This matches exactly with option B when we apply the conditional equivalence: [(p → q) ≡ (~p ∨ q)] and DeMorgan's law: ~(p ∧ q) ≡ ~p ∨ ~q.

Question 15

select the best translation into predicate logic, using the following translation key: 
d: Diego
s: Sean
Dx: x is on the Dean's list 
Px: x is a philosophy major 
Sx: x is a student
Vx: x is a valedictorian
-Of the philosophy majors, only Diego is a valedictorian.
A) Vd • (∀x)[Px ⊃ (~Vx $ \lor $ x=d)]
B) Pd • Vd • (∀x)[(Px • Vx) ⊃ x=d]
C) ~(∃x)(Px • Vx • x≠d)
D) (∀x)[(Px • Vx) ⊃ x=d]
E) (∃x)(∀y)[x=d • Vd • Pd • (Py ⊃ y=d)]

Accepted Answer

Option B correctly translates the statement "Of the philosophy majors, only Diego is a valedictorian" into predicate logic. It asserts that Diego is a philosophy major and a valedictorian, and for any x, if x is a philosophy major and a valedictorian, then x must be Diego.

Question 16

select the best translation into predicate logic, using the following translation key: 
d: Diego
s: Sean
Dx: x is on the Dean's list 
Px: x is a philosophy major 
Sx: x is a student
Vx: x is a valedictorian
-At most two philosophy majors are on the Dean's list.
A) (∃x)(∃y){Px • Dx • Py • Dy • (∀z)[(Pz • Dz) ⊃ (x=z $ \lor $ y=z)]}
B) (∃x)(∃y)[Px • Dx • Py • Dy • ~(∃z)(Pz • Dz • x=z • y=z)]
C) (∀x)(∀y)(∀z)(Px • Dx • Py • Dy • Pz • Dz • x≠y • x≠z • y≠z)
D) ~(∃x)(∃y)(∃z)(Px • Dx • Py • Dy • Pz • Dz • x≠y • x≠z • y≠z)
E) (∀x)(∀y)(∀z)[(Px • Dx • Py • Dy • Pz • Dz) ⊃ (x=y $ \lor $ x=z $ \lor $ y=z)]

Accepted Answer

The statement "At most two philosophy majors are on the Dean's list" means that if there are three philosophy majors on the Dean's list, then at least two of them must be the same person. This is captured by option E, which states that if x, y, and z are all philosophy majors on the Dean's list, then at least two of them must be the same (x=y or x=z or y=z).

Question 17

select the best translation into predicate logic, using the following translation key: 
d: Diego
s: Sean
Dx: x is on the Dean's list 
Px: x is a philosophy major 
Sx: x is a student
Vx: x is a valedictorian
-Exactly one student is a valedictorian.
A) (∃x)(Sx • Vx) • ~(∃y)(Sy • Vy)
B) (∀x)[(Sx • Vx) ⊃ ~(∃y)(Sy • Vy • x≠y)]
C) ~(∃x)(∃y)(Sx • Vx • Sy • Vy • x≠y)
D) (∃x)[Sx • Vx • (∃y)(Sy • Vy • x=y)]
E) (∃x){Sx • Vx • (∀y)[(Sy • Vy) ⊃ y=x]}

Accepted Answer

E) is the correct translation because it states that there exists exactly one student who is a valedictorian, and for all other students y, if y is a student and a valedictorian, then y must be the same as x, ensuring uniqueness.

Question 18

select the best translation into predicate logic, using the following translation key: 
d: Diego
s: Sean
Dx: x is on the Dean's list 
Px: x is a philosophy major 
Sx: x is a student
Vx: x is a valedictorian
-The valedictorian is Diego.
A) (∃x){Vx • (∀y)[(Vy ⊃ y=x) • x=d]}
B) (∀x)(Vx ⊃ x=d)
C) (∀x)(Vx ≡ x=d)
D) Vd • ~(∃x)(Vx • x=d)
E) (∃x)[Vx • (∃y)(Vy • x=y) • x=d]

Accepted Answer

The correct answer is A because it correctly translates the statement "The valedictorian is Diego" into predicate logic using the given translation key.

Question 19

select the best translation into predicate logic, using the following translation key: 
d: Diego
s: Sean
Dx: x is on the Dean's list 
Px: x is a philosophy major 
Sx: x is a student
Vx: x is a valedictorian
-Only Sean and Diego are both on the Dean's list and philosophy majors.
A) (∀x)[(Dx • Px) ⊃ (x=s • x=d)]
B) (∀x)[(Dx • Px) ⊃ ~(x=s • x=d)]
C) Ds • Ps • Dd • Pd • [(∃x)(Px • Dx) ⊃ ~(x=s • x=d)]
D) Ds • Dd • Ps • Pd • (∀x)[(Dx • Px) ⊃ (x=s $ \lor $ x=d)]
E) Ds • Ps • Dd • Pd • (∀x)[(Dx • Px) ⊃ ~(x=s • x=d)]

Accepted Answer

The correct translation states that Sean and Diego are both on the Dean's list and philosophy majors, and if anyone is on the Dean's list and a philosophy major, then that person must be either Sean or Diego.

Question 20

select the best translation into predicate logic, using the following translation key: 
d: Diego
s: Sean
Dx: x is on the Dean's list 
Px: x is a philosophy major 
Sx: x is a student
Vx: x is a valedictorian
-The philosophy major on the Dean's list is a valedictorian.
A) (Ps • Ds • Vs) $ \lor $ (Pd • Dd • Vd)
B) (∀x)[Vx ⊃ (Px • Dx)]
C) (∃x){Px • Dx • (∀y)[(Py • Dy) ⊃ y=x] • Vx}
D) (∃x){Px • Vx • (∀y)[(Py • Vy) ⊃ y=x] • Dx}
E) (∃x){Dx • Vx • (∀y)[(Dy • Vy) ⊃ y=x] • Px}

Accepted Answer

The correct translation is "There exists an x such that x is a philosophy major, x is on the Dean's list, for all y if y is a philosophy major and y is on the Dean's list then y is x, and x is a valedictorian." This captures the uniqueness of the philosophy major on the Dean's list being a valedictorian.

Quiz 5: Full First-Order Logic

determine whether the given argument is valid or invalid. If it is invalid, select a counterexample. -1. (∀x) (Ax ⊃ Dex) 2) (∃x) (Bx • Dxe) 3) (∀x) (Bx ⊃ Ax) / (∃x) [Ax • (Dxe • Dex) ]

determine whether the given argument is valid or invalid. If it is invalid, select a counterexample. -1. (?x) (?y) (Pxy • ?Pyx) 2) (?x) [(?y) Pxy ? (?y) Qxy] / (?x) (?y) (Qxy • ?Pyx)

determine whether the given argument is valid or invalid. If it is invalid, select a counterexample. -1. (?x) [Tx ? (?y) (Sy • Wxy) ] 2) (?x) (Sx • Vx) 3) (?x) (Tx • Rx) / (?x) [(Rx • Vx) • (?y) Wxy]

determine whether the given argument is valid or invalid. If it is invalid, select a counterexample. -1. (?x) [Lx ? (?y) (My • Nxy) ] 2) (?x) [Lx • (?y) Nxy] / (?x) [Mx • (?y) Nyx]

select the best translation into predicate logic, using the following translation key: d: Diego s: Sean Dx: x is on the Dean's list Px: x is a philosophy major Sx: x is a student Vx: x is a valedictorian -There is exactly one philosophy major on the Dean's list.

select the best translation into predicate logic, using the following translation key: d: Diego s: Sean Dx: x is on the Dean's list Px: x is a philosophy major Sx: x is a student Vx: x is a valedictorian -There are exactly two philosophy majors on the Dean's list.

select the best translation into predicate logic, using the following translation key: d: Diego s: Sean Dx: x is on the Dean's list Px: x is a philosophy major Sx: x is a student Vx: x is a valedictorian -At least four philosophy majors are on the Dean's list.

select the best translation into predicate logic, using the following translation key: d: Diego s: Sean Dx: x is on the Dean's list Px: x is a philosophy major Sx: x is a student Vx: x is a valedictorian -All philosophy majors except Sean are on the Dean's list.

select the best translation into predicate logic, using the following translation key: d: Diego s: Sean Dx: x is on the Dean's list Px: x is a philosophy major Sx: x is a student Vx: x is a valedictorian -Of the philosophy majors, only Diego is a valedictorian.

select the best translation into predicate logic, using the following translation key: d: Diego s: Sean Dx: x is on the Dean's list Px: x is a philosophy major Sx: x is a student Vx: x is a valedictorian -At most two philosophy majors are on the Dean's list.

select the best translation into predicate logic, using the following translation key: d: Diego s: Sean Dx: x is on the Dean's list Px: x is a philosophy major Sx: x is a student Vx: x is a valedictorian -Exactly one student is a valedictorian.

select the best translation into predicate logic, using the following translation key: d: Diego s: Sean Dx: x is on the Dean's list Px: x is a philosophy major Sx: x is a student Vx: x is a valedictorian -The valedictorian is Diego.

select the best translation into predicate logic, using the following translation key: d: Diego s: Sean Dx: x is on the Dean's list Px: x is a philosophy major Sx: x is a student Vx: x is a valedictorian -Only Sean and Diego are both on the Dean's list and philosophy majors.

select the best translation into predicate logic, using the following translation key: d: Diego s: Sean Dx: x is on the Dean's list Px: x is a philosophy major Sx: x is a student Vx: x is a valedictorian -The philosophy major on the Dean's list is a valedictorian.