Question 1

1. (∀x)(∀y)(∀z)[(Pxy • Pyz) ⊃ Pxz]
2. (∀x)Pxf(x)
-Which of the following propositions is an immediate (one-step) consequence in FF of the given premises?
A) Pxx
B) Paa
C) Pf(x)f(x)
D) (∀y)(∀z)[(Pay • Pyz) ⊃ Paz]
E) (∀y)(∀z){[Pf(x)y • Pyz] ⊃ Pxz}

Accepted Answer

The first premise is a universal instantiation of the transitive property, and the second premise allows for substitution of 'f(x)' for 'x'. However, only choice D directly follows in one step by applying the transitive property specified in the first premise to any instance, including 'a'. Choices A, B, C, and E either require additional steps or do not directly follow from the premises given.

Question 2

1. (∀x)(∀y)(∀z)[(Pxy • Pyz) ⊃ Pxz]
2. (∀x)Pxf(x)
-Which of the following propositions is derivable from the given premises in FF?
A) Pf(a)a
B) Pf(a)f(a)
C) (∀x)Pf(x)x
D) (∀x)Pf(x)f(x)
E) (∀x)Pxf(f(x))

Accepted Answer

The answer of 1. (∀x)(∀y)(∀z)[(Pxy • Pyz) ⊃ Pxz]
2. (∀x)Pxf(x)
-Which...

Question 3

1. (∀x)(∀y)[(Px • Py) ⊃ Pf(x,y)]
2. (∃x)[Px • (∃y)(Py • x≠y • Rxy)]
-Which of the following propositions is an immediate (one-step) consequence in FF of the given premises?
A) Pf(a) • (∃y)[Py • f(a)≠x • Rf(a)y]
B) Pa • (∃y)(Py • a≠y • Ray)
C) Pf(a)
D) (∀y)[(Pa • Py) ⊃ Pf(a)]
E) (∀y)[(Pf(a) • Py) ⊃ Pf(a,y)]

Accepted Answer

From premise 2, we know that there exists an x such that Px is true and there exists a y such that Py, x is not equal to y and Rxy is true. Let us call this x "a". Then, from premise 1, we know that for all x and y such that Px and Py are true, Pf(x,y) is true. Since Pa and Py are both true for the particular values of x and y we found in premise 2, we can conclude that Pf(a,y) is true for all y. Therefore, we can immediately infer proposition B.

Question 4

1. (∀x)(∀y)[(Px • Py) ⊃ Pf(x,y)]
2. (∃x)[Px • (∃y)(Py • x≠y • Rxy)]
-Which of the following propositions is derivable from the given premises in FF?
A) (∃x)(∃y)[Pf(x,y) • Rxy • x≠y]
B) (∃x)(∃y)[Pf(x) • Rxy • x≠y]
C) (∃x)(∃y)[Pf(f(x)) • Rxy • x≠y]
D) (∀x)(∀y)Pf(x,y)
E) (∀x)Pf(x,x)

Accepted Answer

From the first premise, we can derive (∀x)(∀y)[Px ⊃ (Py ⊃ Pf(x,y))] using conditional proof.
Then, from the second premise, we can derive (∃x)(Px • (∃y)(Py • x≠y • Rxy)) using existential instantiation.
Next, using (∀x)(∀y)[Px ⊃ (Py ⊃ Pf(x,y))] and (∃x)(Px • (∃y)(Py • x≠y • Rxy)), we can derive (∃y)[Py • x≠y • Rxy • Pf(x,y)] using universal instantiation and modus ponens.
Finally, using (∃y)[Py • x≠y • Rxy • Pf(x,y)], we can use existential instantiation twice to get (∃x)(∃y)[Pf(x,y) • Rxy • x≠y].

Question 5

1. (∀x){Px ⊃ (∃y)[Py • f(x)=y]}
2. Pa • (∀x)f(x)≠a
-Which of the following propositions is an immediate (one-step) consequence in FF of the given premises?
A) Pf(a)
B) f(a)=a
C) Pa ⊃ (∃y)[Py • f(a)=y]
D) Pa ⊃ (∃y)[Pa • a=y]
E) Pa ⊃ [Pb • f(b)=a]

Accepted Answer

From premise 1, we know that for any x that satisfies Px, there exists a y such that Py and f(x) = y. From premise 2, we know that f(a) is not equal to a. Combining these two premises, if Pa is true, then there exists a y such that Py and f(a) = y, which is exactly what choice C states.

Question 6

1. (∀x){Px ⊃ (∃y)[Py • f(x)=y]}
2. Pa • (∀x)f(x)≠a
-Which of the following propositions is derivable from the given premises in FF?
A) Pf(a)
B) (∃x)Pf(x)
C) (∃x)(∀y)f(x)=y
D) (∃x)[(∃y)f(x)=y • ∼(∃y)f(y)=x]
E) f(a)=a

Accepted Answer

The answer of 1. (∀x){Px ⊃ (∃y)[Py • f(x)=y]}
2. Pa...

Question 7

1. (∀x)[(Px • Qx) ⊃ Rf(x)]
2. (∀x)[Rx ⊃ (∃y)Pxy]
3. ∼(∀x)(Px ⊃ ∼Qx)
-Which of the following propositions is an immediate (one-step) consequence in FF of the given premises?
A) (∃x)∼(Px ⊃ ∼Qx)
B) Px ⊃ ∼Qx
C) ∼(Px ⊃ ∼Qx)
D) Pf(a) ⊃ ∼Qf(a)
E) (Pa • Qa) ⊃ Ra

Accepted Answer

From premise 3, we can infer that there exists at least one object x such that Px ∧ Qx. From premise 1, if Px ∧ Qx, then Rf(x). Thus, there exists an object x such that Px ∧ Qx ⊃ Rf(x). Using the rule of inference called existential generalization, we can introduce an existential quantifier to obtain (∃x)(Px ∧ Qx ⊃ Rf(x)). Next, we can use the rule of inference called contraposition on premise 2 to obtain (∀x)(∀y)(¬Pxy ⊃ ¬Rx). Finally, we can use the rule of inference called existential instantiation to obtain ∼(Pa ⊃ ∼Qa). By applying De Morgan's law to this proposition, we arrive at (∃x)∼(Px ⊃ ∼Qx).

Question 8

1. (∀x)[(Px • Qx) ⊃ Rf(x)]
2. (∀x)[Rx ⊃ (∃y)Pxy]
3. ∼(∀x)(Px ⊃ ∼Qx)
-Which of the following propositions is derivable from the given premises in FF?
A) (∀x)[Pf(x) ⊃ Qf(x)]
B) (∀x)Rf(x)
C) (∀x)Rf(f(x))
D) (∃x)(∃y)Pf(x)f(y)
E) (∃x)(∃y)Pxy

Accepted Answer

The third premise negates the implication that Px implies not Qx, suggesting there exists some x for which Px and Qx are both true. This aligns with the existential quantifier in option E.

Question 9

1. (∀x)(∀y)(∃z)Sf(x)yz
2. (∀x)(∀y)(∀z)[Sxyz ⊃ ∼(Cxyz $ \lor $ Mzyx)]
-Which of the following propositions is an immediate (one-step) consequence in FF of the given premises?
A) (∀y)(∃z)Sf(x)yz
B) (∀y)(∃z)Sxyz
C) (∀y)(∃z)Sf(x)f(y)f(z)
D) Sabc
E) Sf(a)f(b)f(c)

Accepted Answer

BA) is an immediate consequence of the first premise by existential instantiation, removing the universal quantifier for x and replacing it with a specific instance, f(x).B) is also an immediate consequence of the first premise by existential instantiation, removing the universal quantifiers for x and y, and introducing an existential quantifier for z, without specifying the function f.

Question 10

1. (∀x)(∀y)(∃z)Sf(x)yz
2. (∀x)(∀y)(∀z)[Sxyz ⊃ ∼(Cxyz $ \lor $ Mzyx)]
-Which of the following propositions is derivable from the given premises in FF?
A) (∃x)(∃y)(∃z)∼Cxf(y)g(f(y),z)
B) (∃x)(∃y)(∃z)∼Cxf(x)f(f(x))
C) (∃x)(∃y)(∃z)∼Mf(x)yz
D) (∃x)(∃y)(∃z)∼Mzg(y)f(g(x))
E) (∃x)(∃y)(∃z)∼Mzg(x)f(g(x))

Accepted Answer

The answer of 1. (∀x)(∀y)(∃z)Sf(x)yz
2. (∀x)(∀y)(∀z)[Sxyz ⊃ ∼(Cxyz \( \lor...

Question 11

Translate each sentence into predicate logic, using the given translation keys.
use: 
h: Hume
l: Locke
Px: x is a philosopher
Rx: x is a rationalist 
Ixy: x influenced y
Sxy: x is more skeptical than y
-Hume is more skeptical than Locke.

Accepted Answer