Question 1

1. (∀x)[Ax ⊃ (Bx ⊃ Cx)]
2. ∼(∀x)(Bx ⊃ Dx)
-Consider assuming '(∀x)Ax' for conditional proof. Which of the following propositions is an immediate (one-step) consequence in M of the given premises with that further assumption for conditional proof?
A) Bx ⊃ Cx
B) ∼(Bx ⊃ Dx)
C) Ax
D) ∼(∃x)∼(Bx ⊃ Dx)
E) Ax ⊃ (Bc ⊃ Cx)

Accepted Answer

The answer of 1. (∀x)[Ax ⊃ (Bx ⊃ Cx)]
2. ∼(∀x)(Bx...

Question 2

1. (∀x)[Ax ⊃ (Bx ⊃ Cx)]
2. ∼(∀x)(Bx ⊃ Dx)
-Which of the following propositions is derivable in M from the given premises?
A) (∀x)(Ax ⊃ Dx)
B) (∀x)Ax ⊃ (∃x)(Cx • ∼Dx)
C) (∀x)Ax ⊃ (∀x)Dx
D) (∀x)[Ax ⊃ (Cx • ∼Dx)]
E) (∀x)Ax ⊃ (∀x)(Cx • ∼Dx)

Accepted Answer

From the premises, we cannot derive that for all x, if A then D (A), or that for all x, if A then for all x, D (C), directly. The negation of the universal conditional in premise 2 suggests there is at least one x for which Bx does not imply Dx, which can be interpreted as there exists an x such that Bx and not Dx. Given the first premise, if Ax then Bx implies Cx, it follows that if Ax, then it's possible to have Cx and not Dx for at least one x, leading to the correct answer (B) (∀x)Ax ⊃ (∃x)(Cx • ∼Dx), which correctly captures the logical relationship implied by the premises.

Question 3

1. (∀x)[(Px $ \lor $ Qx) ⊃ (Rx • ∼Sx)]
2. (∀x)[Rx ⊃ (Tx ⊃ ∼Sx)]
-Consider assuming 'Px' for conditional proof. Which of the following propositions is an immediate (one-step) consequence in M of the given premises with that further assumption for conditional proof?
A) Ra ⊃ (Tb ⊃ ∼Sc)
B) (∀x)Px
C) Px $ \lor $ Qx
D) Tx ⊃ ∼Sx
E) Rx • ∼Sx

Accepted Answer

Given the assumption 'Px' for conditional proof, the immediate one-step consequence in M of the given premises is 'Px $ \lor $ Qx'. This follows directly from the rules of disjunction introduction, where assuming 'Px' allows us to infer 'Px $ \lor $ Qx' without needing to reference the other premises.

Question 4

1. (∀x)[(Px $ \lor $ Qx) ⊃ (Rx • ∼Sx)]
2. (∀x)[Rx ⊃ (Tx ⊃ ∼Sx)]
-Which of the following propositions is derivable in M from the given premises?
A) (∀x)[Px ⊃ (Sx ⊃ ∼Tx)]
B) (∀x)[Px ⊃ (Qx • ∼Tx)]
C) (∀x)Qx
D) (∀x)[Px ⊃ (Qx • ∼Sx)]
E) (∀x)[Px ⊃ (Qx ⊃ Tx)]

Accepted Answer

From the given premises, we can derive that if Px, then not Sx, and if Rx then not Sx implies not Tx, leading to the conclusion that if Px, then Sx implies not Tx.

Question 5

1. (∃x)Qx ⊃ (∀x)(Rx ⊃ Sx)
2. (∀x)∼Qx ⊃ (∃x)Sx
3. (∀x)Rx / (∃x)Sx
-Which of the following propositions is an appropriate assumption for an indirect proof of the conclusion of the given argument?
A) ∼Sx
B) (∃x)Sx
C) ∼(∃x)Sx
D) ∼(∃x)∼Sx
E) (∃x)∼Sx

Accepted Answer

An indirect proof (also known as proof by contradiction) involves assuming the negation of the conclusion and showing that this assumption leads to a contradiction. The conclusion of the given argument is (∃x)Sx, so the appropriate assumption for an indirect proof is its negation, which is ∼(∃x)Sx.

Question 6

1. (∃x)Qx ⊃ (∀x)(Rx ⊃ Sx)
2. (∀x)∼Qx ⊃ (∃x)Sx
3. (∀x)Rx / (∃x)Sx
-Which of the following propositions is not a likely last line of the indented sequence for an indirect proof of the given argument?
A) Rx • ∼Rx
B) Qx • ∼Qx
C) Rb • ∼Rb
D) Qb • ∼Qb
E) Sx • ∼Sx

Accepted Answer

The answer of 1. (∃x)Qx ⊃ (∀x)(Rx ⊃ Sx)
2. (∀x)∼Qx...

Question 7

(∃x)(Px • Qx) ⊃ [(∃x)Px • (∃x)Qx]
-Consider assuming '(∃x)(Px • Qx)' for a conditional proof of the above logical truth. Which
Of the following propositions is a legitimate second step in that proof?
A) (Px • Qx) ⊃ [(∃x)Px • (∃x)Qx]
B) Px • Qx
C) (∃x)Px
D) Pb • Qb
E) (∃x)Qx

Accepted Answer

The answer of (∃x)(Px • Qx) ⊃ [(∃x)Px • (∃x)Qx]
-Consider...

Question 8

(∃x)(Px • Qx) ⊃ [(∃x)Px • (∃x)Qx]
-Which of the following propositions is also derivable in M given that same assumption?
A) Qx
B) Px • Qx
C) (∀x)(Px • Qx)
D) (∀x)(Px ⊃ Qx)
E) Qb

Accepted Answer

The answer of (∃x)(Px • Qx) ⊃ [(∃x)Px • (∃x)Qx]
-Which...

Question 9

∼(∃x)(Px • Qx) ≡ (∀x)(Px ⊃ ∼Qx)
-Consider assuming '∼(∃x)(Px • Qx)' for a conditional proof of one direction of the biconditional in the above logical truth. Which of the following propositions is a legitimate second step in that proof?
A) (∀x) ∼ (Px • Qx)
B) Px ⊃ ∼Qx
C) Px • Qx
D) Pc • Qc
E) ∼(∀x)(Px ⊃ ∼Qx)

Accepted Answer

The negation of an existential quantifier (∃x) can be converted into a universal quantifier (∀x) with the negation of the predicate, turning ∼(∃x)(Px • Qx) into (∀x)∼(Px • Qx). This is a direct application of the logical rules concerning quantifiers and negation.

Question 10

∼(∃x)(Px • Qx) ≡ (∀x)(Px ⊃ ∼Qx)
-Which of the following propositions is also derivable in M given that same assumption?
A) ∼Qx
B) ∼Pa $ \lor $ ∼Qa
C) Px
D) ∼Qx ⊃ Px
E) Pa • Qa

Accepted Answer

In order to derive proposition B, we can first use the equivalence given to rewrite the original proposition as (∀x)(Px ⊃ ∼Qx). We can then instantiate this with a to get Pa ⊃ ∼Qa. From this, we can use De Morgan's law to get ∼(Pa • Qa), which is logically equivalent to ∼Pa $ \lor $ ∼Qa. Therefore, proposition B is derivable in M.

Question 11

Things are pleasant if, and only if, they are not too crowded. Everything too crowded is noisy. So if something isn't noisy, then something is pleasant.
-Which of the following is the best translation into M of this argument?
A) (∀x)(Px ⊃ ∼Cx) (∀x)(Cx ⊃ Nx) / (∃x)(∼Nx • Px)
B) (∀x)(Px ≡ ∼Cx) (∀x)(Cx ⊃ Nx) / ∼(∃x)Nx ⊃ (∃x)Px
C) (∀x)(Px ⊃ ∼Cx) (∀x)(Cx ⊃ Nx) / (∃x)∼Nx ⊃ (∃x)Px
D) (∀x)(Px ≡ ∼Cx) (∀x)(Cx ⊃ Nx) / (∃x)∼Nx ⊃ (∃x)Px
E) (∀x)(Px ≡ ∼Cx) (∀x)(Cx ⊃ Nx) / (∃x)(∼Nx • Px)

Accepted Answer

The argument structure is best captured by choice D, which translates the premises and conclusion accurately into logical form. The first premise states that things are pleasant if and only if they are not too crowded, which is represented as (∀x)(Px ≡ ∼Cx). The second premise states that everything too crowded is noisy, represented as (∀x)(Cx ⊃ Nx). The conclusion drawn is that if something isn't noisy, then something is pleasant, which is logically represented as (∃x)∼Nx ⊃ (∃x)Px. This choice correctly captures the bi-conditional relationship in the first premise and the conditional relationship in the conclusion.

Question 12

Things are pleasant if, and only if, they are not too crowded. Everything too crowded is noisy. So if something isn't noisy, then something is pleasant.
-Which of the following propositions is a good assumption for conditional proof for the above argument?
A) (∃x)∼Nx
B) ∼(∃x)Nx
C) ∼(∃x)∼Nx
D) ∼Nx
E) Nx

Accepted Answer

The argument essentially states that if something is not noisy, it can be considered pleasant. Option A, which translates to "There exists something that is not noisy," directly supports the conclusion that something can be pleasant, aligning with the argument's logic.

Question 13

Things are pleasant if, and only if, they are not too crowded. Everything too crowded is noisy. So if something isn't noisy, then something is pleasant.
-Which of the following propositions can also be proved from the premises of the above argument?
A) (∀x)Nx $ \lor $ (∀x)Px
B) (∀x)(Nx ≡ Cx)
C) (∀x)(Nx $ \lor $ Px)
D) (∀x)(Nx $ \lor $ Cx)
E) (∀x)[Nx ⊃ (∼Px • Cx)]

Accepted Answer

The argument states that things are pleasant if they are not too crowded, which can be represented as "¬C ⇒ P" (if not crowded, then pleasant). This can be simplified using the contrapositive to "¬P ⇒ C" (if not pleasant, then crowded). This is logically equivalent to "¬C ∨ ¬P" (not crowded or not pleasant). Using De Morgan's Law, this can be rewritten as "(C ∧ P)" (crowded and pleasant) is false for everything. This means that "¬(C ∧ P)" is true for everything, which can be written as "¬C ∨ ¬P" again using De Morgan's Law. This statement is equivalent to choice C, which states that everything is either not crowded or noisy.

Question 14

Everything is material, or ideal or transcendental. All atoms are not ideal and not transcendental. And nothing is material. So there are no atoms.
-Which of the following is the best translation into M of this argument?
A) (∀x)[Mx $ \lor $ (Ix $ \lor $ Tx)] (∀x)[Ax ⊃ (∼Ix • ∼Tx)]
∼(∃x)Mx / ∼(∃x)Ax
B) (∀x)[Mx $ \lor $ (Ix $ \lor $ Tx)] (∀x)[Ax ⊃ (∼Ix • ∼Tx)]
∼(∀x)Mx / ∼(∀x)Ax
C) (∀x)[Mx $ \lor $ (Ix $ \lor $ Tx)] (∀x)[Ax ⊃ ∼(Ix • Tx)]
∼(∃x)Mx / ∼(∃x)Ax
D) (∀x)[Mx $ \lor $ (Ix $ \lor $ Tx)] (∀x)[Ax ⊃ ∼(Ix • Tx)]
∼(∀x)Mx / ∼(∃x)Ax
E) (∀x)[Mx $ \lor $ (Ix $ \lor $ Tx)] (∀x)[Ax ⊃ (∼Ix • ∼Tx)]
∼(∃x)Mx / ∼(∀x)Ax

Accepted Answer

The correct translation reflects the structure of the argument: everything is either material, ideal, or transcendental; atoms are neither ideal nor transcendental; nothing is material; thus, there are no atoms. Option A accurately captures this logic using symbolic notation.

Question 15

Everything is material, or ideal or transcendental. All atoms are not ideal and not transcendental. And nothing is material. So there are no atoms.
-Which of the following propositions is a good assumption for indirect proof for the above argument?
A) ∼(∃x)Ax
B) (∃x)Ax
C) ∼(∀x)Ax
D) (∀x)Ax
E) (∀x)∼Ax

Accepted Answer

The argument concludes that there are no atoms based on the premises given. For an indirect proof, we assume the opposite of what we want to prove and show that it leads to a contradiction. Here, assuming that there exists at least one atom (∃x)Ax is the opposite of the conclusion that there are no atoms, making it a good starting point for an indirect proof.

Question 16

Everything is material, or ideal or transcendental. All atoms are not ideal and not transcendental. And nothing is material. So there are no atoms.
-Which of the following propositions can also be proved from the premises of the above argument?
A) (∃x)(Ax • Mx)
B) (∃x)Tx
C) (∀x)[∼Ax • (Ix • Tx)]
D) (∃x)Ix
E) (∀x)[∼Ax • (Ix $ \lor $ Tx)]

Accepted Answer

The premises state that nothing is material, and all atoms are neither ideal nor transcendental. Therefore, for all x, if x is not an atom, then x is either ideal or transcendental.

Question 17

consider the following domain, assignment of objects in the domain, and assignments sets to predicates.
Domain = {Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, Neptune, Pluto} 
P = {Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, Neptune}
I = {Mercury, Venus}
O = {Mars, Jupiter, Saturn, Uranus, Neptune}
a = Mercury b = Jupiter c = Saturn
d = Pluto
-Given the customary truth tables, which of the following theories is modeled by the above interpretation?
A) Pa • ∼Pb
Ia • ∼Id
B) Pa • ∼Pd
∼Ia • ∼Id
C) Pa • ∼Pd
Oa • ∼Id
D) Pa • ∼Pd
Ob • Od
E) Pa • ∼Pd
Ia • ∼Id

Accepted Answer

In option E, Pa • ∼Pd means that Mercury is present and Pluto is absent. Ia • ∼Id means that both Mercury and Venus are present and Pluto is absent, which is true based on the given assignments. Thus, option E is the correct theory modeled by the interpretation.

Question 18

consider the following domain, assignment of objects in the domain, and assignments sets to predicates.
Domain = {Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, Neptune, Pluto} 
P = {Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, Neptune}
I = {Mercury, Venus}
O = {Mars, Jupiter, Saturn, Uranus, Neptune}
a = Mercury b = Jupiter c = Saturn
d = Pluto
-Given the customary truth tables, which of the following theories is modeled by the above interpretation?
A) (∃x)(∼Px • Ix)
(∃x)∼Px
B) (∃x)(Px • ∼Ix) (∃x)∼Px
C) (∃x)(Px • Ix)
(∀x)(Px ⊃ Ix)
D) (∃x)(Ox • Ix)
(∃x)∼Px
E) (∃x)(Ox • ∼Px) (∀x)(Ix ⊃ Px)

Accepted Answer

B) is modeled by the given interpretation because there exists at least one object in the domain (e.g., Mars) that is P but not I, and there exists at least one object (Pluto) that is not P.

Question 19

consider the following domain, assignment of objects in the domain, and assignments sets to predicates.
Domain = {Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, Neptune, Pluto} 
P = {Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, Neptune}
I = {Mercury, Venus}
O = {Mars, Jupiter, Saturn, Uranus, Neptune}
a = Mercury b = Jupiter c = Saturn
d = Pluto
-Given the customary truth tables, which of the following theories is modeled by the above interpretation?
A) (∃x)(Px • Ix) • (∃x)(Px • ∼Ix) (∀x)[Px ⊃ (Ix • Ox)]
B) (∃x)(Ox • Ix) • (∃x)(Px • ∼Ix) (∀x)[Px ⊃ (Ix $ \lor $ Ox)]
C) (∃x)(Px • Ix) • (∃x)(Px • ∼Ix) (∀x)[(Ix $ \lor $ Ox) ⊃ Px]
D) (∃x)(Px • Ix) • (∃x)(Ix • ∼Px) (∀x)[Ix ⊃ (Px $ \lor $ Ox)]
E) (∃x)(Px • Ix) • (∃x)(Ox • ∼Px) (∀x)[Ox ⊃ (Ix $ \lor $ Px)]

Accepted Answer

- (∃x)(Px • Ix) - there exists an object that is both a planet and an inner planet
- (∃x)(Px • ∼Ix) - there exists an object that is a planet but not an inner planet
- (∀x)[(Ix $ \lor $ Ox) ⊃ Px] - for any object, if it is either an inner planet or an outer planet, then it is a planet
These assignments satisfy the conditions of choice C.

Question 20

consider the following domain, assignment of objects in the domain, and assignments sets to predicates.
Domain = {Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, Neptune, Pluto} 
P = {Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, Neptune}
I = {Mercury, Venus}
O = {Mars, Jupiter, Saturn, Uranus, Neptune}
a = Mercury b = Jupiter c = Saturn
d = Pluto
-Given the customary truth tables, which of the following theories is modeled by the above interpretation?
A) (∃x)[Px • (Ix • ∼Ox)] (∃x)[Px • (Ox • ∼Ix)] (∀x)[Px ⊃ (Ix ≡ ∼Ox)]
B) (∃x)[Px • (Ix • ∼Ox)] (∃x)[Px • (Ox • ∼Ix)] (∀x)[Px ⊃ (Ix ≡ Ox)]
C) (∃x)[Px • (Ix • Ox)] (∃x)[Px • (Ox • ∼Ix)] (∀x)[Px ⊃ (Ix ≡ ∼Ox)]
D) (∃x)[Px • (Ix • ∼Ox)] (∃x)[Px • (∼Ox • ∼Ix)]
(∀x)[Px ⊃ (Ix ≡ ∼Ox)]
E) (∃x)[Px • (Ix • ∼Ox)] (∃x)[Px • (Ox • ∼Ix)] (∀x)[Ix ⊃ (Px ≡ ∼Ox)]

Accepted Answer

The answer of consider the following domain, assignment of objects...

Quiz 4: Monadic Predicate Logic

1. (∀x) [Ax ⊃ (Bx ⊃ Cx) ] 2. ∼(∀x) (Bx ⊃ Dx) -Consider assuming '(∀x) Ax' for conditional proof. Which of the following propositions is an immediate (one-step) consequence in M of the given premises with that further assumption for conditional proof?

1. (∀x) [Ax ⊃ (Bx ⊃ Cx) ] 2. ∼(∀x) (Bx ⊃ Dx) -Which of the following propositions is derivable in M from the given premises?

1. (∀x) [(Px $\lor$ Qx) ⊃ (Rx • ∼Sx) ] 2. (∀x) [Rx ⊃ (Tx ⊃ ∼Sx) ] -Consider assuming 'Px' for conditional proof. Which of the following propositions is an immediate (one-step) consequence in M of the given premises with that further assumption for conditional proof?

1. (∀x) [(Px $\lor$ Qx) ⊃ (Rx • ∼Sx) ] 2. (∀x) [Rx ⊃ (Tx ⊃ ∼Sx) ] -Which of the following propositions is derivable in M from the given premises?

1. (∃x) Qx ⊃ (∀x) (Rx ⊃ Sx) 2. (∀x) ∼Qx ⊃ (∃x) Sx 3. (∀x) Rx / (∃x) Sx -Which of the following propositions is an appropriate assumption for an indirect proof of the conclusion of the given argument?

1. (∃x) Qx ⊃ (∀x) (Rx ⊃ Sx) 2. (∀x) ∼Qx ⊃ (∃x) Sx 3. (∀x) Rx / (∃x) Sx -Which of the following propositions is not a likely last line of the indented sequence for an indirect proof of the given argument?

(∃x) (Px • Qx) ⊃ [(∃x) Px • (∃x) Qx] -Consider assuming '(∃x) (Px • Qx) ' for a conditional proof of the above logical truth. Which Of the following propositions is a legitimate second step in that proof?

(∃x) (Px • Qx) ⊃ [(∃x) Px • (∃x) Qx] -Which of the following propositions is also derivable in M given that same assumption?

∼(∃x) (Px • Qx) ≡ (∀x) (Px ⊃ ∼Qx) -Consider assuming '∼(∃x) (Px • Qx) ' for a conditional proof of one direction of the biconditional in the above logical truth. Which of the following propositions is a legitimate second step in that proof?

∼(∃x) (Px • Qx) ≡ (∀x) (Px ⊃ ∼Qx) -Which of the following propositions is also derivable in M given that same assumption?

Things are pleasant if, and only if, they are not too crowded. Everything too crowded is noisy. So if something isn't noisy, then something is pleasant. -Which of the following is the best translation into M of this argument?

Things are pleasant if, and only if, they are not too crowded. Everything too crowded is noisy. So if something isn't noisy, then something is pleasant. -Which of the following propositions is a good assumption for conditional proof for the above argument?

Things are pleasant if, and only if, they are not too crowded. Everything too crowded is noisy. So if something isn't noisy, then something is pleasant. -Which of the following propositions can also be proved from the premises of the above argument?

Everything is material, or ideal or transcendental. All atoms are not ideal and not transcendental. And nothing is material. So there are no atoms. -Which of the following is the best translation into M of this argument?

Everything is material, or ideal or transcendental. All atoms are not ideal and not transcendental. And nothing is material. So there are no atoms. -Which of the following propositions is a good assumption for indirect proof for the above argument?

Everything is material, or ideal or transcendental. All atoms are not ideal and not transcendental. And nothing is material. So there are no atoms. -Which of the following propositions can also be proved from the premises of the above argument?