Question 1

determine whether the given argument is valid or invalid. If it is invalid, select a counterexample.
-1. (∃x)(Jx • ∼Kx) ⊃ (∃x)(Lx • Mx)
2) (∃x)(Jx • Lx)
3) ∼(∃x)(Jx • Kx) / ∼(∀x)(Lx ⊃ Mx)
A) Valid
B) Invalid. Counterexample in a domain of 1 member, in which:
Ja: True Ka: False La: True Ma: True
C) Invalid. Counterexample in a domain of 1 member, in which:
Ja: False Ka: False La: True Ma: True
D) Invalid. Counterexample in a domain of 1 member, in which:
Ja: True Ka: True La: True Ma: True
E) Invalid. Counterexample in a domain of 1 member, in which:
Ja: True Ka: False La: False Ma: False

Accepted Answer

Assuming a domain of 1 member where Ja is true, Ka is false, La is true, and Ma is false, the premise (∃x)(Jx • ∼Kx) is true because there exists an x (in this case, the only member in the domain) that satisfies both Jx and ∼Kx. However, the conclusion (∃x)(Lx • Mx) is false because no x in the domain satisfies both Lx and Mx. Therefore, the argument is invalid and the correct answer is B.

Question 2

determine whether the given argument is valid or invalid. If it is invalid, select a counterexample.
-1. (?x)(Ax ? Bx) ? (?x)Cx
2) (?x)(Ax • ?Bx)
3) (?x)(Dx ? Bx) / (?x)(Dx ? Cx)
A) Valid
B) Invalid. Counterexample in a domain of 2 members, in which:
$$\begin{array} { l l l l } 
\text { Aa: True } & \text { Ba: False } & \text { Ca: False } & \text { Da: False } \
\text { Ab: False} &\text { Bb: True } &\text {Cb: False }&\text { Db: True } \
\end{array}$$
C) Invalid. Counterexample in a domain of 2 members, in which:
$$\begin{array} { l l l l } 
\text { Aa: True } & \text { Ba: False } & \text { Ca: False } & \text { Da: False } \
\text { Ab: True} &\text { Bb: True } &\text {Cb: True }&\text { Db: True } \
\end{array}$$
D) Invalid. Counterexample in a domain of 2 members, in which:
$$\begin{array} { l l l l } 
\text { Aa: False} & \text { Ba: False } & \text { Ca: False } & \text { Da: False } \
\text { Ab: False} &\text { Bb: True } &\text {Cb: False }&\text { Db: True } \
\end{array}$$
E) Invalid. Counterexample in a domain of 2 members, in which:
Aa: True Ba: False Ca: False Da: False Ab: True Bb: False Cb: False Db: True

Accepted Answer

The argument is invalid because the premises do not logically guarantee the conclusion. The counterexample provided shows a scenario where all premises are true, but the conclusion is false. This demonstrates that it's possible for the premises to be true without the conclusion necessarily being true, thus invalidating the argument.

Question 3

determine whether the given argument is valid or invalid. If it is invalid, select a counterexample.
-1. (?x)(Gx ? Hx)
2) (?x)(Gx • Ix)
3) (?x)(Gx • Jx)
4) (?x)(Jx ? Hx) / (?x)(Ix • Jx)
A) Valid
B) Invalid. Counterexample in a domain of 1 member, in which:
$$\begin{array} { l l l l } 
\text { Ga: True } & \text { Ha: True  } & \text { Ia: True  } & \text { Ja: True } \
\text { Gb: True} &\text { Hb: True } &\text {Ib: True }&\text { Jb: True } \
\end{array}$$
C) Invalid. Counterexample in a domain of 1 member, in which:
$$\begin{array} { l l l l } 
\text { Ga: False } & \text { Ha: True  } & \text { Ia: True  } & \text { Ja: True } \
\text { Gb: True} &\text { Hb: True } &\text {Ib: True }&\text { Jb: False } \
\end{array}$$
D) Invalid. Counterexample in a domain of 1 member, in which:
$$\begin{array} { l l l l } 
\text { Ga: True } & \text { Ha: True  } & \text { Ia: True  } & \text { Ja: True } \
\text { Gb: True} &\text { Hb: True } &\text {Ib: True }&\text { Jb: False } \
\end{array}$$
E) Invalid. Counterexample in a domain of 1 member, in which:
$$\begin{array} { l l l l } 
\text { Ga: False } & \text { Ha: True  } & \text { Ia: False   } & \text { Ja: False  } \
\text { Gb: False } &\text { Hb: True } &\text {Ib: True }&\text { Jb: False } \
\end{array}$$

Accepted Answer

The argument is invalid because the premises do not logically lead to the conclusion. In the counterexample provided in option D, the premises are true, but the conclusion is false, demonstrating the invalidity of the argument.

Question 4

determine whether the given argument is valid or invalid. If it is invalid, select a counterexample.
-1. (∃x)(Px • Qx)
2) (∃x)(Px • Rx)
3) (∀x)(Qx ⊃ Sx)
4) (∀x)(Rx ⊃ Tx) / (∃x)(Px • Sx) • (∃x)(Px • Tx)
A) Valid
B) Invalid. Counterexample in a domain of 1 member, in which:
Pa: True Qa: True Ra: True Sa: True Ta: True
C) Invalid. Counterexample in a domain of 1 member, in which:
Pa: False Qa: True Ra: True Sa: True Ta: True
D) Invalid. Counterexample in a domain of 1 member, in which:
Pa: True Qa: True Ra: False Sa: True Ta: True
E) Invalid. Counterexample in a domain of 1 member, in which:
Pa: True Qa: True Ra: True Sa: False Ta: True

Accepted Answer

The argument is valid because from the premises, it logically follows that there exists at least one x such that Px and Sx are true (due to the first and third premises), and there exists at least one x such that Px and Tx are true (due to the second and fourth premises).

Question 5

determine whether the given argument is valid or invalid. If it is invalid, select a counterexample.
-1. (∀x)[Fx ⊃ (Gx $ \lor $ Hx)]
2) (∀x)(Gx ⊃ ∼Ix)
3) ∼(∃x)(Hx • Ix)
4) Fg / ∼Ig
A) Valid
B) Invalid. Counterexample in a domain of 1 member, in which: Fg: True Gg: True Hg: True Ig: True
C) Invalid. Counterexample in a domain of 1 member, in which: Fg: True Gg: False Hg: True Ig: True
D) Invalid. Counterexample in a domain of 1 member, in which: Fg: True Gg: False Hg: False Ig: True
E) Invalid. Counterexample in a domain of 1 member, in which: Fg: True Gg: True Hg: False Ig: True

Accepted Answer

The argument is valid because if Fg is true, then Gg or Hg must be true (from 1). If Gg is true, then Ig must be false (from 2), satisfying the conclusion. If Hg is true, given that there cannot exist an x such that both Hx and Ix are true (from 3), and since Fg implies Gg or Hg, the conclusion ∼Ig follows logically.

Question 6

determine whether the given argument is valid or invalid. If it is invalid, select a counterexample.
-1. (∃x)[Ax • (Bx $ \lor $ Cx)]
2) (∀x)(Bx ⊃ ∼Cx)
3) (∃x)Bx
4) Ca / (∃x)(Ax • Cx)
A) Valid
B) Counterexample in a domain of 2 members, in which:
Aa: False Ba: False Ca: True
Ab: False Bb: False Cb: False
C) Counterexample in a domain of 2 members, in which: Aa: False Ba: False Ca: True
Ab: True Bb: True Cb: False
D) Counterexample in a domain of 2 members, in which: Aa: True Ba: True Ca: True
Ab: True Bb: True Cb: False
E) Counterexample in a domain of 2 members, in which: Aa: True Ba: True Ca: False
Ab: False Bb: True Cb: False

Accepted Answer

The argument is invalid. In the counterexample provided by option C, the premises are true, but the conclusion is false, demonstrating the invalidity of the argument.

Question 7

determine whether the given formula is a logical truth of M or not. If it is not a logical truth, select a false valuation.
-(∀x)(Ax ⊃ ∼Bx) $ \lor $ (∃x)(Ax • Bx)
A) Logical truth
B) Not a logical truth. False valuation in a domain of 2 members, in which: Aa: True Ba: False
Ab: False Bb: True
C) Not a logical truth. False valuation in a domain of 2 members, in which: Aa: True Ba: True
Ab: False Bb: True
D) Not a logical truth. False valuation in a domain of 2 members, in which: Aa: True Ba: True
Ab: False Bb: False
E) Not a logical truth. False valuation in a domain of 2 members, in which: Aa: False Ba: False
Ab: True Bb: True

Accepted Answer

The formula is a logical truth because it is a disjunction of two formulas, each of which is a logical truth.

Question 8

determine whether the given formula is a logical truth of M or not. If it is not a logical truth, select a false valuation.
-(∃x)(Cx • Dx) $ \lor $ (∃x)(Cx • ∼Dx)
A) Logical truth
B) Not a logical truth. False valuation in a domain of 1 member, in which:
Ca: True Da: False
C) Not a logical truth. False valuation in a domain of 1 member, in which: Ca: False Da: False
D) Not a logical truth. False valuation in a domain of 2 members, in which: Ca: True Da: True
Cb: True Db: False
E) Not a logical truth. False valuation in a domain of 2 members, in which: Ca: True Da: False
Cb: True Db: True

Accepted Answer

The answer of determine whether the given formula is a...

Question 9

determine whether the given formula is a logical truth of M or not. If it is not a logical truth, select a false valuation.
-[(∃x)Ex • (∃x)∼Ex)] ⊃ (∀x)(Ex $ \lor $ ∼Ex)
A) Logical truth
B) Not a logical truth. False valuation in a domain of 2 members, in which: Ea: True
Ea: False
C) Not a logical truth. False valuation in a domain of 2 members, in which: Ea: True
Ea: True
D) Not a logical truth. False valuation in a domain of 2 members, in which: Ea: False
Ea: False
E) Not a logical truth. False valuation in a domain of 2 members, in which: Ea: False
Ea: True

Accepted Answer

The formula is a logical truth because it states that if there exists an x for which E is true and there exists an x for which E is not true, then for all x, E is either true or not true. This is logically true by the principle of bivalence, which asserts that every proposition is either true or false. The antecedent (left side of the implication) does not need to be true for the consequent (right side) to be a logical truth, as the consequent is a tautology expressing the law of excluded middle.

Question 10

determine whether the given formula is a logical truth of M or not. If it is not a logical truth, select a false valuation.
-[(∀x)(Fx ⊃ Gx) • (∀x)(Gx ⊃ Hx)] ⊃ (∃x)(Fx • Hx)
A) Logical truth
B) Not a logical truth. False valuation in a domain of 1 member, in which: Fa: False Ga: False Ha: False
C) Not a logical truth. False valuation in a domain of 1 member, in which: Fa: True Ga: True Ha: False
D) Not a logical truth. False valuation in a domain of 1 member, in which: Fa: True Ga: True Ha: True
E) Not a logical truth. False valuation in a domain of 1 member, in which: Fa: True Ga: False Ha: True

Accepted Answer

To prove that it is not a logical truth, we need to find a false valuation. In this case, the domain has only one member, and we can assign false to all three predicates. With this valuation, the left side of the formula evaluates to false and the right side evaluates to false. Therefore, the implication is also false, which means the formula is not logically true.

Question 11

determine whether the given formula is a logical truth of M or not. If it is not a logical truth, select a false valuation.
-(∃x)(Ix • Jx) ⊃ [(∀x)Ix ⊃ (∀x)Jx]
A) Logical truth
B) Not a logical truth. False valuation in a domain of 2 members, in which: Ia: False Ja: True
Ib: True Jb: False
C) Not a logical truth. False valuation in a domain of 2 members, in which: Ia: True Ja: True
Ib: True Jb: False
D) Not a logical truth. False valuation in a domain of 2 members, in which: Ia: True Ja: True
Ib: False Jb: False
E) Not a logical truth. False valuation in a domain of 2 members, in which: Ia: True Ja: False
Ib: True Jb: False

Accepted Answer

The formula is not a logical truth because there exists a valuation where the antecedent is true, but the consequent is false. In option C, (∃x)(Ix • Jx) is true because there is an x (specifically 'a') for which both Ix and Jx are true. However, (∀x)Ix ⊃ (∀x)Jx is false because while (∀x)Ix is true, (∀x)Jx is false due to Jb being false.

Question 12

Translate each of the following sentences into predicate logic, using the given constants and predicates.
-Abhishek loves ice cream and pizza. (a: Abhishek; Ix: x loves ice cream; Px: x loves pizza)

Accepted Answer