Question 2 In the following exercises, you are given an argument in quantified relational logic. Your task is to use models to check whether the arguments are valid. Take your time to remind yourself the definition of a valid argument. So the idea here is that if an argument is valid, then it is impossible to find a model that makes all the premises true and the conclusion false. But if an argument is invalid, then there is at least one model that makes the all of the premises true and the conclusion false. For example the argument: (1) ∀x(Fx→Gx) (2) ¬∃x(Fx∧¬Gx) is valid. It is impossible to find a model that makes (1) true and the conclusion (2) false. The first step then is to negate the conclusion (2) which results in ¬¬∃x(F x ∧ ¬Gx), which is equivalent to ∃x(F x ∧ ¬Gx). The second step is to see if it is possible to find a model that makes both the premise (1) true and the negated conclusion true. But in this example, it is impossible. Here’s why. The negated conclusion ∃x(F x ∧ ¬Gx) is true if for some a ∈ U, (Fa∧¬Ga). But (Fa∧¬Ga) and ∀x(Fx → Gx) are contradictory. So it is impossible to find a model that makes both ¬¬∃x(F x ∧ ¬Gx) and ∀x(F x → Gx) true, which means the argument is valid. On the other hand, the argument: (3) ∀x(Gx → Hx) (4) ¬∃x(Fx∧Gx) (5) ¬∃x(Hx→Fx) is invalid. Here is a model that makes (3) and (4) true. • U = {a}, • a = a, • [F]={a}, • [G] = {}, i.e., [G] is empty, • [H]={a}. Since there’s no counterexample to (3), (3) is true. Similarly, (4) is true in the model because (4) is equivalent to ∀x¬(F x ∧ Gx) and there is no counterexample to ∀x¬(F x ∧ Gx). But (5) is false in the model because (5) is equivalent to ∀x(Hx ∧ ¬Fx) and a is a counterexample. Since it possible to find a model that makes the premises true and the conclusion false, the argument is invalid. In the following exercises determine which of the following arguments are valid. If an argument is valid, say “VALID” and explain why. If an argument is invalid, say “INVALID” and find at least one model that makes all the premises true and the conclusion false. Since different students can come up with different models, there are multiple possible answers when an argument is “INVALID”. So, be creative but start with simple models first! (a) 1. ∀x(Fx→Gx) 2. ∀x(Fx→Hx) ∴ 3. ∀x(Gx → Hx) (b) 1. ∃xFx 2. ∃xGx 3. ∃x(Fx∧Gx) (c) 1. (Fa∧¬Ga) 2. ∀x(Fx∧Gx) (d) 1. ∀x(Fx→Gx) 2. ∃x(¬Fx∧¬Gx)

Question 2 In the following exercises, you are given an argument in quantified relational logic. Your task is to use models to check whether the arguments are valid. Take your time to remind yourself the definition of a valid argument. So the idea here is that if an argument is valid, then it is impossible to find a model that makes all the premises true and the conclusion false. But if an argument is invalid, then there is at least one model that makes the all of the premises true and the conclusion false. For example the argument: (1) ∀x(Fx→Gx) (2) ¬∃x(Fx∧¬Gx) is valid. It is impossible to find a model that makes (1) true and the conclusion (2) false. The first step then is to negate the conclusion (2) which results in ¬¬∃x(F x ∧ ¬Gx), which is equivalent to ∃x(F x ∧ ¬Gx). The second step is to see if it is possible to find a model that makes both the premise (1) true and the negated conclusion true. But in this example, it is impossible. Here’s why. The negated conclusion ∃x(F x ∧ ¬Gx) is true if for some a ∈ U, (Fa∧¬Ga). But (Fa∧¬Ga) and ∀x(Fx → Gx) are contradictory. So it is impossible to find a model that makes both ¬¬∃x(F x ∧ ¬Gx) and ∀x(F x → Gx) true, which means the argument is valid. On the other hand, the argument: (3) ∀x(Gx → Hx) (4) ¬∃x(Fx∧Gx) (5) ¬∃x(Hx→Fx) is invalid. Here is a model that makes (3) and (4) true. • U = {a}, • a = a, • [F]={a}, • [G] = {}, i.e., [G] is empty, • [H]={a}. Since there’s no counterexample to (3), (3) is true. Similarly, (4) is true in the model because (4) is equivalent to ∀x¬(F x ∧ Gx) and there is no counterexample to ∀x¬(F x ∧ Gx). But (5) is false in the model because (5) is equivalent to ∀x(Hx ∧ ¬Fx) and a is a counterexample. Since it possible to find a model that makes the premises true and the conclusion false, the argument is invalid. In the following exercises determine which of the following arguments are valid. If an argument is valid, say “VALID” and explain why. If an argument is invalid, say “INVALID” and find at least one model that makes all the premises true and the conclusion false. Since different students can come up with different models, there are multiple possible answers when an argument is “INVALID”. So, be creative but start with simple models first! (a) 1. ∀x(Fx→Gx) 2. ∀x(Fx→Hx) ∴ 3. ∀x(Gx → Hx) (b) 1. ∃xFx 2. ∃xGx 3. ∃x(Fx∧Gx) (c) 1. (Fa∧¬Ga) 2. ∀x(Fx∧Gx) (d) 1. ∀x(Fx→Gx) 2. ∃x(¬Fx∧¬Gx)

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Question 2

In the following exercises, you are given an argument in quantified relational logic. Your task is to use models to check whether the arguments are valid. Take your time to remind yourself the definition of a valid argument.

So the idea here is that if an argument is valid, then it is impossible to find a model that makes all the premises true and the conclusion false. But if an argument is invalid, then there is at least one model that makes the all of the premises true and the conclusion false.

For example the argument:

(1) ∀x(Fx→Gx)

(2) ¬∃x(Fx∧¬Gx)

is valid. It is impossible to find a model that makes (1) true and the conclusion (2) false. The first step then is to negate the conclusion (2) which results in ¬¬∃x(F x ∧ ¬Gx), which is equivalent to ∃x(F x ∧ ¬Gx). The second step is to see if it is possible to find a model that makes both the premise (1) true and the negated conclusion true. But in this example, it is impossible. Here’s why. The negated conclusion ∃x(F x ∧ ¬Gx) is true if for some a ∈ U, (Fa∧¬Ga). But (Fa∧¬Ga) and ∀x(Fx → Gx) are contradictory. So it is impossible to find a model that makes both ¬¬∃x(F x ∧ ¬Gx) and ∀x(F x → Gx) true, which means the argument is valid.

On the other hand, the argument:

(3) ∀x(Gx → Hx)
(4) ¬∃x(Fx∧Gx)

(5) ¬∃x(Hx→Fx)
is invalid. Here is a model that makes (3) and (4) true.

• U = {a},
• a = a,
• [F]={a},
• [G] = {}, i.e., [G] is empty, • [H]={a}.

Since there’s no counterexample to (3), (3) is true. Similarly, (4) is true in the model because (4) is equivalent to ∀x¬(F x ∧ Gx) and there is no counterexample to ∀x¬(F x ∧ Gx). But (5) is false in the model because (5) is equivalent to ∀x(Hx ∧ ¬Fx) and a is a counterexample. Since it possible to find a model that makes the premises true and the conclusion false, the argument is invalid.

In the following exercises determine which of the following arguments are valid. If an argument is valid, say “VALID” and explain why. If an argument is invalid, say “INVALID” and find at least one model that makes all the premises true and the conclusion false. Since different students can come up with different models, there are multiple possible answers when an argument is “INVALID”. So, be creative but start with simple models first!

(a)
1. ∀x(Fx→Gx)

2. ∀x(Fx→Hx) ∴ 3. ∀x(Gx → Hx)

(b)

1. ∃xFx

2. ∃xGx

3. ∃x(Fx∧Gx)

2. ∀x(Fx∧Gx)

(d)
1. ∀x(Fx→Gx)

2. ∃x(¬Fx∧¬Gx)

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