Let C(x) be the statement "x has a cat," let D(x) be the statement "x has a dog," and let F(x) be the statement "x has a ferret." Express each of these statements in terms of C(x), D(x), F(x), quantifiers, and logical connectives. Let the domain consist of all students in your class. a) A student in your class has a cat, a dog, and a ferret. b) All students in your class have a cat, a dog, or a ferret. c) Some student in your class has a cat and a ferret, but not a dog. d) No student in your class has a cat, a dog, and a ferret. e) For each of the three animals, cats, dogs, and ferrets, there is a student in your class who has this animal as a pet.

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Answer:

[tex]\mathbf{a)} \left( \exists x \in X\right) \; C(x) \; \wedge \; D(x) \; \wedge \; F(x)\\\mathbf{b)} \left( \forall x \in X\right) \; C(x) \; \vee \; D(x) \; \vee \; F(x)\\\mathbf{c)} \left( \exists x \in X\right) \; C(x) \; \wedge \; F(x) \; \wedge \left(\neg \; D(x) \right)\\\mathbf{d)} \left( \forall x \in X\right) \; \neg C(x) \; \vee \; \neg D(x) \; \vee \; \neg F(x)\\\mathbf{e)} \left((\exists x\in X)C(x) \right) \wedge \left((\exists x\in X) D(x) \right) \wedge \left((\exists x\in X) F(x) \right)[/tex]

Step-by-step explanation:

Let [tex]X[/tex] be a set of all students in your class. The set [tex]X[/tex] is the domain. Denote

                                        [tex]C(x) - ' \text{$x $ has a cat}'\\D(x) - ' \text{$x$ has a dog}'\\F(x) - ' \text{$x$ has a ferret}'[/tex]

[tex]\mathbf{a)}[/tex]

Consider the statement 'A student in your class has a cat, a dog, and a ferret'. This means that [tex]\exists x \in X[/tex] so that all three statements C(x), D(x) and F(x) are true. We can express that in terms of C(x), D(x) and F(x) using quantifiers, and logical connectives as follows

                         [tex]\left( \exists x \in X\right) \; C(x) \; \wedge \; D(x) \; \wedge \; F(x)[/tex]

[tex]\mathbf{b)}[/tex]

Consider the statement 'All students in your class have a cat, a dog, or a ferret.' This means that [tex]\forall x \in X[/tex] at least one of the statements C(x), D(x) and F(x) is true. We can express that in terms of C(x), D(x) and F(x) using quantifiers, and logical connectives as follows

                        [tex]\left( \forall x \in X\right) \; C(x) \; \vee \; D(x) \; \vee F(x)[/tex]

[tex]\mathbf{c)}[/tex]

Consider the statement 'Some student in your class has a cat and a ferret, but not a dog.' This means that [tex]\exists x \in X[/tex] so that the statements C(x), F(x) are true and the negation of the statement D(x) . We can express that in terms of C(x), D(x) and F(x) using quantifiers, and logical connectives as follows

                      [tex]\left( \exists x \in X\right) \; C(x) \; \wedge \; F(x) \; \wedge \left(\neg \; D(x) \right)[/tex]

[tex]\mathbf{d)}[/tex]

Consider the statement 'No student in your class has a cat, a dog, and a ferret..' This means that [tex]\forall x \in X[/tex] none of  the statements C(x), D(x) and F(x) are true. We can express that in terms of C(x), D(x) and F(x) using quantifiers, and logical connectives as a negation of the statement in the part a), as follows

[tex]\neg \left( \left( \exists x \in X\right) \; C(x) \; \wedge \; D(x) \; \wedge \; F(x)\right) \iff \left( \forall x \in X\right) \; \neg C(x) \; \vee \; \neg D(x) \; \vee \; \neg F(x)[/tex]

[tex]\mathbf{e)}[/tex]

Consider the statement ' For each of the three animals, cats, dogs, and ferrets, there is a student in your class who has this animal as a pet.'

This means that for each of the statements C, F and D there is an element from the domain [tex]X[/tex] so that each statement holds true.

We can express that in terms of C(x), D(x) and F(x) using quantifiers, and logical connectives as follows

           [tex]\left((\exists x\in X)C(x) \right) \wedge \left((\exists x\in X) D(x) \right) \wedge \left((\exists x\in X) F(x) \right)[/tex]

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