# Thread: Predicate Logic converted to English Statements

1. ## Predicate Logic converted to English Statements

domain: all students in your class
C(x): "x has a cat"
D(x): "x has a dog"
F(x): "x has a ferret"

Some student in your class has a cat and a ferret, but not a dog.
$\displaystyle \exists{x[C(x)\wedge}{F(x)\wedge}{\neg}{D(x)]}$

No student in your class has a cat, a dog, and a ferret.
$\displaystyle \forall{\neg}{[C(x)\wedge}{D(x)\wedge}{F(x)]}$

For each of the three animals, cats, dogs, and ferrets, there is a student in your class who has one of these animals as pets.
$\displaystyle \forall{x}\exists{x[C(x)\vee}{D(x)\vee}{F(x)]}$

The next question:

P(x): "x is a duck"
Q(x): "x is one of my poultry"
R(x): "x is an officer"
S(x): "x is willing to waltz"

No ducks are willing to waltz.
$\displaystyle \forall{x[P(x)\rightarrow}{\neg}{S(x)]}$

No officers ever decline to waltz.
$\displaystyle \forall{x[}{\neg}{R(x)\rightarrow}{\neg}{S(x)]}$

All my poultry are ducks.
$\displaystyle \forall{x[Q(x)\rightarrow}{P(x)]}$

My poultry are not officers.
$\displaystyle \forall{x[Q(x)\rightarrow}{\neg}{R(x)]}$

2. ## Predicate Logic

Hello aaronrj
Originally Posted by aaronrj

domain: all students in your class
C(x): "x has a cat"
D(x): "x has a dog"
F(x): "x has a ferret"

Some student in your class has a cat and a ferret, but not a dog.
$\displaystyle \exists{x[C(x)\wedge}{F(x)\wedge}{\neg}{D(x)]}$
Correct.

No student in your class has a cat, a dog, and a ferret.
$\displaystyle \forall\color{red}x\color{black}{\neg}{[C(x)\wedge}{D(x)\wedge}{F(x)]}$
Correct, except that you missed out the $\displaystyle x$.

For each of the three animals, cats, dogs, and ferrets, there is a student in your class who has one of these animals as pets.
$\displaystyle \forall{x}\exists{x[C(x)\vee}{D(x)\vee}{F(x)]}$
No. You can re-phrase it as "Someone owns a cat AND someone owns a dog AND someone owns a ferret". So you get:

$\displaystyle \exists x[C(x)] \land \exists x ...$

I'm sure you can complete it now.

The next question:

P(x): "x is a duck"
Q(x): "x is one of my poultry"
R(x): "x is an officer"
S(x): "x is willing to waltz"

No ducks are willing to waltz.
$\displaystyle \forall{x[P(x)\rightarrow}{\neg}{S(x)]}$
Correct.

No officers ever decline to waltz.
$\displaystyle \forall{x[}{\neg}{R(x)\rightarrow}{\neg}{S(x)]}$
No. This means "If you're not an officer, then you're not willing to waltz."

"No officers ever decline to waltz" simply means "Every officer is willing to waltz". So

$\displaystyle \forall x [R(x)\rightarrow S(x)]$

All my poultry are ducks.
$\displaystyle \forall{x[Q(x)\rightarrow}{P(x)]}$

My poultry are not officers.
$\displaystyle \forall{x[Q(x)\rightarrow}{\neg}{R(x)]}$
Both of these are correct!

,

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### what is the logical statement of some student in your class has a cat and ferret but not a dog

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