# 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.
$
\exists{x[C(x)\wedge}{F(x)\wedge}{\neg}{D(x)]}
$

No student in your class has a cat, a dog, and a ferret.
$
\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.
$
\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.
$
\forall{x[P(x)\rightarrow}{\neg}{S(x)]}
$

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

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

My poultry are not officers.
$
\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.
$
\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.
$
\forall\color{red}x\color{black}{\neg}{[C(x)\wedge}{D(x)\wedge}{F(x)]}
$
Correct, except that you missed out the $x$.

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.
$
\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:

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

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.
$
\forall{x[P(x)\rightarrow}{\neg}{S(x)]}
$
Correct.

No officers ever decline to waltz.
$
\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

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

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

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