Single Quantifier
In this section, we will see how to translate simpler statements between English to predicate logic. These translations will involve a single quantifier.
Example: Predicate logic to English
Suppose our domain is animals and that we have the following two predicates:
isMouse(x)
: whether animalx
is a mouseinHouse(x)
: whether animalx
is in the house
Suppose we also have that Squeaky
is an individual in our domain.
We will practice translating from predicate logic to English. Think about what the following propositions mean, and click to reveal each answer:

isMouse(Squeaky) ∧ ¬inHouse(Squeaky)

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> "Squeaky is a mouse, and Squeaky is not in the house."


∃ x isMouse(x)

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> "There is a mouse".


¬(∃ x isMouse(x))

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> "There is not a mouse."


∃ x ¬isMouse(x)

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> "There is an animal that is not a mouse".


∀ x isMouse(x)

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> "All animals are mice."


¬(∀ x isMouse(x))

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> "Not all animals are mice."


∀ x ¬isMouse(x)

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> "All animals are not mice."


∀ x (isMouse(x) → inHouse(x))

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> "All mice are in the house."


∀ x (isMouse(x) ∧ inHouse(x))

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> "Every animal is a mouse and is in the house." (We usually don't want ∧ with ∀.)


¬(∀ x (isMouse(x) → inHouse(x)))

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> "Not all mice are in the house."


∀ x (inHouse(x) → isMouse(x))

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> "Everything in the house is a mouse."


¬(∀ x (inHouse(x) → isMouse(x)))

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> "Not everything in the house is a mouse."


∃ x (isMouse(x) ∧ inHouse(x))

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> "There is a mouse in the house."


∃ x (isMouse(x) → inHouse(x))

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> "There exists an animal, and if that animal is a mouse, then it is in the house." Recall that this statement will be true if there is an animal that is NOT a mouse (since the → would be vacuously true) as well as being true if there is a mouse in the house.


¬(∃ x (isMouse(x) ∧ inHouse(x)))

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> "There is not a mouse in the house."

Translation guide
When translating from English to predicate logic, you can look for particular wording in your sentences to see how to choose a quantifier and/or negation placement. We will also see that certain phrases can be translated multiple (equivalent) ways.

Every/all/each/any is translated as:
∀ x ...

Some/at least one/there exists/there is is translated as:
∃ x ...

None/no/there does not exist can be translated as either
¬(∃ x ...)
or∀ x ¬(...)

Not every/not all can be translated as either
¬(∀ x ...)
or∃ x ¬(...)

Some Pish thing is a Qish thing is translated as:
∃ x (P(x) ∧ Q(x))

All Pish things are Qish things is translated as:
∀ x (P(x) → Q(x))

No Pish thing is a Qish thing can be translated as either
¬(∃ x (P(x) ∧ Q(x)))
or∀ x (P(x) → ¬Q(x))

Not all Pish things are Qish things can be translated as either
¬(∀ x (P(x) → Q(x)))
or∃ x (P(x) ∧ ¬Q(x))
DeMorgan’s laws for quantifiers
In the translation guide above, we saw that we could often translate the same statement two different ways – one way using an existential quantifier and one way using a universal quantifier. These equivalencies are another iteration of DeMorgan’s laws, this time applied to predicate logic.
Suppose we have some domain, and that P(x)
is a predicate for individuals in that domain. DeMorgan’s laws give us the following equivalencies:
¬(∃ x P(x))
is equivalent to∀ x ¬P(x)
¬(∀ x P(x))
is equivalent to∃ x ¬P(x)
In Chapter 6, we will learn to prove that these translations are indeed equivalent.
Example: English to predicate logic
Suppose our domain is people and that we have the following two predicates:
K(x)
: whether personx
is a kidM(x)
: whether personx
likes marshmallows
We will practice translating from English to predicate logic. Think about what the following sentences mean, and click to reveal each answer:
 No kids like marshmallows.

Click here for solution
¬(∃ x (K(x) ∧ M(x))
, or equivalently,∀ x (K(x) → ¬M(x))

 Not all kids like marshmallows.

Click here for solution
¬(∀ x (K(x) → M(x))
, or equivalently,∃ x (K(x) ∧ ¬M(x))

 Everyone who likes marshmallows is a kid.

Click here for solution
∀ x (M(x) → K(x))

 Some people who like marshmallows are not kids.

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∃ x (M(x) ∧ ¬K(x))

 Some kids don’t like marshmallows.

Click here for solution
∃ x (K(x) ∧ ¬M(x))

 Anyone who doesn’t like marshmallows is not a kid.

Click here for solution
∀ x (¬M(x) → ¬K(x))

Evaluating predicate logic statements on a toy domain
Suppose we have the following toy domain of people with the following characteristics:
 Bob, age 10, lives in Kansas, has siblings, has brown hair
 Jane, age 25, lives in Delaware, has no siblings, has blonde hair
 Alice, age 66, lives in Kansas, has siblings, has gray hair
 Joe, age 50, lives in Nebraska, has siblings, has black hair
Now suppose that we have the following predicates for individuals in our domain:
Ad(x)
: whether personx
is an adult (adults are age 18 and older)KS(x)
: whether personx
lives in KansasSib(x)
: whether personx
has siblingsRed(x)
: whether personx
has red hair
We will practice evaluating predicate logic statements on our domain of people. Think about whether the following propositions would be true or false over our domain, and then click to reveal each answer:

∀ x Ad(x)

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This proposition translates as, "All people are adults". This is false for our domain, as we have one person (Bob) who is not an adult.


∀ x ¬Ad(x)

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This proposition translates as, "All people are not adults". This is false for our domain, as we have three people (Jane, Alice, and Joe) are are adults.


¬(∀ x Ad(x))

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This proposition translates as, "Not all people are adults". This is true for our domain, as we can find a person (Bob) who is not an adult.


∀ x (KS(x) → Sib(x))

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This proposition translates as, "Everyone who lives in Kansas has siblings". This is true for our domain, as we have two people who live in Kansas (Bob and Alice), and both of them have siblings.


∃ x (¬KS(x) ∧ Sib(x))

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This proposition translates as, "There is a person who doesn't live in Kansas and has siblings". This is true for our domain, as Joe lives in Nebraska and has siblings.


¬(∃ x (KS(x) ∧ ¬Ad(x))

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This proposition translates as, "There does not exist a person who lives in Kansas and is not an adult". This is false for our domain, as Bob lives in Kansas and is not an adult.


¬(∃ x (Sib(x) ∧ Red(x))

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This proposition translates as, "There does not exist a person with siblings who has red hair". This is true for our domain, as no one with siblings (Bob, Alice, or Joe) has red hair.


∀ x (Red(x) → Sib(x))

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This proposition translates as, "All people with red hair have siblings". This is true for our domain, as no one has red hair. This means that the implies statement is vacuously true for every person (since `Red(x)` is false for each person), which makes the overall proposition true.


∀ x (KS(x) ∨ Sib(x))

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This proposition translates as, "Everyone lives in Kansas and/or has siblings". This is false for our domain  there is one person, Jane, who doesn't live in Kansas and also doesn't have siblings.
