Negation of Quantifiers

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Discrete Structures

Predicate Logic 2

Dr. Muhammad Humayoun

Assistant Professor

COMSATS Institute of Computer Science, Lahore. mhumayoun@ciitlahore.edu.pk

https://sites.google.com/a/ciitlahore.edu.pk/dstruct/

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Negation of Quantifiers

• ¬∀𝑥 𝑃 𝑥 ≡ ∃¬𝑃 𝑥

• ¬∃𝑥 𝑃 𝑥 ≡ ∀¬𝑃 𝑥

• ¬∀𝑥 (𝑃 𝑥 ∧ 𝑄 𝑥 ) ≡

???

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Negation of Quantifiers

• ¬∀𝑥 𝑃 𝑥 ≡ ∃¬𝑃 𝑥

• ¬∃𝑥 𝑃 𝑥 ≡ ∀¬𝑃 𝑥

• ¬∀𝑥 (𝑃 𝑥 ∧ 𝑄 𝑥 ) ≡

???

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Negation of Quantifiers

• ¬∀𝑥 𝑃 𝑥 ≡ ∃¬𝑃 𝑥

• ¬∃𝑥 𝑃 𝑥 ≡ ∀¬𝑃 𝑥

• ¬∀𝑥 (𝑃 𝑥 ∧ 𝑄 𝑥 ) ≡

???

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Exercise

B(x): “x is a baby”

ignorant(x): “x is ignorant”

vain(x): “x is vain”

Universe: The set of all people.

• Babies are ignorant.

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Exercise

• Babies are ignorant. (Ambiguous)

• All/Some babies are ignorant

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Exercise

• Babies are ignorant. (Ambiguous)

• All babies are ignorant

• ∀𝒙 (𝑩 𝒙 → 𝒊𝒈𝒏𝒐𝒓𝒂𝒏𝒕 𝒙 )

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Exercise

professor(x): “x is a professor”

• No professors are ignorant.

• It is not the case that there exists an x such that x is a professor and x is ignorant.

• ¬∃𝑥(𝑝𝑟𝑜𝑓𝑒𝑠𝑠𝑜𝑟 𝑥 ∧ 𝑖𝑔𝑛𝑜𝑟𝑎𝑛𝑡 𝑥 )

• It is not the case that all professors are ignorant.

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Exercise

• [There is no such professor who is ignorant]

• [It is not the case that there is an x such that x is a professor and x is ignorant.]

• It is not the case that all professors are ignorant.

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Exercise

professors are ignorant.

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Exercise

• ¬∃𝑥(𝑝𝑟𝑜𝑓𝑒𝑠𝑠𝑜𝑟 𝑥 ∧ 𝑖𝑔𝑛𝑜𝑟𝑎𝑛𝑡 𝑥 ) • All professors are not ignorant

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Exercise

• All (and all of them) professors are not ignorant.

• ∀𝑥 𝑝𝑟𝑜𝑓𝑒𝑠𝑠𝑜𝑟 𝑥 → ¬𝑖𝑔𝑛𝑜𝑟𝑎𝑛𝑡 𝑥

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Exercise

professor(x): “x is a professor” ignorant(x): “x is ignorant”

vain(x): “x is vain”

• All ignorant people are vain.

• For all people x, if x is ignorant then x is vain. • ∀𝑥(𝑖𝑔𝑜𝑟𝑎𝑛𝑡 𝑥 → 𝑣𝑎𝑖𝑛 𝑥 )

• It is logically equivalent to ¬[¬(∀𝑥 𝑖𝑔𝑜𝑟𝑎𝑛𝑡 𝑥 → 𝑣𝑎𝑖𝑛 𝑥 ) ]

• ≡ ¬[¬(∀𝑥 ¬𝑖𝑔𝑜𝑟𝑎𝑛𝑡 𝑥 ∨ 𝑣𝑎𝑖𝑛 𝑥 ]

• ≡ ¬ ∃𝑥 ¬¬𝑖𝑔𝑜𝑟𝑎𝑛𝑡 𝑥 ∧ ¬𝑣𝑎𝑖𝑛 𝑥

• ≡ ¬ ∃𝑥 𝑖𝑔𝑜𝑟𝑎𝑛𝑡 𝑥 ∧ ¬𝑣𝑎𝑖𝑛 𝑥

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Exercise

• For all people x, if x is ignorant then x is vain.

• ∀𝑥(𝑖𝑔𝑜𝑟𝑎𝑛𝑡 𝑥 → 𝑣𝑎𝑖𝑛 𝑥 )

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Exercise

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Exercise

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Exercise

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Exercise

professor(x): “x is a professor” ignorant(x): “x is ignorant”

vain(x): “x is vain”

• _{No professors are vain}

• It is not the case that there is an x such that x is professor and x is vain.

• ¬[∃𝑥(𝑝𝑟𝑜𝑓𝑒𝑠𝑠𝑜𝑟 𝑥 ∧ 𝑣𝑎𝑖𝑛(𝑥))]

• [∀𝑥(¬𝑝𝑟𝑜𝑓𝑒𝑠𝑠𝑜𝑟 𝑥 ∨ ¬𝑣𝑎𝑖𝑛(𝑥))]

• ∀𝑥 𝑝𝑟𝑜𝑓𝑒𝑠𝑠𝑜𝑟 𝑥 → ¬𝑣𝑎𝑖𝑛 𝑥

• _{For all people x, if x is a professor then x not vain.}

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Exercise

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Exercise

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Precedence of Quantifiers

• _{The quantifiers}

∀

and

∃

have higher

precedence then all logical operators from

propositional calculus.

• e.g.

∀𝒙 (𝑷 𝒙 ∨ 𝑸 𝒙 )

is the disjunction

of

∀𝒙 𝑷 𝒙

and

∀𝒙 𝑸(𝒙)

.

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Quantifiers with Restricted Domain

• ∀𝑥 𝑥2 > 0

– ∀ 𝑥 < 0 𝑥2 > 0

– ≡ ∀𝑥 (𝑥 < 0) → (𝑥2> 0)

• ∀𝑦 𝑦3 ≠ 0

– ∀ 𝑦 ≠ 0 𝑦3 ≠ 0

– ≡ ∀ 𝑦 ≠ 0 → 𝑦3 ≠ 0

• ∃𝑧 (𝑧2 = 2)

– ∃𝑧 > 0 (𝑧2 = 2)

– ≡ ∃𝑧(𝑧 > 0 → 𝑧2 = 2)

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Quantifiers with Restricted Domain

• ∀𝑥 𝑥2 > 0

– ∀ 𝑥 < 0 𝑥2 > 0

– ≡ ∀𝑥 (𝑥 < 0) → (𝑥2> 0)

• ∀𝑦 𝑦3 ≠ 0

– ∀ 𝑦 ≠ 0 𝑦3 ≠ 0

– ≡ ∀ 𝑦 ≠ 0 → 𝑦3 ≠ 0

• ∃𝑧 (𝑧2 = 2)

– ∃𝑧 > 0 (𝑧2 = 2)

– ≡ ∃𝑧(𝑧 > 0 → 𝑧2 = 2)

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Quantifiers with Restricted Domain

• ∀𝑥 𝑥2 > 0

– ∀ 𝑥 < 0 𝑥2 > 0

– ≡ ∀𝑥 (𝑥 < 0) → (𝑥2> 0)

• ∀𝑦 𝑦3 ≠ 0

– ∀ 𝑦 ≠ 0 𝑦3 ≠ 0

– ≡ ∀ 𝑦 ≠ 0 → 𝑦3 ≠ 0

• ∃𝑧 (𝑧2 = 2)

– ∃𝑧 > 0 (𝑧2 = 2)

– ≡ ∃𝑧(𝑧 > 0 → 𝑧2 = 2)

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Quantifiers with Restricted Domain

• ∀𝑥 𝑥2 > 0

– ∀ 𝑥 < 0 𝑥2 > 0

– ≡ ∀𝑥 (𝑥 < 0) → (𝑥2> 0)

• ∀𝑦 𝑦3 ≠ 0

– ∀ 𝑦 ≠ 0 𝑦3 ≠ 0

– ≡ ∀ 𝑦 ≠ 0 → 𝑦3 ≠ 0

• ∃𝑧 (𝑧2 = 2)

– ∃𝑧 > 0 (𝑧2 = 2)

– ≡ ∃𝑧(𝑧 > 0 → 𝑧2 = 2)

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Nested Quantifiers

∀𝑥∃𝑦 𝑃(𝑥, 𝑦)

 “For all 𝑥, there exists a 𝑦 such that 𝑃(𝑥, 𝑦)”.

 Example:

 ∀𝑥∃𝑦 (𝑥 + 𝑦 = 0) where 𝑥 and 𝑦 are integers

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Nested Quantifiers

 ∀𝑥∃𝑦 𝑃(𝑥, 𝑦)

 “For all 𝑥, there exists a 𝑦 such that 𝑃(𝑥, 𝑦)”.

 Example:

 ∀𝑥∃𝑦 (𝑥 + 𝑦 = 0) where 𝑥 and 𝑦 are integers

 𝑥𝑦 𝑃(𝑥, 𝑦)

 There exists an x such that for all 𝑦, 𝑃(𝑥, 𝑦) is true”

 Example: 𝑥𝑦 (𝑥 × 𝑦 = 0)

• ∀𝑥∀𝑦∀𝑧 𝑥 + 𝑦 + 𝑧 = 𝑥 + 𝑦 + 𝑧

• THINK QUANTIFICATION AS LOOPS

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Meanings of multiple quantifiers

Suppose 𝑷(𝒙, 𝒚) = “x likes y.”

Domain of x: {St1, St2}; Domain of y: {DS, Calculus}

• 𝒙𝒚 𝑷(𝒙, 𝒚)

– 𝑃(𝑥, 𝑦) true for all x, y pairs. • 𝒙𝒚 𝑷(𝒙, 𝒚)

– 𝑃(𝑥, 𝑦) true for at least one x, y pair. • 𝒙𝒚 𝑷(𝒙, 𝒚)

– For every value of x we can find a (possibly different) y so that P(x,y) is true.

• 𝒙𝒚 𝑷(𝒙, 𝒚)

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Meanings of multiple quantifiers

• 𝒙𝒚 𝑷(𝒙, 𝒚)

– 𝑃(𝑥, 𝑦) true for all x, y pairs.

• 𝒙𝒚 𝑷(𝒙, 𝒚)

– 𝑃(𝑥, 𝑦) true for at least one x, y pair. • 𝒙𝒚 𝑷(𝒙, 𝒚)

• 𝒙𝒚 𝑷(𝒙, 𝒚)

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Meanings of multiple quantifiers

• 𝒙𝒚 𝑷(𝒙, 𝒚)

• 𝒙𝒚 𝑷(𝒙, 𝒚)

– 𝑃(𝑥, 𝑦) true for at least one x, y pair.

• 𝒙𝒚 𝑷(𝒙, 𝒚)

• 𝒙𝒚 𝑷(𝒙, 𝒚)

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Meanings of multiple quantifiers

• 𝒙𝒚 𝑷(𝒙, 𝒚)

• 𝒙𝒚 𝑷(𝒙, 𝒚)

– 𝑃(𝑥, 𝑦) true for at least one x, y pair.

• 𝒙𝒚 𝑷(𝒙, 𝒚)

• 𝒙𝒚 𝑷(𝒙, 𝒚)

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• Quantification order is not commutative

𝒙𝒚 𝑷 𝒙, 𝒚 ≠ 𝒙𝒚 𝑷(𝒙, 𝒚)

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Example

𝑄 (𝑥 , 𝑦, 𝑧): x + y = z

Domain: Real numbers

• ∀𝑥∀𝑦∃𝑧 𝑄(𝑥, 𝑦, 𝑧) True/False???

• For all real numbers x and for all real numbers y there is a real number z such that 𝑥 + 𝑦 = 𝑧.

• True

• ∃𝑧∀𝑥∀𝑦 𝑄(𝑥, 𝑦, 𝑧) True/False???

• There is a real number z such that for all real

numbers x and for all real numbers y it is true that

𝑥 + 𝑦 = 𝑧.

• False

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Example

𝑄 (𝑥 , 𝑦, 𝑧): x + y = z

• True

numbers x and for all real numbers y it is true that

𝑥 + 𝑦 = 𝑧.

• False

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Example

𝑄 (𝑥 , 𝑦, 𝑧): x + y = z

• True

numbers x and for all real numbers y it is true that 𝑥 + 𝑦 = 𝑧.

• False

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Example

𝑄 (𝑥 , 𝑦, 𝑧): x + y = z

• True

numbers x and for all real numbers y it is true that 𝑥 + 𝑦 = 𝑧.

• False

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From Nested Quantifiers to English

• F (a, b): “a and b are friends”

• Domain: All students in COMSATS.

∃𝑥∀𝑦∀𝑧 𝐹 𝑥, 𝑦 ∧ 𝐹 𝑥, 𝑧 ∧ 𝑦 ≠ 𝑧 → ¬𝐹 𝑦, 𝑧

• There is a student x such that for all students y and all

students z other than y, if x and y are friends and x and z are friends, then y and z are not friends.

• There is a student none of whose friends are also

friends with each other.

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From Nested Quantifiers to English

• F (a, b): “a and b are friends”

• Domain: All students in COMSATS.

∃𝑥∀𝑦∀𝑧 𝐹 𝑥, 𝑦 ∧ 𝐹 𝑥, 𝑧 ∧ 𝑦 ≠ 𝑧 → ¬𝐹 𝑦, 𝑧

• There is a student x such that for all students y and all students z other than y, if x and y are friends and x and z are friends, then y and z are not friends.

• There is a student none of whose friends are also friends with each other.

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From English to Nested Quantifiers

• "If a person is female and is a parent, then this person is someone's mother“

• For every person x , if person x is female and person x is a parent, then there exists a person y such that person x is the mother of person y.“

– F(x): “x is female”

– P(x): “x is a parent“

– M(x, y) : “x is the mother of y”

• ∀𝑥[(𝑃 𝑥 ∧ 𝐹(𝑥)) → ∃𝑦𝑀 𝑥, 𝑦 ]

• ∀𝑥∃𝑦[(𝑃 𝑥 ∧ 𝐹(𝑥)) → 𝑀 𝑥, 𝑦 ]

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From English to Nested Quantifiers

• "If a person is female and is a parent, then this person is someone's mother“

• For every person x , if person x is female and person x is a parent, then there exists a person y such that person x is the mother of person y.“

– F(x): “x is female”

– P(x): “x is a parent“

– M(x, y) : “x is the mother of y”

• ∀𝑥[(𝑃 𝑥 ∧ 𝐹(𝑥)) → ∃𝑦𝑀 𝑥, 𝑦 ]

• ∀𝑥∃𝑦[(𝑃 𝑥 ∧ 𝐹(𝑥)) → 𝑀 𝑥, 𝑦 ]

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• The sum of two positive integers is always positive.

• ∀𝑥 ∀𝑦 [ 𝑥 > 0 ∧ 𝑦 > 0 → 𝑥 + 𝑦 > 0 ]

• ∀𝑥 ∀𝑦 [𝑝𝑜𝑠 𝑥 ∧ 𝑝𝑜𝑠 𝑦 → 𝑝𝑜𝑠 𝑥 + 𝑦 ]

• What is domain above? • Integers

• If domain is “+ve integers”

• ∀𝑥 ∀𝑦 𝑥 + 𝑦 > 0

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• ∀𝑥 ∀𝑦 [ 𝑥 > 0 ∧ 𝑦 > 0 → 𝑥 + 𝑦 > 0 ] • ∀𝑥 ∀𝑦 [𝑝𝑜𝑠 𝑥 ∧ 𝑝𝑜𝑠 𝑦 → 𝑝𝑜𝑠 𝑥 + 𝑦 ] • What is domain above?

• Integers

• ∀𝑥 ∀𝑦 𝑥 + 𝑦 > 0

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• ∀𝑥 ∀𝑦 [ 𝑥 > 0 ∧ 𝑦 > 0 → 𝑥 + 𝑦 > 0 ] • ∀𝑥 ∀𝑦 [𝑝𝑜𝑠 𝑥 ∧ 𝑝𝑜𝑠 𝑦 → 𝑝𝑜𝑠 𝑥 + 𝑦 ] • What is domain above?

• Integers

• ∀𝑥 ∀𝑦 𝑥 + 𝑦 > 0

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• Everyone has exactly one best friend

• For every person x , person x has exactly one best friend.

• ∀𝑥 [person x has exactly one best friend]

• B(x,y): “x has best friend y”