03 - Predicate logic

Predicatelogic

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Boolean algebra Propositional logic0, 1 True, False

boolean variables: a∈{0, 1} atomic formulas: p∈{True,False}

boolean operators: ¬ ∧ ∨ ⇒ ⇔ logical connectives: ¬ ∧ ∨ ⇒ ⇔

boolean functions:

boolean variables

if a is a boolean function, then ¬a is a boolean function

if a and b are boolean functions, then a∧b, a∨b, a⇒b, a⇔b are boolean

functions

propositional formulas (propositions):

True and False

atomic formulas

if a is a propositional formula, then ¬a is a propositional formula

if a and b are propositional formulas, then a∧b, a∨b, a⇒b, a⇔b are

propositional formulas

truth value of a boolean function (truth tables)

truth value of a propositional formula (truth tables)

Boolean algebra Propositional logic0, 1 True, False

boolean variables: a∈{0, 1} atomic formulas: p∈{True,False}

boolean operators: ¬ ∧ ∨ ⇒ ⇔ logical connectives: ¬ ∧ ∨ ⇒ ⇔

boolean functions:

boolean variables

if a is a boolean function, then ¬a is a boolean function

if a and b are boolean functions, then a∧b, a∨b, a⇒b, a⇔b are boolean

functions

propositional formulas (propositions):

True and False

atomic formulas

if a is a propositional formula, then ¬a is a propositional formula

if a and b are propositional formulas, then a∧b, a∨b, a⇒b, a⇔b are

propositional formulas

truth value of a boolean function (truth tables)

truth value of a propositional formula (truth tables)

The Russian spy problem

There are three persons: Stirlitz, Mueller, and Eismann. It is known that exactly one of them is Russian, while the other two are Germans.

Moreover, every Russian must be a spy.

When Stirlitz meets Mueller in a corridor, he makes the following joke: “you know, Mueller, you are as German as I am Russian”. It is known that Stirlitz always says the truth when he is joking.

Establish that Eismann is not a Russian spy.

X ∈ {R, G, S} (denoting Russian, German, Spy)Y ∈ {S,M,E} (denoting Stirlitz, Müller, Eismann)

SE : Eismann is a SpyRS : Stirlitz is Russian

There are three persons: Stirlitz, Mueller, and Eismann. It is known that exactly one of them is Russian, while the other two are Germans.(RS ∧ GM ∧ GE) ∨ (GS ∧ RM ∧ GE) ∨ (GS ∧ GM ∧ RE).(RS ⇔ ¬GS) ∧ (RM ⇔ ¬GM) ∧ (RE ⇔ ¬GE).

Moreover, every Russian must be a spy.(RS ⇒ SS) ∧ (RM ⇒ SM) ∧ (RE ⇒ SE).

When Stirlitz meets Mueller in a corridor, he makes the following joke: “you know, Mueller, you are as German as I am Russian”. It is known that Stirlitz always says the truth when he is joking.RS ⇔ GM.

Establish that Eismann is not a Russian spy.RE ∧ SE.

computerinformation information

computation

2 is greater than 5

2 is greater than 5 False

2 is greater than 5

6 is greater than 5

2 is greater than 5

6 is greater than 5

2 is greater than 5

6 is greater than 5

x is greater than 5

2 is greater than 5

6 is greater than 5

x is greater than 5 ?

2 is greater than 5

6 is greater than 5

x is greater than 5 ?predicate

x is greater than 5

2 is greater than 5

6 is greater than 5

predicate

x is greater than 5F(x) =

2 is greater than 5

6 is greater than 5

predicate

2 is greater than 5

6 is greater than 5F(6) =

F(2) = False

predicate

2 is greater than 5

F(2) = False

for all natural numbers x,x is greater than 5

predicate

2 is greater than 5

F(2) = False

for all natural numbers x,x is greater than 5 False

predicate

2 is greater than 5

F(2) = False

for some natural numbers x,x is greater than 5

predicate

2 is greater than 5

F(2) = False

predicate

2 is greater than 5

F(2) = False

predicate

quantifier

2 is greater than 5

F(2) = False

predicate

quantifier

2 is greater than 5

F(2) = False

predicate

quantifier

universe

Existential quantifiers ∃ :Universal quantifier ∀ :

∀x: U(x): F(x) ∃x: U(x): F(x)

∀x:x∈N:x>5 ∃x:x∈N:x>5for all natural numbers x,x is greater than 5

Gottlob Frege1848—1925

no appeal to intuition

Distributivity of ∀ ∧∀x: U(x): (P(x) ∧ Q(x))

Distributivity of ∃ ∨∃x: U(x): (P(x) ∨ Q(x))

Non-distributivity of ∀ ∨(∀x: U(x): P(x)) ∨ (∀x: U(x): Q(x))

Non-distributivity of ∃ ∧∃x: U(x): (P(x) ∧ Q(x))

⇔ (∀x: U(x): P(x)) ∧ (∀ x: U(x): Q(x))

⇔ (∃x: U(x): P(x)) ∨ (∃ x: U(x): Q(x))

⇒ ∀x: U(x): (P(x) ∨ Q(x))

⇒ (∃x: U(x): P(x)) ∧ (∃x: U(x): Q(x))

Examples

∀ i: 0≤i<N: ai=0

Examples

∀ i: 0≤i<N: ai=0all elements ai of array a (of length N) are zero

Examples

∀ i: 1≤i<N: ai-1 ≤ ai

Examples

∀ i: 1≤i<N: ai-1 ≤ aielements of array a are monotonically increasing

Examples

∀ i,j: 0≤i<j<N: ai ≤ aj

Examples

∀ i,j: 0≤i<j<N: ai ≤ ajelements of array a are monotonically increasing

Examples

∀ i,j: 0≤i,j<N: ai = aj

Examples

∀ i,j: 0≤i,j<N: ai = ajall elements of array a are equal

Examples

∀ i: 0<i<N: ai ≤ a0

Examples

∀ i: 0<i<N: ai ≤ a0a0 is the greatest element of array a

Examples

∃ i: 0≤i<N: ai ≠ 0

Examples

∃ i: 0≤i<N: ai ≠ 0at least one element of array a (of length N) is not zero

Examples

∃ i: 1≤i<N: ai-1 > ai

Examples

∃ i: 1≤i<N: ai-1 > aielements of array a are not monotonically increasing

Examples

∃ i,j: 0≤i<j<N: ai > aj

Examples

∃ i,j: 0≤i<j<N: ai > ajelements of array a are not monotonically increasing

Examples

∃ i,j: 0≤i<j<N: ai ≠ aj

Examples

∃ i,j: 0≤i<j<N: ai ≠ ajnot all elements of array a are equal

Examples

∃ i: 0<i<N: ai > a0

Examples

∃ i: 0<i<N: ai > a0a0 is not the greatest element of array a

Examples

∃ i: 0<i<N: ai > a0a0 is not the greatest element of array a

Examples

¬(∃ x: U(x): F(x)) ⇔ ∀ x: U(x): ¬ F(x)

¬(∀ x: U(x): F(x)) ⇔ ∃ x: U(x): ¬ F(x)

De Morgan’s Axiom ¬∃ :

De Morgan’s Axiom ¬∀ :

Example

What is the negation of the statement“Always, when he drives then he is sober.”

Example

D(t): He drives at moment t.

Example

D(t): He drives at moment t.S(t): He is sober at moment t.

Example

D(t): He drives at moment t.S(t): He is sober at moment t.∀t : D(t)->S(t)

Example

¬(∀t :: D(t)->S(t)) ⇔

Example

¬(∀t :: D(t)->S(t)) ⇔¬(∀t :: ¬D(t)∨S(t)) ⇔

Example

¬(∀t :: D(t)->S(t)) ⇔¬(∀t :: ¬D(t)∨S(t)) ⇔ ∃t :: ¬(¬D(t)∨S(t)) ⇔

Example

¬(∀t :: D(t)->S(t)) ⇔¬(∀t :: ¬D(t)∨S(t)) ⇔ ∃t :: ¬(¬D(t)∨S(t)) ⇔∃t :: D(t)∧¬S(t))

computation

True, False are WFFs.

Propositional variables are WFFs.

Predicates with variables are WFFs.

If A and B are WFFs, then ¬A, A∧B, A∨B, A⇒B, A⇔B are also WFFs.

If x is a variable in universe U and A is a WFF, then∀x:U(x):A(x) and ∃x:U(x):A(x) are also WFFs.

Well formed formulas (WFF)

Bounded variables

(∃j: 0<j<N: P(i,j) ∨ Q(i,j))

Free variablesIf a variable is not bound in a WFF thenit is free.

An appearance of variable is bound in a WFF ifa specific value is assigned to it, or it is quantified.

∃i: 0<i<N: (

∃i,j: 0<i,j<N: P(i,j) ∨ Q(i,j)

Bounded variables

(∃j: 0<j<N: P(i,j) ∨ Q(i,j))

Free variablesIf a variable is not bound in a WFF thenit is free.

An appearance of variable is bound in a WFF ifa specific value is assigned to it, or it is quantified.

∃i: 0<i<N: ( )

∃i,j: 0<i,j<N: P(i,j) ∨ Q(i,j)

∃x: U1(x): (∀y: U2(y): Q(x,y)) ⇔ ∀y: U2(y): (∃x: U1(x): Q(x,y))∃x: U1(x): (∀y: U2(y): Q(x,y)) ⇒ ∀y: U2(y): (∃x: U1(x): Q(x,y))∀y: U2(y): (∃x: U1(x): Q(x,y)) ⇒ ∃x: U1(y): (∀y: U2(y): Q(x,y))

Swapping ∀ and ∃∀x: U1(x): (∀y: U2(y): Q(x,y)) ⇔ ∀y: U2(y): (∀x: U1(x): Q(x,y))∃x: U1(x): (∃y: U2(y): Q(x,y))⇔ ∃y: U2(y): (∃x: U1(x): Q(x,y))

However

∃x: U1(x): (∀y: U2(y): Q(x,y)) ⇔ ∀y: U2(y): (∃x: U1(x): Q(x,y))∃x: U1(x): (∀y: U2(y): Q(x,y)) ⇒ ∀y: U2(y): (∃x: U1(x): Q(x,y))∀y: U2(y): (∃x: U1(x): Q(x,y)) ⇒ ∃x: U1(y): (∀y: U2(y): Q(x,y))

Swapping ∀ and ∃∀x: U1(x): (∀y: U2(y): Q(x,y)) ⇔ ∀y: U2(y): (∀x: U1(x): Q(x,y))∃x: U1(x): (∃y: U2(y): Q(x,y))⇔ ∃y: U2(y): (∃x: U1(x): Q(x,y))

However

Example

Every boy loves some girl does not imply thatsome girl is loved by every boy.

U1 - the universe of boys.U2 - the universe of girls.Q(x,y) - boy x loves girl y.

∀y: U2(y): (∃x: U1(x): Q(x,y)) ⇒ ∃x: U1(y): (∀y: U2(y): Q(x,y))

∀x: U(x): (Q(x) ∨ P) ⇔ (∀x: U(x): Q(x)) ∨ P∀x: U(x): (Q(x) ∧ P) ⇔ (∀x: U(x): Q(x)) ∧ P

∃x: U(x): (Q(x) ∨ P) ⇔ (∃x: U(x): Q(x)) ∨ P∃x: U(x): (Q(x) ∧ P) ⇔ (∃x: U(x): Q(x)) ∧ P

If P does not contain x as a free variable

Example

Given: an array a[0..N-1] of length N≥3

Find: i, j, k such that:

i is the index of the greatest element of a;

j is the index of the 2nd greatest element of a;

k is the index of the 3rd greatest element of a.

Example

Given: an array a[0..N-1] of length N≥3

Find: i, j, k such that:

i is the index of the greatest element of a;

j is the index of the 2nd greatest element of a;

k is the index of the 3rd greatest element of a.

∃ i,j,k: (0≤i,j,k<N)∧(i≠j)∧(i≠k)∧(j≠k): ( (ai≥aj≥ak) ∧ (∀m: (0≤m<N)∧(m≠i)∧(m≠j)∧(m≠k): ak≥am) )

Example

All elements of an array a of length N are either zero or one

Example

∀i: 0≤i<N: ( ai=0 ∨ ai=1 )

Example

∀i: 0≤i<N: ( ai=0 ∨ ai=1 )

(∀i: 0≤i<N: ai=0) ∨ (∀i: 0≤i<N: ai=1)

Example

∀i: 0≤i<N: ( ai=0 ∨ ai=1 )

(∀i: 0≤i<N: ai=0) ∨ (∀i: 0≤i<N: ai=1)

ambiguous

Example

The elements of arrays a and b are the same (both arrays of length N).

Example

∀i: 0≤i<N: (ai=bi)(1)

Example

∀i: 0≤i<N: (ai=bi)

∀i: 1≤i<N: ((ai=ai-1) ∧ (bi=bi-1))

Example

∀i: 0≤i<N: (ai=bi)

∀i: 1≤i<N: ((ai=ai-1) ∧ (bi=bi-1))

(∀i: 1≤i<N: (ai=ai-1)) ∧ (∀i: 0≤i<N: (ai=bi))(3)

Example

∀i: 0≤i<N: (ai=bi)

∀i: 1≤i<N: ((ai=ai-1) ∧ (bi=bi-1))

(∀i: 1≤i<N: (ai=ai-1)) ∧ (∀i: 0≤i<N: (ai=bi))(3)

∀i: 0≤i<N: (∃ j: 0≤j<N: (ai=bi) )(4)

Example

∀i: 0≤i<N: (ai=bi)

∀i: 1≤i<N: ((ai=ai-1) ∧ (bi=bi-1))

ambiguous

(∀i: 1≤i<N: (ai=ai-1)) ∧ (∀i: 0≤i<N: (ai=bi))(3)

∀i: 0≤i<N: (∃ j: 0≤j<N: (ai=bi) )(4)

Example

They drank two cups of tea because they were warm.

Example

They drank two cups of tea because they were warm.

They drank two cups of tea because they were cold.

Example

Go to the shop and buy bread.If they have eggs, buy 10.

Example

Go to the shop and buy bread.If they have eggs, buy 10.

...Do you have eggs?Yes.Then give me 10 pieces of bread.

Example

Min x: U(x): T(x)gives the “smallest value of terms T(x)for all objects x from the universe U“

Min x: False: T(x) is +∞.

( m = (Min i: 0≤i<N: a[i]) ) ⇔

( m = (Min i: 0≤i<N: a[i]) ) ⇔((∃ i: 0≤i<N: m=a[i]) ∧ (∀ i: 0≤i<N: m≤a[i]))

Max x: U(x): T(x)gives the “greatest value of terms T(x)for all objects x from the universe U“

Max x: False: T(x) is -∞.

Max x: U(x): T(x) = - (Min x: U(x): - T(x))

Max x: U(x): T(x) = - (Min x: U(x): - T(x))Min x: U(x): T(x) = - (Max x: U(x): - T(x))

Anz x: U(x): F(x)gives the “number of all those objects xfrom the universe U for which F(x) holds“

Anz x: False: F(x) is 0.

((Anz x: U(x): F(x)) = 0) ⇔ ∀x: U(x):¬F(x)

((Anz x: U(x): F(x)) = 0) ⇔ ∀x: U(x):¬F(x)((Anz x: U(x): F(x)) ≥ 0) ⇔ ∃x: U(x): F(x)

Given arrays a and b.Both arrays are of length N.The elements of both arrays are natural numbers.

Example

(Anz i: 0≤i<N: ai=0) = 0

Example

(Anz i: 0≤i<N: ai=0) = 0all elements ai of a are different than 0

Example

(Anz i: 0≤i<N: ai=x) > 0

Example

(Anz i: 0≤i<N: ai=x) > 0at least one element ai has the value x (x∈N)

Example

(Anz i: 0≤i<N: ai<x) = (Anz i: 0≤i<N: ai>x)

Example

(Anz i: 0≤i<N: ai<x) = (Anz i: 0≤i<N: ai>x)x is the median value of a

Example

∀x:: ( (Anz i: 0≤i<N: ai=x) = (Anz i: 0≤i<N: bi=x) )

Example

∀x:: ( (Anz i: 0≤i<N: ai=x) = (Anz i: 0≤i<N: bi=x) )a is a permutation of b

Example

computation

Tudor Gîrbawww.tudorgirba.com

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03 - Predicate logic

spy rs

rs gs rm gm

denoting russian

russian spy problem

rs gm ge gs rm ge gs

denoting stirlitz

rs ss rm sm

boolean operators

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