http://en.wikipedia.org/wiki/Gottlob_Frege
http://en.wikipedia.org/wiki/First-order_logic
http://www.cs.odu.edu/~toida/nerzic/content/logic/pred_logic/intr_to_pred_logic.html
Predicate
logic
While propositional logic deals with simple declarative propositions, first-order logic additionally
covers predicates and quantification.
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Boolean algebra
Propositional logic
0, 1
True, False
boolean variables: a∈{0, 1}
atomic formulas: p∈{True,False}
boolean operators: ¬ ∧ ∨
logical connectives: ¬ ∧ ∨
boolean functions:
propositional formulas (propositions):
0, 1
True and False
boolean variables
atomic formulas
if a is a boolean function, then ¬a is a
boolean function
if a is a propositional formula, then ¬a
is a propositional formula
if a and b are boolean functions, then
a∧b, a∨b, a b, a b are boolean
if a and b are propositional formulas,
then a∧b, a∨b, a b, a b are
functions
truth value of a boolean function (truth
tables)
Boolean algebra and propositional logic are different ways of interpreting the same thing. Usage:
computer processors (circuits, switches).
Atomic formulas are also called atoms.
Propositions are called “Aussagen” in German
propositional formulas
truth value of a propositional formula (truth
tables)
What is propositional logic good for?
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.
lem
py prob
ssian s
The Ru
Consider the problem. How can we formalize it?
There are three persons: Stirlitz, Mueller, and Eismann. It is known that exactly
one of them is Russian, while the other two are Germans.
Introduce propositions (atomic formulas, boolean variables) X Y ,
where X ∈ {R, G, S} (denoting Russian, German, Spy)
Y ∈ {S,M,E} (denoting Stirlitz, Mu ̈ller, Eismann)
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.
For example,
SE : Eismann is a Spy
RS : Stirlitz is Russian
X ∈ {R, G, S} (denoting Russian, German, Spy)
Y ∈ {S,M,E} (denoting Stirlitz, Müller, Eismann)
SE : Eismann is a Spy
RS : 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).
If this set of formulas is satisfiable in some situation I, then
1. All conditions of the puzzle are satisfied by I.
2. I also satisfies the property that Eismann is a Russian spy.
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
If we demonstrate that the problem is unsatisfiable (it is a contradiction), then we prove that
Eismann cannot be a Russian spy.
GM.
Establish that Eismann is not a Russian spy.
RE ∧ SE.
Why formalisms?
Formalisms can describe computation and information.
computation
information
computer
information
Propositional logic not enough to reason about concepts like of “greater than 5”.
2 is greater than 5
False
6 is greater than 5
True
x is greater than 5
?
predicate
The value of “x is greater than 5” depends on the value of variable x.
This construct speaks about many simple propositions. Thus, it is not a propositional formula. We
need predicate logic.
Usage: software systems in financial services, automotive industry, medicine.
“ x is greater than 5” is a predicate, denoted it by F.
F(2) = 2 is greater than 5
False
F(6) = 6 is greater than 5
True
F(x) = x is greater than 5
A predicate describes a property of objects represented by variables from an universe (domain,
Bereichsangabe).
Any predicate can be transformed into a proposition by replacing variables with actual values.
predicate
universe
quantifier for all natural numbers x,
x is greater than 5
False
quantifier for some natural numbers x,
x is greater than 5
True
Universal quantifier ∀ :
Existential quantifiers ∃ :
∀x: U(x): F(x)
∃x: U(x): F(x)
∀x:x∈N:x>5
∃x:x∈N:x>5
for all natural numbers x,
x is greater than 5
for some natural numbers x,
x is greater than 5
Universal quantifier: ∀ ∀x: U(x): F(x)
“for every object x in the universe U, F(x) holds”, “for all (any) object x in the universe U, F(x)
holds”
Example: ∀x:x∈N:x>5
Note: ∀x: False: F(x) is True
Existential quantifier: ∃ ∃x: U(x): F(x)
“for some object x in the universe U, F(x) holds”, “there exist an object x in the universe U such
that F(x) holds”
Example: ∃ x:x∈N:x>5
Note: ∃x: False: F(x) is False
Gottlob Frege
1848—1925
no appeal to intuition
Distributivity of ∀ ∧
∀x: U(x): (P(x) ∧ Q(x))
(∀x: U(x): P(x)) ∧ (∀ x: U(x): Q(x))
Distributivity of ∃ ∨
∃x: U(x): (P(x) ∨ Q(x))
(∃x: U(x): P(x)) ∨ (∃ x: U(x): Q(x))
Non-distributivity of ∀ ∨
(∀x: U(x): P(x)) ∨ (∀x: U(x): Q(x))
∀x: U(x): (P(x) ∨ Q(x))
Non-distributivity of ∃ ∧
∃x: U(x): (P(x) ∧ Q(x))
(∃x: U(x): P(x)) ∧ (∃x: U(x): Q(x))
A couple of examples.
∀ i: 0≤i<N: ai=0
∃ i: 0≤i<N: ai ≠ 0
all elements ai of array a (of length N) are
zero
at least one element of array a (of length N) is
not zero
∀ i: 1≤i<N: ai-1 ≤ ai
∃ i: 1≤i<N: ai-1 > ai
elements of array a are monotonically
increasing
elements of array a are not monotonically
increasing
∀ i,j: 0≤i<j<N: ai ≤ aj
∃ i,j: 0≤i<j<N: ai > aj
elements of array a are monotonically
increasing
elements of array a are not monotonically
increasing
∀ i,j: 0≤i,j<N: ai = aj
∃ i,j: 0≤i<j<N: ai ≠ aj
all elements of array a are equal
not all elements of array a are equal
∀ i: 0<i<N: ai ≤ a0
∃ i: 0<i<N: ai > a0
a0 is the greatest element of array a
a0 is not the greatest element of array a
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∀ i: 0≤i<N: ai=0
∃ i: 0≤i<N: ai ≠ 0
all elements ai of array a (of length N) are
zero
at least one element of array a (of length N) is
not zero
∀ i: 1≤i<N: ai-1 ≤ ai
∃ i: 1≤i<N: ai-1 > ai
elements of array a are monotonically
increasing
elements of array a are not monotonically
increasing
∀ i,j: 0≤i<j<N: ai ≤ aj
∃ i,j: 0≤i<j<N: ai > aj
elements of array a are monotonically
increasing
elements of array a are not monotonically
increasing
∀ i,j: 0≤i,j<N: ai = aj
∃ i,j: 0≤i<j<N: ai ≠ aj
all elements of array a are equal
not all elements of array a are equal
∀ i: 0<i<N: ai ≤ a0
∃ i: 0<i<N: ai > a0
a0 is the greatest element of array a
a0 is not the greatest element of array a
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De Morgan’s Axiom ¬∃ :
¬(∃ x: U(x): F(x))
∀ x: U(x): ¬ F(x)
De Morgan’s Axiom ¬∀ :
¬(∀ x: U(x): F(x))
∃ x: U(x): ¬ F(x)
What is the negation of the statement
“Always, when he drives then he is sober.”
D(t): He drives at moment t.
S(t): He is sober at moment t.
∀t : D(t)->S(t)
¬(∀t :: D(t)->S(t))
¬(∀t :: ¬D(t)∨S(t))
∃t :: ¬(¬D(t)∨S(t))
∃t :: D(t)∧¬S(t))
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Examples:
- Rewrite (∃ i: 0≤i<N: ai≠0) into an equivalent WFF with ∀
- Rewrite (∀ i: 1≤i<N: ai-1<ai ) into an equivalent WFF with ∃
- Rewrite (∃ i,j: 0≤i<j<N: ai≠aj) into an equivalent WFF with ∀
Formalisms can describe computation.
We have talked about the computation part. Letʼs talk about the information part.
computation
information
computer
information
We say that x is a variable in Universe when it represents objects of the Universe U.
Well formed formulas (WFF)
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.
Bounded variables
An appearance of variable is bound in a WFF if
a specific value is assigned to it, or it is quantified.
∃i,j: 0<i,j<N: P(i,j) ∨ Q(i,j)
Free variables
If a variable is not bound in a WFF then
it is free.
∃i: 0<i<N: ( (∃j: 0<j<N: P(i,j) ∨ Q(i,j)) )
In the first proposition, it does not matter logically whether we use x or some other letter.
However, it could be confusing to use the same letter again elsewhere in some compound
proposition. That is, free variables become bound, and then in a sense retire from being available
as stand-in values for other values in the creation of formulae.
An appearance (Auftreten) of variable is bound (gebunden) in a WFF if
- a specific value is assigned to it or
- it is quantified.
If a variable is not bound in a WFF then it is free (frei).
The extent of the effect of a quantifier is called the scope (Einflussbereich) of the quantifier:
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))
Every boy loves some girl does not imply that
some 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))
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If P does not contain x as a free variable
∀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
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) )
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ambiguous
All elements of an array a of length N are either zero or one
(1)
∀i: 0≤i<N: ( ai=0 ∨ ai=1 )
(2)
(∀i: 0≤i<N: ai=0) ∨ (∀i: 0≤i<N: ai=1)
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ambiguous
The elements of arrays a and b are the same (both arrays of length N).
(1)
∀i: 0≤i<N: (ai=bi)
(2)
∀i: 1≤i<N: ((ai=ai-1) ∧ (bi=bi-1))
(3)
(∀i: 1≤i<N: (ai=ai-1)) ∧ (∀i: 0≤i<N: (ai=bi))
(4)
∀i: 0≤i<N: (∃ j: 0≤j<N: (ai=bi) )
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They drank two cups of tea because they were warm.
They drank two cups of tea because they were cold.
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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.
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Quantifiers for Math Reasoning.
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]) )
((∃ i: 0≤i<N: m=a[i]) ∧ (∀ i: 0≤i<N: m≤a[i]))
Quantifiers for Math Reasoning.
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))
Min x: U(x): T(x) = - (Max x: U(x): - T(x))
Quantifiers for Math Reasoning.
Anz x: U(x): F(x)
gives the “number of all those objects x
from 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)
Quantifiers for Math Reasoning.
Given arrays a and b.
Both arrays are of length N.
The elements of both arrays are natural numbers.
(Anz i: 0≤i<N: ai=0) = 0
all elements ai of a are different than 0
(Anz i: 0≤i<N: ai=x) > 0
at least one element ai has the value x (x∈N)
(Anz i: 0≤i<N: ai<x) = (Anz i: 0≤i<N: ai>x)
x is the median value of a
∀x:: ( (Anz i: 0≤i<N: ai=x) = (Anz i: 0≤i<N: bi=x) )
a is a permutation of b
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Formalisms can describe computation.
computation
information
computer
Tudor Gîrba
www.tudorgirba.com
creativecommons.org/licenses/by/3.0/
information