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Today’s topics:• Associative and distributive properties of composition• Powers of a relation• Inverse relation

Homework #3 is posted (due on June 26)

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Theorem. Let R AB, S BC and T CD denote three binaryrelations. Then relation compositions satisfy the following associative law:

(RS)T = R(ST)

Properties of composition

RS

A C B

DT R S

ST

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Theorem. Let R AB, S BC and T CD denote three binaryrelations. Then relation compositions satisfy the following associative law:

(RS)T = R(ST)

Proof. First note that both sides define a relation from A to D. We also note the following:(RS)T = {(a, d) | aA and dD and cC [(a, c)RS cTd ]}

(by the definition of composition T )

= {(a, d) |… (cC, bB) [(aRb bSc ) cTd ]}(by definition of RS )

= {(a, d) |… (cC, bB) [aRb (bSc cTd )]}(by associative property of )

= {(a, d) |… (bB) [aRb (b, d ) ST ]}(by definition of ST )

= R(ST) (by the definition of composition R )

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Associative property of composition means, thatno ambiguity arises if we simply write RST .

We can also define powers of a relation RAA recursivelyas R1=R and Rn+1=RnR (n 1)

Definition. A path in a relation R is a sequence a0, … , ak with k 0 such that (ai, ai+1) R for every i<k . We call k the length of the path.

5•

1•

7•

6•

4 •

3•

2•

A

a0

a1

a2

a3

1, 5, 7, 6 – a path of length 3

5

a1

a0 a2

a3

R (parent-of)

a1

RR= R2 (grandparent-of)

a0 a2

a3

RRR= R3

a1

a0 a2

a3

R4=

a1

a0 a2

a3

Powers of a relation

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Theorem. (a, b) Rn iff there is a path of length n in R.(will be proved later by induction)

Ra

c

d

b

R2

b

a d

c

R3

a

bc

d

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Proof 1) If RAA is reflexive, then for any aA, (a, a) R. To prove that R2 is reflexive, we need to show that for any aA, (a, a) R2, i. e. you can ‘return back to a in two steps’.This is obviously true, because you can simply repeat two loops.

Theorem. 1) If a relation R AA is reflexive, then R2 is also reflexive 2) If a relation R AA is symmetric, then R2 is also symmetric3) If a relation R AA is transitive, then R2 is also transitive.

Note, that we can prove in general, that if (a, a) R,then (a, a) Rn for any n > 0 (repeat loop n times)

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Proof. 2) R AA is symmetric R2 is symmetric

a b

R2

?

assume (a, b) R2 cA, (a, c) R and (c, b) R

c

a b

R

(b, c) R and (c, a) R

a b

Rc

(b, a) R2

a

b

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Proof 3) Assume R is transitive. To prove that R2 is transitive, assume that there exist (a, b) R2 and (b, c) R2 (otherwise R2

is ‘vacuously’ transitive) to show that (a, c) R2 .

Sketch of the proof

a bR2

c

a b

x A

c y A

a b

(a, b) R x

(b, c) R

yc

a b

x

yc(a, c) R2

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It is interesting also to see how the composition distributes over union and intersection.

Theorem. Let R AB, S BC and T BC denote three binary relations. Then

1) R(ST) = (RS)(RT)2) R(ST) (RS)(RT)

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1) R(ST) = (RS)(RT)

R = {(a1, b1), (a1, b2), (a2, b4)}

S = {(b1, c2), (b4, c1)} T ={(b4, c4)}

ST = {(b1, c2), (b4, c1), (b4, c4)}

AB C

R S

T

a1

a2

a3

c1

c2

c3

c4

b1

b2b3

b4

R(ST) = {(a1, c2), (a2, c1), (a2, c4)}

RS = {(a1, c2), (a2, c1)}, RT = {(a2, c4)},

(RS)(RT) ={(a1, c2), (a2, c1), (a2, c4)}

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Proof (1) R(ST)={(a, c)|aA and cC and

bB [aRb (b, c) ST]}… …..dfn of R

= {(a, c)|… bB [aRb ( bSc bT c)]}……………dfn of

= {(a, c)|… bB [(aRb bSc) (aRb bT c)]}……distributive law

= {(a, c)|… [bB (aRb bSc)] [bB (aRb bT c)]}

by x[p(x) q(x)] [x p(x)] [x q(x)]

= {(a, c)|… [(a, c)RS ] [(a, c)RT ]}

by dfn of composition

= {(a, c)|… (a, c)RS RT }

by dfn of

= RS RT

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Proof (2) R(ST) (RS)(RT)

Take an element (pair) (a, c) R(ST)

aA, cC and bB[aRb (b, c)(ST)]

dfn of composition

… bB[aRb (bSc bTc)]

dfn of ST

… bB[(aRb bSc)(aRbbTc)]

associative and idempotent laws

… [bB(aRb bSc)][bB(aRbbTc)]

by x[p(x) q(x)] [x p(x)] [x q(x)]

… [(a, c) RS] [(a, c) RT]

by dfn of composition

(a, c) (RS)(RT)

by dfn of

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The converse subset relation (RS)(RT) R(ST) can be disproved by the following counterexample:

R

R

S

T

a

b

c

d

R = {(a, b), (a, c)}

S = {(b, d )}

T = {(c, d )}

RS = {(a, d )} RT = {(a, d )} (RS)(RT)= {(a, d )}

ST = R(ST)= (RS)(RT) R(ST)/

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Definition. Inverse of relation R is defined as:R-1 = {(a, b) | (b, a)R} .

•Inverse of “parent-of” is “child of”.

•If R AB , then R-1 BA

• It is just relation “turned backward”.

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Prove or disprove that RR-1 is reflexive.

A B

R AB a1 b1

R-1 BA

a2 b2

A

a1

a2

RR-1 AA

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Theorem. Let R be a binary relation from A to B and S be a binary relation from B to C, so that RS is a composite relation from A to C. Then the inverse of this relation

(RS)-1 = S-1 R-1

Proof. (RS)-1 is a binary relation from C to A. (RS)-1 = {(c, a) | (a, c) RS }…….. by the definition of inverse

= {(c, a) | b B, aRb and bSc}…. by dfn of composition= {(c, a) | b B, (b, a) R-1 and (c, b) S-1 }

by dfn of inverse = {(c, a) | b B, (c, b) S-1 and (b, a)R-1}

by commutative property of and = {(c, a) | b B, (c, a) S-1R-1}

by dfn of composition= S-1 R-1

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Closures of relations

A closure “extends” a relation to satisfy some property.But extends it as little as possible.

Definition. The closure of relation R with respect to property P is the relation S that

i) contains Rii) satisfies property Piii) is contained in any relation satisfying i) and ii). That is

S is the “smallest” relation satisfying i) and ii).

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Just a reminder:

• reflexive: aRa for any a A• symmetric: aRb bRa• transitive: aRbbRc aRc• anti-symmetric: aRbbRa a = b,

i. e. a b, aRb bRa

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Lemma 1. The reflexive closure of R is S = R {(a, a) | aA } =r(R).

S contains R and is reflexive by design. Furthermore, any relation satisfying i) must contain R, any satisfying ii) mustcontain the pairs (a, a), so any relation satisfying both i) and ii)must contain S

Proof. In accordance to the definition of a closure, we need to prove three things to show that S is reflexive closure:

i) S contains R ii) S is reflexive iii) S is the smallest relation satisfying i) and ii).

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Example: R={(a, b), (a, c), (b, d), (d, e)}

a b c d ea 0 1 1 0 0 b 0 0 0 1 0c 0 0 0 0 0d 0 0 0 0 1e 0 0 0 0 0

R

a b c d ea 1 1 1 0 0 b 0 1 0 1 0c 0 0 1 0 0d 0 0 0 1 1e 0 0 0 0 1

S = r (R)

a •

b • c

•

••e d

R

a •

•c •

••e d

b

r (R)

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Sometimes the relation on A that consists of all loops is called identity relation on A, i. e.

IA={(a, a)|aA}Then reflexive closure of a relation RAA is

r(R)=RIA

Example of a reflexive closure. Let A be any set and consider the relation on Power(A)

R = {(x, y)Power(A) Power(A) | xy}

The reflexive closure of R would be the relation: RIPower(A)={(x, y)Power(A)Power(A)|(x, y)R or (x, y) IPower(A)}

={(x, y)Power(A)Power(A)| xy or x=y} ={(x, y)Power(A)Power(A)| x y}

Thus, the reflexive closure of the ‘proper subset’ relation is the ‘subset’ relation.

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Consider the relation ‘less then’ on A={1, 2, 3, 4, 5}

R = {(1, 2), (1, 3), (1, 4), (1, 5), (2, 3),(2, 4), (2, 5), (3, 4), (3, 5), (4, 5)}

What is the reflexive closure of R?

1 2

3

4

5

R

r(R)

1 2

3

4

5

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Lemma 2. The symmetric closure of R is S = R R-1 =s(R)

Proof. S is symmetric and contains R. It is also the smallest such .For suppose we have some symmetric relation T with R T.To show that T contains S we need to show that R-1 T. Take any (a, b) R-1 , it implies (b, a) R , and (b, a) T, since R T. But then (a, b) T as well, because T is symmetric by assumption. So, S = R R-1 T .

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Example: R={(a, b), (a, c), (b, d), (d, e)}

a b c d ea 0 1 1 0 0 b 0 0 0 1 0c 0 0 0 0 0d 0 0 0 0 1e 0 0 0 0 0

R a b c d ea 0 1 1 0 0 b 1 0 0 1 0c 1 0 0 0 0d 0 1 0 0 1e 0 0 0 1 0

s (R)

a b

c

d

e

R

a

b

c

d

e

s(R)

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Theorem. The transitive closure of a relation R is the setS = {(a, b) | there is a path from a to b in R}= t (R)

Proof. Let’s show that S is transitive. Suppose (a, b) S and (b, c) S. This means there is a path from a to b and a path from b to c in R. If we “concatenate” them (attach the end of the (a, b)-path to the start of the (b, c)-path) we get a path from a to c. Thus, (a, c) S and S is transitive. We need to show, that any transitive relation T containing R contains S .Let’s prove by contradiction. Assume that there exists a transitive relation T containing R, that is smaller then S.

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More formally: assume by the way of contradiction that thereexists a transitive relation T, such that RT, but S T.We want to show that it results to contradiction, that willprove the claim, that S is the smallest transitive relation,that includes R.

/

(x, y)S but (x, y)T /

‘a path’ from x to y in R

xy

…

R

xy

…

T

since RT and T is transitive(x, y) T (contradiction)

S T

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How to build transitive closure?

An alternative way to represent transitive closure isas the union of paths with different length (if |A| = n, the longest path has length n).

1

2 3

4R = {(1, 1), (1, 2), (2, 3), (3, 1), (4, 1)}

t(R) = {(1, 1), (1, 2), (2, 3), (3, 1), (4, 1), (1, 3), (2, 1), (4, 2), (2, 2), (4, 3)}

Lemma 3. The transitive closure of the relation R is t(R) = R R2 R3…

(left for you to prove)

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We can also define a composition of closures, e. g. tr(R) = t(r(R)), rs(R)=r(s(R)), etc.

It turns out, that the order of composition does matter.

Theorem. For any binary relation R AA st(R) ts(R).By the theorem about symmetric and transitive closures

st(R) = s(t(R)) = t(R)(t(R))-1

= (R R2 R3 …)(R R2 R3 …)-1

ts(R) = t(s(R)) = t(R R-1) = (R R-1)(R R-1)2 (R R-1)3…

By the distributive property of composition over union(R R-1)2= (R R-1)(R R-1) = R(RR-1)R-1(R R-1)

= RR RR-1R-1R R-1R-1

= R2 RR-1R-1R R-2, where R-1R-1= R-2

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In general, ts(R) is the union of terms of the form(R R-1)n = (R R-1)(R R-1)… (R R-1)

=R… R RR… R-1 … R-1R-1… R-1

Are RR-1 and R-1R the same?

Not at all!R-1={(b, a)}

a b

a

RR-1={(a, a)}

b

R={(a, b)}

ba

a

R-1R={(b, b)}b

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We can observe, that ts(R) includes compositions of all possible sequences of R and R-1, while st(R) is a union of positive and negative powers of R (but not mixtures). So, it can be proved that st(R)ts(R), but ts(R) st(R).

Proof. Take arbitrary (x, y) st(R) to show that (x, y)ts(R). By the Lemma 2 (x, y) st(R) implies that (x, y)t(R)(t(R))-1. We have two cases. Case 1: (x, y)t(R), i. e. there exists a path from x to y in R. It implies that there exist a path from x to y in s(R), since R s(R) = R R-1. By the definition of transitive closureit means that (x, y) t(s(R)). Case 2: (x, y)(t(R))-1 , i. e. (y, x)t(R), and similar to the case 1,there exists a path from y to x in R, that implies the path in s (R). It implies there exists a path from x to y, or (x, y) ts(R).

Try to find a counterexample to disproves that ts(R) st(R).