Proof that $ 1_A cup s cdot s cdot s $ is equivalence relation
$begingroup$
I have problem with this task:
Proof that $$ 1_A cup s cdot s cdot s $$ is equivalence relation
where $r subset A times A$ and $r$ is a relation such that
$$ forall_{x,y,z} r(x,y) wedge r(x,z) rightarrow r(y,z) $$ and
$$ s = r cup r^{-1} $$
Reflexivity results directly from the definition and symmetry results from the symmetry of the relationship $s$. This part was not difficult for me, but I am trying to show that this relation is transitive and I have no idea how to show that...
proof-writing relations equivalence-relations
asked Jan 19 at 19:01
VirtualUserVirtualUser
865114
$endgroup$
add a comment |
$begingroup$
I have problem with this task:
Proof that $$ 1_A cup s cdot s cdot s $$ is equivalence relation
where $r subset A times A$ and $r$ is a relation such that
$$ forall_{x,y,z} r(x,y) wedge r(x,z) rightarrow r(y,z) $$ and
$$ s = r cup r^{-1} $$
Reflexivity results directly from the definition and symmetry results from the symmetry of the relationship $s$. This part was not difficult for me, but I am trying to show that this relation is transitive and I have no idea how to show that...
proof-writing relations equivalence-relations
asked Jan 19 at 19:01
VirtualUserVirtualUser
865114
$endgroup$
$begingroup$
What is maneuverability?
$endgroup$
– William Elliot
Jan 20 at 0:10
$begingroup$
reflexivity @WilliamElliot
$endgroup$
– VirtualUser
Jan 20 at 0:38
add a comment |
$begingroup$
I have problem with this task:
Proof that $$ 1_A cup s cdot s cdot s $$ is equivalence relation
where $r subset A times A$ and $r$ is a relation such that
$$ forall_{x,y,z} r(x,y) wedge r(x,z) rightarrow r(y,z) $$ and
$$ s = r cup r^{-1} $$
Reflexivity results directly from the definition and symmetry results from the symmetry of the relationship $s$. This part was not difficult for me, but I am trying to show that this relation is transitive and I have no idea how to show that...
proof-writing relations equivalence-relations
asked Jan 19 at 19:01
VirtualUserVirtualUser
865114
$endgroup$
I have problem with this task:
Proof that $$ 1_A cup s cdot s cdot s $$ is equivalence relation
where $r subset A times A$ and $r$ is a relation such that
$$ forall_{x,y,z} r(x,y) wedge r(x,z) rightarrow r(y,z) $$ and
$$ s = r cup r^{-1} $$
Reflexivity results directly from the definition and symmetry results from the symmetry of the relationship $s$. This part was not difficult for me, but I am trying to show that this relation is transitive and I have no idea how to show that...
proof-writing relations equivalence-relations
proof-writing relations equivalence-relations
asked Jan 19 at 19:01
VirtualUserVirtualUser
865114
asked Jan 19 at 19:01
VirtualUserVirtualUser
865114
edited Jan 21 at 22:32
VirtualUser
asked Jan 19 at 19:01
VirtualUserVirtualUser
865114
asked Jan 19 at 19:01
VirtualUserVirtualUser
865114
asked Jan 19 at 19:01
VirtualUserVirtualUser
865114
865114
$begingroup$
What is maneuverability?
$endgroup$
– William Elliot
Jan 20 at 0:10
$begingroup$
reflexivity @WilliamElliot
$endgroup$
– VirtualUser
Jan 20 at 0:38
add a comment |
$begingroup$
What is maneuverability?
$endgroup$
– William Elliot
Jan 20 at 0:10
$begingroup$
reflexivity @WilliamElliot
$endgroup$
– VirtualUser
Jan 20 at 0:38
$begingroup$
What is maneuverability?
$endgroup$
– William Elliot
Jan 20 at 0:10
$begingroup$
What is maneuverability?
$endgroup$
– William Elliot
Jan 20 at 0:10
$begingroup$
reflexivity @WilliamElliot
$endgroup$
– VirtualUser
Jan 20 at 0:38
$begingroup$
reflexivity @WilliamElliot
$endgroup$
– VirtualUser
Jan 20 at 0:38
add a comment |
1 Answer
1
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oldest
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$begingroup$
Let R be a reflexive, symmetrical relation with six elements in a chain: xRa, aRb, bRz, zRc, cRd, dRz.
Thus R = S and $1_A$ $cup$ S•S•S is not transitive
where A = { x,a,b,y,c,d,z }
answered Jan 22 at 22:21
William ElliotWilliam Elliot
8,1622720
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add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let R be a reflexive, symmetrical relation with six elements in a chain: xRa, aRb, bRz, zRc, cRd, dRz.
Thus R = S and $1_A$ $cup$ S•S•S is not transitive
where A = { x,a,b,y,c,d,z }
answered Jan 22 at 22:21
William ElliotWilliam Elliot
8,1622720
$endgroup$
add a comment |
$begingroup$
Let R be a reflexive, symmetrical relation with six elements in a chain: xRa, aRb, bRz, zRc, cRd, dRz.
Thus R = S and $1_A$ $cup$ S•S•S is not transitive
where A = { x,a,b,y,c,d,z }
answered Jan 22 at 22:21
William ElliotWilliam Elliot
8,1622720
$endgroup$
add a comment |
$begingroup$
Let R be a reflexive, symmetrical relation with six elements in a chain: xRa, aRb, bRz, zRc, cRd, dRz.
Thus R = S and $1_A$ $cup$ S•S•S is not transitive
where A = { x,a,b,y,c,d,z }
answered Jan 22 at 22:21
William ElliotWilliam Elliot
8,1622720
$endgroup$
Let R be a reflexive, symmetrical relation with six elements in a chain: xRa, aRb, bRz, zRc, cRd, dRz.
Thus R = S and $1_A$ $cup$ S•S•S is not transitive
where A = { x,a,b,y,c,d,z }
answered Jan 22 at 22:21
William ElliotWilliam Elliot
8,1622720
edited Jan 23 at 22:42
answered Jan 22 at 22:21
William ElliotWilliam Elliot
8,1622720
answered Jan 22 at 22:21
William ElliotWilliam Elliot
8,1622720
answered Jan 22 at 22:21
William ElliotWilliam Elliot
8,1622720
8,1622720
add a comment |
add a comment |
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$begingroup$
What is maneuverability?
$endgroup$
– William Elliot
Jan 20 at 0:10
$begingroup$
reflexivity @WilliamElliot
$endgroup$
– VirtualUser
Jan 20 at 0:38