Unformatted text preview: 3.9 1. IF R ⊆ A × B , then id A ◦ R = R = R ◦ id B . 2. ◦ is associative: that is, For arbitrary relations R ⊆ A × B and S ⊆ B × C and T ⊆ C × D , then R ◦ ( S ◦ T ) = ( R ◦ S ) ◦ T. Proof The prooF oF part 1 is simple and leFt as an exercise. We prove part 2. Let R , S and T be relations specifed in the proposition, and let ( x,u ) be an arbitrary member oF ( R ◦ S ) ◦ T . We show that ( x,u ) ∈ R ◦ ( S ◦ T ) using the Following reasoning: x ( R ◦ S ) ◦ T u ⇒ ∃ z. x ( R ◦ S ) z ∧ z T u ⇒ ∃ z. ( ∃ y. xRy ∧ y S z ) ∧ z T u ⇒ ∃ z,y. ( xRy ∧ y S z ∧ z T u ) ⇒ ∃ y. ( xRy ) ∧ ( ∃ z.y S z ∧ z T u ) ⇒ ∃ y. ( xRy ) ∧ ( y S ◦ T u ) ⇒ xR ◦ ( S ◦ T ) u 20...
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 Spring '09
 Koskesh
 Math, Logic, Exponentiation, Composition of relations, Haskell notation, relational composition

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