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4.11. Proving and Disproving Set Statements. 4.11.1. Proof by Exhaustion To prove set results for finite sets, the method of exhaustion is used. That is every element in the set is tested to ensure it satisfies the condition. Example: • Let A = {1, 2} , B = {1, 2, 3, 4}. Prove A ⊆ A ∪ B . To prove the statement, we must show every element in A is in A ∪ B . Now A ∪ B = {1, 2, 3, 4} 1 ∈ A, 1 ∈ A ∪ B 2 ∈ A, 2 ∈ A ∪ B Thus all elements in A are in A ∪ B , and so by exhaustion A ⊆ A ∪ B. WUCT121 Logic 162 Exercise: • Let A = {1, 2} , B = {1, 2, 3, 4}. Prove A = A ∩ B . To prove the statement, we must show every element in A is in A ∩ B and every element in A ∩ B is in A. Thus all elements in A are in A ∩ B and vice versa, and so by exhaustion A = A ∩ B . Exercise: • Give an example of three sets A, B and C such that C ⊆ A∩ B. WUCT121 Logic 163 4.11.2. Disproof by Counterexample. A set result can be disproven by giving a counterexample. To find a counterexample often creating a Venn diagram will be of benefit. Example: • Disprove A ⊆ A ∩ B . To disprove the statement, we must give a counterexample. Let A = {1, 2} , B = {3, 4} Now A ∩ B = 1 ∈ A, however 1 ∉ A ∩ B = Thus by counterexample A⊄A ∩ B . Exercise: • Disprove A ⊆ A − B . To disprove the statement, we must show a counterexample. WUCT121 Logic 164 4.11.3. Proof by Typical Element. To prove set results for infinite sets, generalised methods must be used. The typical element method considers a particular but arbitrary element of the set and by applying knows laws, rules and definitions prove the result. It is the method for proving subset relationships. So prove that A ⊆ B , we must show that ∀x, ( x ∈ A ⇒ x ∈ B ) Begin by letting x ∈ A , that is, we take x to be a particular but arbitrary element of A. Using the definitions, we prove that x ∈ B . As long as we use no special properties of the element x, we can conclude that A ⊆ B , which is what we wanted to prove. This method can be used to prove set equalities. By using the definition A = B ⇔ ( A ⊆ B ∧ B ⊆ A) and showing A ⊆ B ∧ B ⊆ A , that is proving ∀x, ( x ∈ A ⇒ x ∈ B ) and ∀x, ( x ∈ B ⇒ x ∈ A) , the result A = B follows. Using this definition is sometimes called a “double containment” proof. WUCT121 Logic 165 Examples: • Let U be a set and let A and B be elements of (U ) . Prove A ⊆ A ∪ B . Need to prove ∀x, ( x ∈ A ⇒ x ∈ A ∪ B ) Let x ∈ A , then x∈ A ⇒ x∈ A∨ x∈B ⇒ x∈ A∪ B ∴ A ⊆ A∪ B see note definition of ∪ Note: Appling rules of logic, we know P ⇒ P ∨ Q is a tautology. Let P( x ) : x ∈ A, Q( x ) : x ∈ B . Thus x ∈ A ⇒ x ∈ A ∨ x ∈ B is a tautology in the proof above. • Let U be a set and let A and B be elements of (U ) . Prove A ⊆ B ⇔ A ∪ B = B . Need to prove two parts: 1. A ⊆ B ⇒ A ∪ B = B 2. A ∪ B = B ⇒ A ⊆ B WUCT121 Logic 166 • Proof of 1: KNOW: A ⊆ B , that is ∀x, ( x ∈ A ⇒ x ∈ B ) K (1) PROVE: A ∪ B = B . Need to prove two parts: i. A ∪ B ⊆ B ii. B ⊆ A ∪ B Proof of i.: Let x ∈ A ∪ B then x∈ A∪ B ⇒ x∈ A∨ x∈B ⇒ x∈B ∨ x∈B ⇒ x∈B ∴A∪ B ⊆ B Proof of ii.: definition of ∪ by (1) Logic rule P ∨ P ≡ P Let x ∈ B then x∈B ⇒ x∈ A∨ x∈B ⇒ x∈ A∪ B ∴B ⊆ A∪ B see previous example definition of ∪ Since A ∪ B ⊆ B and B ⊆ A ∪ B , A ∪ B = B Thus A ⊆ B ⇒ A ∪ B = B . WUCT121 Logic 167 Proof of 2: KNOW: A ∪ B = B , that is ∀x , ( x ∈ A ∪ B ⇔ x ∈ B ) K( 2) PROVE: A ⊆ B . Let x ∈ A then x∈ A ⇒ ⇒ ⇒ ∴A⊆ B x∈ A∨ x∈B x∈ A∪ B x∈B see previous example definition of ∪ by (2) Thus A ∪ B = B ⇒ A ⊆ B Since A ⊆ B ⇒ A ∪ B = B and A ∪ B = B ⇒ A ⊆ B it is proven that A ⊆ B ⇔ A ∪ B = B . Exercise: • Let U be a set and let A and B be elements of (U ) . Prove A ∩ B ⊆ A . That is, prove ∀x, ( x ∈ A ∩ B ⇒ x ∈ A) WUCT121 Logic 168 4.11.4. Proof by Equivalence of Statements. If A can be written as A = { x ∈ U : P( x )} and B = { x ∈ U : Q ( x )}, the equality of specification theorem can be used to show that A = B by showing that P( x ) ≡ Q( x ) , that is, by showing that P( x ) ⇔ Q ( x ) is a tautology. Examples: • Let A = { x ∈ : x 2 ≤ 1} and B = { x ∈ : −1 ≤ x ≤ 1} . Prove A = B Let P( x ) : x 2 ≤ 1 and Q( x ) : −1 ≤ x ≤ 1 . Now x 2 ≤ 1 ⇔ −1 ≤ x ≤ 1 ∴ P( x ) ⇔ Q( x ) ∴A= B • Let U be a set and let A and B be elements of (U ) . Prove that ( A ∩ B ) = A ∪ B . We need to show that the statements defining the sets ( A ∩ B ) and A ∪ B are equivalent. ( A ∩ B ) = { x ∈ U :~ ( x ∈ A ∩ B )} definition of A A ∪ B = { x ∈ U : x ∈ A ∪ B} WUCT121 Logic 169 Let P( x ) :~ ( x ∈ A ∩ B ) , and Q( x ) : x ∈ A ∪ B x∈ A∪ B ≡ x∈ A ∨ x∈B by definition of ∪ ≡ ~ ( x ∈ A) ∨ ~ ( x ∈ B ) definition of A ≡ ~ (x ∈ A ∧ x ∈ B ) ≡ ~ (x ∈ A ∩ B ) by Logic De Morgan' s by definition of ∩ ∴ Q( x ) ≡ P( x ) ∴ (A ∩ B) = A ∪ B Exercise: • Let U be a set and let A and B be elements of (U ) . Prove that ( A ∪ B ) = A ∩ B . We need to show that the statements defining the sets ( A ∪ B ) and A ∩ B are equivalent. ( A ∪ B ) = { x ∈ U :~ ( x ∈ A ∪ B )} axiom of specification A ∩ B = { x ∈ U : x ∈ A ∩ B} axiom of specification WUCT121 Logic 170 • Let U be a set and let A, B and C be elements of P(U). Prove that ( A − B ) − C = ( A − C ) − B . • Let U be a set and let X and Y be elements of (U ) . Prove that X − Y = X ∩ Y . WUCT121 Logic 171 4.11.5. Proof by Set Laws. Set equalities can be proven by using known set laws Examples: • Let U be a set and let A, B and C be elements of P(U). Prove ( A − B ) − C = ( A − C ) − B (A − B) − C = (A − B) ∩ C set difference = (A ∩ B ) ∩ C set difference = A ∩ (B ∩ C ) associativity = A ∩ (C ∩ B ) commutativity = (A ∩ C ) ∩ B associativity = (A − C ) ∩ B = (A − C ) − B WUCT121 set difference set difference Logic 172 4.11.6. Further Examples. Examples: • Let U be a set and let A and B be elements of P(U). Using the following: (i) A ⊆ B ⇔ A ∪ B = B , (ii): A ⊆ B ⇔ A ∩ B = A , (iii): A ∪ B = ( A ∩ B ) . Prove that A ⊆ B ⇔ B ⊆ A . Proof: A ⊆ B ⇔ A∩ B = A • by part (ii) ⇔ (A ∩ B) = A by taking complements ⇔ A∪ B = A by part (iii) ⇔B⊆A by part (i) Let U be a set and let A, B and C be elements of P(U). Disprove that A − (B − C ) = ( A − B ) − C . Let A = {1, 2, 3}, B = {2, 3}, C = {3}. A − (B − C ) = A − {2} = {1, 3} ( A − B ) − C = {1} − C = {1} ≠ A − (B − C ) WUCT121 Logic 173 • Let U be a set and let X and Y be elements of (U ) . Use a typical element argument to prove X − Y = X ∩ Y . Need to prove two parts: 1. X − Y ⊆ X ∩ Y 2. X ∩ Y ⊆ X − Y Proof of 1:Let x ∈ X − Y be a typical element. X − Y ≡ {x ∈ U : x ∈ X − Y } ⇒ {x ∈ U : x ∈ X ∧ x ∉ Y } Axiom of Specification Def of set difference ⇒ {x ∈ U : x ∈ X ∧ x ∈ Y } Def of complement ⇒ {x ∈ U : x ∈ X ∩ Y } Def of intersection ∴ ∀x ∈ U (x ∈ X − Y ⇒ x ∈ X ∩ Y ) ∴ X −Y ⊆ X ∩Y Proof 2: Let x ∈ X ∩ Y be a typical element. X ∩ Y ≡ {x ∈ U : x ∈ X ∩ Y } ⇒ {x ∈ U : x ∈ X ∧ x ∈ Y } Axiom of Specification ⇒ {x ∈ U : x ∈ X ∧ x ∉ Y } Def of intersection Def of complement ⇒ {x ∈ U : x ∈ X − Y } Def of set difference ∴ ∀x ∈ U (x ∈ X ∩ Y ⇒ x ∈ X − Y ) ∴ X ∩Y ⊆ X −Y ∴ ∀x ∈ U (x ∈ X − Y ⇔ x ∈ X ∩ Y ), i.e. X − Y ⊆ X ∩ Y ∧ X ∩ Y ⊆ X − Y , ∴ X −Y = X ∩Y WUCT121 Axiom of extent Logic 174