2019-04-29T04:59:45-04:00
et U be the set of positive integers 1, 2, 3, ...
etc., A be the set of odd positive integers and
B be the set of even positive integers. Verify
De Morgan's laws.
1
2019-05-08T08:07:33-0400
U = { x ∈ Z + } , A = { x ∈ Z + ∣ x i s o d d } , B = { x ∈ Z + ∣ x i s e v e n } U=\{x\isin\Z^+ \}, A=\{x\isin\Z^+|x\ is\ odd \}, B=\{x\isin\Z^+|x\ is\ even \} U = { x ∈ Z + } , A = { x ∈ Z + ∣ x i s o dd } , B = { x ∈ Z + ∣ x i s e v e n }
A ∪ B ‾ = A ‾ ∩ B ‾ \overline{A\cup B}=\overline{A}\cap\overline{B} A ∪ B = A ∩ B
A ∪ B = { x ∈ Z + ∣ x i s o d d ∨ x i s e v e n } = { x ∈ Z + } = U {A\cup B}=\{x\isin\Z^+|x\ is\ odd \vee x\ is\ even \}=\{x\isin\Z^+ \}=U A ∪ B = { x ∈ Z + ∣ x i s o dd ∨ x i s e v e n } = { x ∈ Z + } = U
A ∪ B ‾ = U ‾ = { } \overline{A\cup B}=\overline{U}=\{ \} A ∪ B = U = { }
A ‾ = { x ∈ Z + ∣ x i s n o t o d d } = { x ∈ Z + ∣ x i s e v e n } = B \overline{A}=\{x\isin\Z^+|x\ is\ not\ odd \}=\{x\isin\Z^+|x\ is\ even \}=B A = { x ∈ Z + ∣ x i s n o t o dd } = { x ∈ Z + ∣ x i s e v e n } = B
B ‾ = { x ∈ Z + ∣ x i s n o t e v e n } = { x ∈ Z + ∣ x i s o d d } = A \overline{B}=\{x\isin\Z^+|x\ is\ not\ even \}=\{x\isin\Z^+|x\ is\ odd \}=A B = { x ∈ Z + ∣ x i s n o t e v e n } = { x ∈ Z + ∣ x i s o dd } = A
A ‾ ∩ B ‾ = { x ∈ Z + ∣ x i s e v e n ∧ x i s o d d } = { } \overline{A}\cap\overline{B}=\{x\isin\Z^+|x\ is\ even\ \wedge \ x\ is\ odd\}=\{ \} A ∩ B = { x ∈ Z + ∣ x i s e v e n ∧ x i s o dd } = { } Hence
A ∪ B ‾ = A ‾ ∩ B ‾ \overline{A\cup B}=\overline{A}\cap\overline{B} A ∪ B = A ∩ B
A ∩ B ‾ = A ‾ ∪ B ‾ \overline{A\cap B}=\overline{A}\cup\overline{B} A ∩ B = A ∪ B
A ∩ B = { x ∈ Z + ∣ x i s o d d ∧ x i s e v e n } = { } A\cap B=\{x\isin\Z^+|x\ is\ odd\ \wedge \ x\ is\ even\}=\{ \} A ∩ B = { x ∈ Z + ∣ x i s o dd ∧ x i s e v e n } = { }
A ∩ B ‾ = U − { } = U \overline{A\cap B}=U-\{ \}=U A ∩ B = U − { } = U
A ‾ = { x ∈ Z + ∣ x i s n o t o d d } = { x ∈ Z + ∣ x i s e v e n } = B \overline{A}=\{x\isin\Z^+|x\ is\ not\ odd \}=\{x\isin\Z^+|x\ is\ even \}=B A = { x ∈ Z + ∣ x i s n o t o dd } = { x ∈ Z + ∣ x i s e v e n } = B
B ‾ = { x ∈ Z + ∣ x i s n o t e v e n } = { x ∈ Z + ∣ x i s o d d } = A \overline{B}=\{x\isin\Z^+|x\ is\ not\ even \}=\{x\isin\Z^+|x\ is\ odd \}=A B = { x ∈ Z + ∣ x i s n o t e v e n } = { x ∈ Z + ∣ x i s o dd } = A
A ‾ ∪ B ‾ = B ∪ A = { x ∈ Z + ∣ x i s e v e n ∨ x i s o d d } = { x ∈ Z + } = U \overline{A}\cup\overline{B}=B\cup A=\{x\isin\Z^+|x\ is\ even\vee x\ is\ odd \}=\{x\isin\Z^+ \}=U A ∪ B = B ∪ A = { x ∈ Z + ∣ x i s e v e n ∨ x i s o dd } = { x ∈ Z + } = U Hence
A ∩ B ‾ = A ‾ ∪ B ‾ \overline{A\cap B}=\overline{A}\cup\overline{B} A ∩ B = A ∪ B
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