Prove the identity laws in Table 1 by showing that a) A U 0 A. b) A n U = A.
Solution:
(a) Assume U={1,2,3,4,5,6},A={1,2,3}U=\{1,2,3,4,5,6\}, A=\{1,2,3\}U={1,2,3,4,5,6},A={1,2,3}
To prove A∪ϕ=AA\cup \phi=AA∪ϕ=A
LHS=A∪ϕ={1,2,3}∪ϕ={1,2,3}=A=RHSLHS=A\cup \phi \\=\{1,2,3\} \cup \phi \\=\{1,2,3\} \\=A=RHSLHS=A∪ϕ={1,2,3}∪ϕ={1,2,3}=A=RHS
(b)
To prove A∩U=AA\cap U=AA∩U=A
LHS=A∩U={1,2,3}A∩{1,2,3,4,5,6}={1,2,3}=A=RHSLHS=A\cap U \\=\{1,2,3\}A\cap \{1,2,3,4,5,6\} \\=\{1,2,3\} \\=A=RHSLHS=A∩U={1,2,3}A∩{1,2,3,4,5,6}={1,2,3}=A=RHS
Hence, proved.
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