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Question Number 1744 by Rasheed Ahmad last updated on 13/Sep/15

$$If\mathbf{A}and\mathbf{B}aretwosetsand\mathbb{U}is$$$$auniversalsetprovethat$$$$\mathbf{A}\subseteq \mathbf{B}\Rightarrow \mathbf{B}=\mathbf{A}\cup ({\mathbf{A}}^{\prime }\cap \mathbf{B})$$

Answered by Rasheed Ahmad last updated on 19/Sep/15

$$\mathbf{A}\subseteq \mathbf{B}\Rightarrow \mathbf{B}=\mathbf{A}\cup ({\mathbf{A}}^{\prime }\cap \mathbf{B})$$$$..........\ast \ast \ast .......................$$$$\mathbf{A}\subseteq \mathbf{B}\Rightarrow \mathbf{A}\cup \mathbf{B}=\mathbf{B}$$$$Or\mathbf{B}=\mathbf{A}\cup \mathbf{B}$$$$=\mathbb{U}\cap (\mathbf{A}\cup \mathbf{B})[\because \mathbf{S}=\mathbb{U}\cap \mathbf{S}]$$$$=(\mathbf{A}\cup {\mathbf{A}}^{\prime })\cap (\mathbf{A}\cup \mathbf{B})[\because \mathbb{U}=\mathbf{A}\cup {\mathbf{A}}^{\prime }]$$$$=\mathbf{A}\cup ({\mathbf{A}}^{\prime }\cap \mathbf{B})[\because (\mathbf{A}\cup \mathbf{B})\cap (\mathbf{A}\cup \mathbf{C})$$$$=\mathbf{A}\cup (\mathbf{B}\cap \mathbf{C})]$$

Answered by arvind last updated on 18/Sep/15

Answered by Rasheed Soomro last updated on 18/Sep/15

$$\mathbf{A}\subseteq \mathbf{B}\Rightarrow \mathbf{B}=\mathbf{A}\cup ({\mathbf{A}}^{\prime }\cap \mathbf{B})$$$$RHS:\mathbf{A}\cup ({\mathbf{A}}^{\prime }\cap \mathbf{B})=(\mathbf{A}\cup {\mathbf{A}}^{\prime })\cap (\mathbf{A}\cup \mathbf{B})$$$$[Distributivityof\cup over\cap ]$$$$=\mathbb{U}\cap (\mathbf{A}\cup \mathbf{B})[\mathbf{A}\cup {\mathbf{A}}^{\prime }=\mathbb{U}]$$$$=\mathbf{A}\cup \mathbf{B}[\mathbb{U}\cap \mathbf{S}=\mathbf{S}]$$$$=\mathbf{B}[\because \mathbf{A}\subseteq \mathbf{B}]$$$$=LHS$$

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