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翻訳と辞書　辞書検索 [ 開発暫定版 ] スポンサード リンク consensus theorem ： ウィキペディア英語版 consensus theorem z |- bgcolor="#ddffdd" align="center" | 0 || 0 || 0 || 0 || 0 |- bgcolor="#ddffdd" align="center" | 0 || 0 || 1 || 1 || 1 |- bgcolor="#ddffdd" align="center" | 0 || 1 || 0 || 0 || 0 |- bgcolor="#ddffdd" align="center" | 0 || 1 || 1 || 1 || 1 |- bgcolor="#ddffdd" align="center" | 1 || 0 || 0 || 0 || 0 |- bgcolor="#ddffdd" align="center" | 1 || 0 || 1 || 0 || 0 |- bgcolor="#ddffdd" align="center" | 1 || 1 || 0 || 1 || 1 |- bgcolor="#ddffdd" align="center" | 1 || 1 || 1 || 1 || 1 |} In Boolean algebra, the consensus theorem or rule of consensus〔Frank Markham Brown, ''Boolean Reasoning: The Logic of Boolean Equations'', 2nd edition 2003, p. 44〕 is the identity: :$xy\; \backslash vee\; \backslash barz\; \backslash vee\; yz\; =\; xy\; \backslash vee\; \backslash barz$ The consensus or resolvent of the terms $xy$ and $\backslash barz$ is $yz$. It is the conjunction of all the unique literals of the terms, excluding the literal that appears unnegated in one term and negated in the other. The conjunctive dual of this equation is: :$(x\; \backslash vee\; y)(\backslash bar\; \backslash vee\; z)(y\; \backslash vee\; z)\; =\; (x\; \backslash vee\; y)(\backslash bar\; \backslash vee\; z)$ ==Proof== LHS = $xy\; \backslash vee\; \backslash barz\; \backslash vee\; (x\; \backslash vee\; \backslash bar)yz$ = $xy\; \backslash vee\; \backslash barz\; \backslash vee\; xyz\; \backslash vee\; \backslash baryz$ = $xy\; \backslash vee\; xyz\; \backslash vee\; \backslash barz\; \backslash vee\; \backslash baryz$ = $xy(1\; \backslash vee\; z)\; \backslash vee\; \backslash barz(1\; \backslash vee\; y)$ = $xy\; \backslash vee\; \backslash barz$ = RHS 抄文引用元・出典: フリー百科事典『 ウィキペディア（Wikipedia）』 ■ウィキペディアで「consensus theorem」の詳細全文を読む スポンサード リンク 翻訳と辞書 : 翻訳のためのインターネットリソース Copyright(C) kotoba.ne.jp 1997-2016. All Rights Reserved.