proof of modular law

First we show $C+(B\\cap A)\\subseteq B\\cap(C+A)$ :
Note that $C\\subseteq B,B\\cap A\\subseteq B$ , and therefore $C+(B\\cap A)\\subseteq B$ .
Further, $C\\subseteq C+A$ , $B\\cap A\\subseteq C+A$ , thus $C+(B\\cap A)\\subseteq C+A$ .

Next we show $B\\cap(C+A)\\subseteq C+(B\\cap A)$ :
Let $b\\in B\\cap(C+A)$ . Then $b=c+a$ for some $c\\in C$ and $a\\in A$ .Hence $a=b-c$ , and so $a\\in B$ since $b\\in B$ and $c\\in C\\subseteq B$ .
Hence $a\\in B\\cap A$ , so $b=c+a\\in C+(B\\cap A)$ .

单词	ProofOfModularLaw
释义	proof of modular law First we show $C+(B\\cap A)\\subseteq B\\cap(C+A)$ : Note that $C\\subseteq B,B\\cap A\\subseteq B$ , and therefore $C+(B\\cap A)\\subseteq B$ . Further, $C\\subseteq C+A$ , $B\\cap A\\subseteq C+A$ , thus $C+(B\\cap A)\\subseteq C+A$ . Next we show $B\\cap(C+A)\\subseteq C+(B\\cap A)$ : Let $b\\in B\\cap(C+A)$ . Then $b=c+a$ for some $c\\in C$ and $a\\in A$ .Hence $a=b-c$ , and so $a\\in B$ since $b\\in B$ and $c\\in C\\subseteq B$ . Hence $a\\in B\\cap A$ , so $b=c+a\\in C+(B\\cap A)$ .
随便看	random numbers RandomTuringMachine random Turing machine RandomVariable random variable RandomVector random vector RandomWalk random walk Range range RangeProblem range problem RangeVoting range voting RankLieAlgebra rank (Lie algebra) RanknullityTheorem rank-nullity theorem rank-nullity theorem RanknullityTheorem1 RankOfALinearMapping rank of a linear mapping RankOfAMatrix rank of a matrix