# 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 $b=c+a\in C+(B\cap A)$.

Title | proof of modular law |
---|---|

Canonical name | ProofOfModularLaw |

Date of creation | 2013-03-22 12:50:45 |

Last modified on | 2013-03-22 12:50:45 |

Owner | yark (2760) |

Last modified by | yark (2760) |

Numerical id | 8 |

Author | yark (2760) |

Entry type | Proof |

Classification | msc 16D10 |

Related topic | FirstIsomorphismTheorem |