# modular inequality

In any lattice (http://planetmath.org/lattice) the self-dual *modular inequality ^{}* is true:
if $x\le z$ then $x\vee (y\wedge z)\le (x\vee y)\wedge z$.

###### Proof.

$x\le x\vee y$ and we are given that $x\le z$, so $x\le (x\vee y)\wedge z$. Also, $y\wedge z\le y\le x\vee y$ and $y\wedge z\le z$ imply that $y\wedge z\le (x\vee y)\wedge z$. Therefore, $x\vee (y\wedge z)\le (x\vee y)\wedge z$. ∎

Title | modular inequality |
---|---|

Canonical name | ModularInequality |

Date of creation | 2014-02-01 1:48:21 |

Last modified on | 2014-02-01 1:48:21 |

Owner | ixionid (16766) |

Last modified by | ixionid (16766) |

Numerical id | 10 |

Author | ixionid (16766) |

Entry type | Theorem |

Classification | msc 06C05 |

Related topic | ModularLattice |

Related topic | DistributiveInequalities |

Defines | modular inequality |