# How to prove Bi-residual theorem (Theorem 1)?

It turns out that most frequently set of truth values comprises the so-called residue lattice in other words L is a partiar order set, which includes maximum element(1) and minimum(0), where each element pair $a, b \in L$ has greatest lower limit $a \wedge b$ and smallest upper limit $a \vee b$, and where is defined such binary operations $\odot$( product) and $\rightarrow$ (residue) that $\odot$ is associative, commutative and isotonic( increasing),

$\forall a \in L: a \odot 1 = a$,

$\forall a, b, c \in L: a \odot b \leq c$, if and only if $a \leq b \rightarrow c$.

It can be shown, that residual which corresponds to product-operation and which satisfies previous clauses is unique and defined by formula

$\forall a, b \in L: a \rightarrow b = sup \{x| a \odot x \leq b \}.$

Very important is equivalence relation of many value logic, whose algebraic counterpart is bi-residual $\leftrightarrow$, which is defined by clause

$\forall a, b \in L: a \leftrightarrow b = (a \rightarrow b) \wedge (b \rightarrow a)$

Theorem 1. Bi-residual has following properties

$x \leftrightarrow 1 = x$,

$x = y$ if and only if $x \leftrightarrow y =1$,

$x \leftrightarrow y = y \leftrightarrow x$,

$( x \leftrightarrow y)\odot (y \leftrightarrow z) \leq x \leftrightarrow z$,

where $x, y, z$ are elements of residue lattice.

How to prove Theorem 1?( for example $x \leftrightarrow 1 = x$ and $( x \leftrightarrow y)\odot (y \leftrightarrow z) \leq x \leftrightarrow z$)

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This is not my area, but is it possible that there's a better title for this question? –  Dylan Moreland Jan 31 '12 at 21:32
Well ................... –  laovultai Jan 31 '12 at 21:47
Well what? ${}{}$ –  Mariano Suárez-Alvarez Jan 31 '12 at 21:52
well title has been changed –  laovultai Jan 31 '12 at 22:17
Well... ${}{}{}$ –  Mariano Suárez-Alvarez Jan 31 '12 at 23:37