Can someone explain definition of operation $\rightarrow$

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.

My question is following: I don't understand the definition of operation $\rightarrow$. Can someone explain to me more clearly than it is defined by me above? I understand the definition of bi-residual, but not definition that defines $\rightarrow$, because it seems that it has not been defined clearly. I think it is defined using the definition of operation $\odot$. But I don't understand that if $\odot$ has been defined or not? In this presentation I think it has not been defined. This has been translated from the document here.

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It looks like both of $\odot$ and $\to$ are simply primitive operations which are assumed to satisfy the axioms you quote. (Compare to, for example, ring axioms: the axioms does not define how multiplication works, but a ring must come with some multiplication that satisfies the axioms). –  Henning Makholm Jan 30 '12 at 22:01
Question on axioms( that I quote): where do I quote the axioms? Which axioms? I don't get the definition of $\rightarrow$, because $\odot$ has been defined by using $\rightarrow$, which in turn has been defined by using $\odot$. I think this is cyclical definition. It does not define anything so that it could be understood. –  alvoutila Jan 31 '12 at 7:36
The axioms for $\odot$ would be the two formulas you quote just before "It can be shown ...". –  Henning Makholm Jan 31 '12 at 12:59
All@: How would you prove Theorem 1, for example x $\leftrightarrow$ 1 = x and ( x $\leftrightarrow$ y ) $\odot$ ( y $\leftrightarrow$ z) $\leq$ x $\leftrightarrow$ z, –  alvoutila Jan 31 '12 at 18:28

1 Answer

$a$ implies $b$ if $a\leq b$. $a\rightarrow b$ measures to what extent $a$ implies $b$, or under what conditions $a$ implies $b$. This is the sup-definition of $a\rightarrow b$.

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But I think that $\rightarrow$ is not "imply" operation. How then you get that there are in the beginning "imply" operation? –  alvoutila Jan 31 '12 at 7:23
Well, as I said, $a\rightarrow b$ does not really say that $a$ implies $b$. But I think it measures to what extent $a$ implies $b$. –  Stefan Geschke Feb 5 '12 at 19:05