Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
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. –  laovultai 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, –  laovultai Jan 31 '12 at 18:28
add comment

1 Answer

up vote 1 down vote accepted

$a$ 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 $a\rightarrow b$.

share|improve this answer
But I think that $\rightarrow$ is not "imply" operation. How then you get that there are in the beginning "imply" operation? –  laovultai 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
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.