# How does implication work in Belnap's four-valued logic?

How does implication work in Belnap's four-valued logic? Essentially, I'm trying to better understand the paper, "A Useful Four-Valued Logic." In that paper, page 15 describes how implication works, but I don't quite understand the statement, "...we will say that A entails or implies B just in case for each assignment of one of the four values to variables, the value of A does not exceed (in L4)." Basically, I'm not sure if something like $b\rightarrow b$ gives $b$ or $t$.

Really, if someone could put together a truth table for $A\rightarrow B$ for $\{t,f,b,n\}$, I'd really appreciate it.

For more information the paper is called, "A Useful Four-Valued Logic," by Belnap. I believe Springer has the current rights to it here. There's some general information about the logic system at the Stanford Encyclopedia of Philosophy here.

## Edit 1

Here are what are hopefully the two relevant pages

• Do you have a link to that paper? Do you know its authors? Where it was published? Whenever you mention a paper you should provide bibliographic information and at the very least mention its authors. – Mariano Suárez-Álvarez Mar 7 '17 at 6:55
• The paper is called, "A Useful Four-Valued Logic," by Belnap. I believe Springer has the current rights to it here. There's some general information about the logic system at the Stanford Encyclopedia of Philosophy here. – wyer33 Mar 7 '17 at 6:59
• matwbn.icm.edu.pl/ksiazki/bcp/bcp28/bcp2811.pdf looks useful – Henno Brandsma Mar 7 '17 at 6:59
• Add that information to the body of the question, – Mariano Suárez-Álvarez Mar 7 '17 at 7:00
• I've added it now. – wyer33 Mar 7 '17 at 7:03

This article does not appear to be defining implication as a connective; instead he seems to be using $\to$ for what is more commonly written as $\models$. That is, the phrase $P \to Q$ is not a proposition of the logic, but instead the assertion that "$s(P) \leq s(Q)$ for all truth valuations".

• Thanks for the response. I'm not well-versed in logics, so how should I read $P\rightarrow Q$ or as you assert $P\models Q$? I'm used to simple logics where we only have $\{t,f\}$ and $P\rightarrow Q$ reads as $P$ implies $Q$. As such, I read things like $P\rightarrow Q,P\vdash Q$ as $P$ implies $Q$, $P$ is true, therefore $Q$ is true. Now, we have four different values for logic and I don't fully understand $\models$, so I'm not sure how to interpret things. – wyer33 Mar 7 '17 at 8:25
• You read $P \models Q$ as I said in the post. Put differently it means "you can prove $Q$ from $P$ by truth table", in that a 'proof' is showing that $Q$ is at least as true as $P$ in all cases. – Hurkyl Mar 7 '17 at 8:32

I suppose the truth values have an ordering with false being the lowest and truth being the highest value. So $f < b$ (or $n$) < $n$ (or $b$) $< t$.

Then he seemingly considers $A \rightarrow B$ true iff $t(A) \le t(B)$ where $t$ would denote truth value. This without seeing the paper, post a link to papers you're asking questions about, it helps..

If $\rightarrow$ in $P \rightarrow Q$ is to be considered a logical connective, then he seems above to indicate neither more nor less than that it's truth value is "true" if and only if the truth value of $P$ (in each set up, as he says) is at most the truth value of $Q$. In other words, $\top \vdash P \rightarrow Q$ if and only if $P \vdash Q$. This doesn't really indicate what the truth value would be when the truth value of $P$ is (in some set up) not non-strictly below that of $Q$. So why doesn't he just use $\vdash$? I don't know.

Especially when conjunction and disjunction mean infimum and supremum respectively (when the entailment relation $\vdash$ is considered as giving a preorder), it tends to be useful to define the implication so as to make a Heyting prealgebra, i.e., so that $P \rightarrow Q$ is a largest formula whose conjunction with $P$ entails $Q$ . We know from what Belnap says that the truth value of $P \rightarrow Q$ must be $true$ when $P$ entails $Q$ (nine entries in the truth table), as is the case with logics having an implication that makes a Heyting prealgebra. What would demanding that implication makes a Heyting prealgebra mean in the other cases? Supposing $P$ and $Q$ to have truth values and that the truth value of $Q$ is not false and that $P$ does not entail $Q$ (four cases), this would cause the truth value of $P \rightarrow Q$ to be the truth value of $Q$. As for the remaining three possibilities, the truth value of $none \rightarrow false$ would be $both$ and the truth value of $both \rightarrow false$ would be $none$, and the truth value of $true \rightarrow false$ would be $false$. But it sounds like Belnap doesn't even define $\rightarrow$ as a connective.

Even in classical logic, entailment (denoted by $\vdash$) means something different from implication (denoted by $\rightarrow$). For instance, $(\top \rightarrow x \neq 0) \vee (\top \rightarrow x = 0)$ is a tautology (thus, true), but the metastatement "$\top \vdash x \neq 0$ or $\top \vdash x = 0$" is false; one could argue that the problem with this example results from not distinguishing the operator $\vee$ from the meta disjunction operator "or", but since in oral discourse the same words are used, that is a problem! At best conflating implication and entailment would seem to make it necessary to split hairs many other places--a poor choice that quite possibly it seems to me can't even work coherently or at all elegantly even when hairs are carefully split elsewhere.