discrete mathematics "3 value logic" help

samaha · **Joined:** Fri, 8 Jun 2012 05:00:12 UTC **Posts:** 1

so basically i have a project about 3 valued logic ie truth=1 false = 0, unknown = 1/2

in a previous project I had to come up with formulae for 2 valued logic as follows:

negation
t(~p) = 1-t(p)

Conjunction
T(p^q) = min[t(p), t(q)]

Disjunction
T(p V q) = max[t(p), t(q)]

Conditional
~p -> q === ~p V q => t(p->q) = t[~pVq]
=> max[t(~p), t(q)]
=> max[1-t(p), t(q)]

biconditional
p<->q === (p->q)^(q->p) => t(p->q) = t[(p->q)^(q->p)]
=> min[t(p->q), t(q->p)
=> min[max[1-t(p), t(q)], max[1-t(q), t(p)]]
using this information I have to define the connectives for 3 valued logic. and I dont really know how to do that. this is due tomorrow, please help!!!

outermeasure · **Posted:** Fri, 8 Jun 2012 06:11:24 UTC

samaha wrote:

so basically i have a project about 3 valued logic ie truth=1 false = 0, unknown = 1/2

in a previous project I had to come up with formulae for 2 valued logic as follows:

negation
t(~p) = 1-t(p)

Conjunction
T(p^q) = min[t(p), t(q)]

Disjunction
T(p V q) = max[t(p), t(q)]

Conditional
~p -> q === ~p V q => t(p->q) = t[~pVq]
=> max[t(~p), t(q)]
=> max[1-t(p), t(q)]

biconditional
p<->q === (p->q)^(q->p) => t(p->q) = t[(p->q)^(q->p)]
=> min[t(p->q), t(q->p)
=> min[max[1-t(p), t(q)], max[1-t(q), t(p)]]
using this information I have to define the connectives for 3 valued logic. and I dont really know how to do that. this is due tomorrow, please help!!!

Note that there isn't a complemented lattice of 3 elements, so you necessarily don't have double negation (or excluded middle).

On the other hand, the lattice is pseudocomplemented, so ...

Soroban · **Posted:** Fri, 8 Jun 2012 18:22:34 UTC

Hello, samaha!

I've changed the code somewhat.

Quote:

So basically i have a project about 3-valued logic: .true = T, false = F, unknown = 0.

I would construct the basic truth tables, using common sense or baby-talk.

. . $\begin{array}{|c||c|} p & \sim p \\ \hline T & F \\ 0 & 0 \\ F & T \\ \hline\end{array}$

. . $\begin{array}{|c|c||c|c|c|c|} p & q & p \wedge q & p \vee q & p \to q & p \leftrightarrow q \\ \hline T&T & T&T&T&T \\ T&0 & 0&T&0&0 \\ T&F & F&T&F&F \\ 0&T & 0&T&T&0 \\ 0&0 & 0&0&0&0 \\ 0&F& F&0&0&0 \\ F&T & F&T&T&F \\ F&0 & F&0&T&0 \\ F&F & F&F&T&T \\ \hline \end{array}$

Then apply these to your problems.

outermeasure · **Posted:** Fri, 8 Jun 2012 21:36:42 UTC

Soroban wrote:

$\begin{array}{|c||c|} p & \sim p \\ \hline T & F \\ 0 & 0 \\ F & T \\ \hline\end{array}$

I disagree. $v(\neg p)=\ast v(p)$ , where $v\colon L\to\Lambda$ is our valuation, $\ast\colon\Lambda\to\Lambda$ is our pseudocomplement operator on our pseudocomplemented lattice $\Lambda$ . In particular, the pseudocomplement of the unique nontrivial element in our 3-element bounded lattice is $\bot$ . See, for example, Heyting algebra example in wikipedia (the third one is what the OP asked for).

A way to understand why we necessary have $v(\neg p)=0$ when $v(p)=\frac{1}{2}$ is to examine what would you get if you apply your rules for $v(p\wedge\neg p)$ . Since $v(p\wedge p)=v(p)$ , you must define $v(p\wedge q)=\frac{1}{2}$ when $v(p)=v(q)=\frac{1}{2}$ . Also, $v(p\wedge(\neg p))=v(\bot)=0$ , so $v(\neg p)$ cannot be 1/2 or 1 when v(p)=1/2.

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discrete mathematics "3 value logic" help

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