Thread: constructing truth tables for 3 valued logic help

i have a homework assignment that deals with 3 valued logic, ie, true = 1, false = 0, unknown = 1/2. I dont really understand 3 valued logic all that well, and would like to get some help on the following problems.

making truth tables for these problems, and determine if the propositional forms are a tautology, a contradiction, or neither. If someone could do maybe one or two, I can probably figure out the rest. or do all that would be awesome


a) [(p-> q) -> p] -> p

b) p <-> [p ^ (pVq)]

c) (p -> q) <-> (p ^~q)

d) ~(p V q) <-> (~p ^ ~q)

e) (p <-> q) <-> ~(~p V q) V (~p^q)

Do you know how to construct truth tables for 2-valued logic? What exactly do you not understand concerning truth tables for 3-valued logic?

like here is what i have for a, i just dont really understand the unknown and how to make a table with that involved. is this right?


T(p) T(q) p->q p (p->q)->p [(p->q)->p]->p
1 1 1 1 1 1
1 0 0 1 1 1
1 ½ ½ 1 ½ ½
0 1 0 0 1 0
0 0 1 0 0 1
0 ½ ½ 0 ½ ½
½ 1 ½ ½ ½ ½
½ 0 ½ ½ ½ ½
½ ½ ½ ½ ½ ½

sorry for the alignment

For 2-valued logic there are cases you have to consider: TT, TF, FT, and FF (or 11, 10, 01, and 00). In three valued logic you must consider cases with 2 basic formulae: 11, 1(1/2), 10, (1/2)1, (1/2)(1/2), (1/2)0, 01, 0(1/2), and 00.

i just dont know if my truth table is correct for a). someone confirm it and whether im right if it is a tautology, contradiction, or neither. i just get confused with the 1/2. or unknown.

You can use the [code]...[/code] tag to preserve alignment.

Code:

T(p) T(q) p->q p (p->q)->p [(p->q)->p]->p
 1    1    1   1     1           1
 1    0    0   1     1           1
 1    ½    ½   1     ½           ½
 0    1    0*  0     1*          0*
 0    0    1   0     0           1
 0    ½    ½   0     ½           ½
 ½    1    ½   ½     ½           ½
 ½    0    ½   ½     ½           ½
 ½    ½    ½   ½     ½           ½

Check the values followed by *.

The important thing is that there are many, in fact, , ways to define each binary connective, such as implication. More than one of these ways is reasonable or suitable for a particular application. Since three-valued logics are less common than the regular two-valued logic, these definitions are not standard and are usually given in each source. One cannot proceed without such definitions. My guess concerning the reason for your confusion is that you don't have the definitions of connectives. In your truth table, implication equals ½ if either of the arguments is ½. It is a possible convention; however, I would expect that ½ -> 0 would be 0, for example.

im confused now lol is the * wrong? why would that be wrong?

these are my definitions maybe they re wrong


T(p) T(q) T(~p) T(~q) T(pVq) T(p^q) T(p->q) T(p<->q)
1 1 0 0 1 1 1 1
1 0 0 1 1 0 0 0
1 ½ 0 ½ 1 ½ ½ ½
0 1 1 0 1 0 0 0
0 0 1 1 0 0 1 1
0 ½ 1 ½ ½ 0 ½ ½
½ 1 ½ 0 1 ½ ½ ½
½ 0 ½ 1 ½ 0 ½ ½
½ ½ ½ ½ ½ ½ ½ ½

Hmm, you definitely have non-standard definition because 0 -> 1 = 0. In this case, check the value of (p->q)->p for p = 1 and q = 0. Except that line, the truth table in post #3 seems OK.

Ok i found my mistake. im going to post all the truth tables for a- e, if you could just check to make sure i didnt make any errors, that would be awesome! trying to get 100% on this. here they are:

a) i found this proposition was a contradiction. see above ^^ since i made a mistake pointed out once fixed it made it all 1/2 or 1. which i believe is a contradiction.

Code:

T(p)    T(q)    pVq    P^(pVq)    p    p <-> [p ^ (pVq)]
1    1    1    1    1    1
1    0    1    1    1    1
1    ½    1    1    1    1
0    1    1    0    0    1
0    0    0    0    0    1
0    ½    ½    0    0    1
½    1    1    ½    ½    1
½    0    ½    ½    ½    1
½    ½    ½    ½    ½    1

p <-> [p ^ (pVq)]
b is a tautology

Code:

T(p)    T(q)    ~q    P ^ ~q    P -> q    (p -> q) <-> (p ^~q)
1    1    0    0    1    0
1    0    1    1    0    0
1    ½    ½    ½    ½    1
0    1    0    0    0    1
0    0    1    0    1    0
0    ½    ½    0    ½    ½
½    1    0    0    ½    ½
½    0    1    ½    ½    1
½    ½    ½    ½    ½    1

(p -> q) <-> (p ^~q)
c is neither ***check this one*****

Code:

T(p)    T(q)    ~p    ~q    ~p ^ ~q    pVq    ~(pVq)    ~(p V q) <-> (~p ^ ~q)
1    1    0    0    0    1    0    1
1    0    0    1    0    1    0    1
1    ½    0    ½    0    1    0    1
0    1    1    0    0    1    0    1
0    0    1    1    1    0    1    1
0    ½    1    ½    ½    ½    ½    1
½    1    ½    0    0    1    0    1
½    0    ½    1    ½    ½    ½    1
½    ½    ½    ½    ½    ½    ½    1

~(p V q) <-> (~p ^ ~q)
d is a tautology

Code:

T(p)    T(q)    P ⇔ q    ~p    q    ~p^q    ~pVq    ~(~pVq)    ~(~pVq)V(~p^q)    P ⇔ q    (p <-> q) <-> ~(~p V q) V (~p^q)
1    1    1    0    1    0    1    0    0    1    0
1    0    0    0    0    0    0    1    1    0    0
1    ½    ½    0    ½    0    ½    ½    ½    ½    ½
0    1    0    1    1    1    1    0    1    0    0
0    0    1    1    0    0    1    0    0    1    0
0    ½    ½    1    ½    ½    1    0    ½    ½    ½
½    1    ½    ½    1    ½    1    0    ½    ½    ½
½    0    ½    ½    0    0    ½    ½    ½    ½    ½
½    ½    ½    ½    ½    ½    ½    ½    ½    ½    ½

(p <-> q) <-> ~(~p V q) V (~p^q)
e is a contradiction ***check this one*****

Hello, samaha34!

I've modified the code somewhat . . .

I have a homework assignment that deals with 3-valued logic:
. . true = T, false = F, unknown = 0.
I don't really understand 3-valued logic all that well.

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

. .


. .


Then apply these to your problems.