truth table help

Question

Write the truth table for ~(a ^ b) V (a V b) ?

Answer 1

Philip's answer is correct, but just to break down the concept a little further, I will add a response as well.

I will also use T to mean true, and F to mean false, although you can use 1 for true, and 0 for false as Daniel did depending on your preference.

 

What I would do is to start with a basic table for just a and b where you fill in every possible combination of truth values for each. Since there are 2 letters, then there are 2^2 = 4 possible combinations.

 

a  |  b

-------

T    T

T    F

F    T

F    F

 

Now, you can break the statement up into pieces to make it more manageable, starting with a^b
 and use the table table above to evaluate a^b:

 

 

a |  b  |  a^b
---------------
T    T       T
T    F       F
F    T       F
F    F       F

 

The value in the 3rd column comes from evaluating the expression a^b by using the values for a and b in the first two columns.

 

Then you can now evaluate ~(a^b) since that is part of the original statement you are working towards evaluating.

 

 

a  |  b  |  a^b  | ~(a^b)
--------------------------
T     T       T          F
T     F       F          T
F     T       F          T
F     F       F          T

 

So, in the table above, the values in the 4th column come from looking at the 3rd column, and since we are negating the statement in the 3rd column then it makes sense that each value in the 4th column is the opposite of the value to its left in the 3rd column.

 

In this particular case, you don't need to include the original table for a and b, however to save time, you can just add on to the table from the previous step instead of writing out a whole new table just for ~(a^b).

 

Now we can do the same for a V b, and here I will write it out as its own table using the original table for a and b, but at the end you will see how it can all be done as one whole table, just like Philip did in his response.

 

a  | b  |  a V b
----------------
T    T       T
T    F       T
F    T       T
F    F       F

 

You can see here that there is only one case where a V b is false, because for it to be true it requires that at least one of either a or be to be true. If both are true then that works as well. But if both are false, then you can't say that either a or b are true.

 

now we could put everything into one table, but just for the sake of illustration and to make the point clear without confusing you I will just do a table with ~(a^b) and a v b   so that you can see how their truth values will determine the truth of your original statement: ~(a ^ b) V (a V b) . because all that is required is for either

~(a^b) to be true or for (a V b) to be true. if both are true then that works too.

 

~(a^b)  |  (a V b)  |  ~(a^b) V (a V b)
-------------------------------------------
     F           T                     T

     T           T                     T

     T           T                     T

     T           F                     T

 

So you can see that the statement: ~(a^b) V (a V b) is always true.

 

Here is everything now combined as one table:

 

 

a  |  b  |  a^b  |  ~(a^b) | (a V b) | ~(a^b) V (a V b)
------------------------------------------------------------
T     T      T             F           T                   T

T     F      F             T           T                   T

F     T      F             T           T                   T

F     F      F             T           F                   T

 

 

So in practice, what you would do is create the first line of the table

by starting with a and b. Then you break down the statement into its

pieces so that you can evaluate them each individually in the table.

This makes it easier to evaluate the compound statement.

 

Any questions just ask!

Answer 2

Write the truth table for ~(a∧b)∨(a∨b)

 

a   b  a∧b  ~(a∧b)  (a∨b)  ~(a∧b)∨(a∨b)
0   1    0         1           1              1
0   0    0         1           0              1
1   1    1         0           1              1
1   0    0         1           1              1

 

You could also use DeMorgan's Theorem:

 

~(a∧b) = ~a ∨ ~b

 

Substitute that back into the original logical expression and re-arrange a bit and you get:

 

(~a ∨ a) ∨ (~b ∨ b) 

 

Which, yeah, is pretty much going to be true all the time.