Question 253525
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~(p^q) v (pvq).

Letters stand for sentences.  Some sentences are true and some are false.

Fro example let p = "Today is Monday" and q = "This month is August".

Sometimes p is true and q is true, say on a Monday in August.
Sometimes p is true and q is false, say on a Monday in March.
Sometimes p is false and q is true, say on a Tuesday in August.
Sometimes p is false and q is false, say on a Tuesday in March.

~(p^q) v (pvq).
That says:

It's not both Monday and August or it's Monday or August.

That's a compound sentence and it is hard to analyze, just reading it. 
That's why we need truth tables to sort out the various possibilities 

There are only four possibilities to consider.

1.  p is true and q is true
2.  q is true and q is false
3.  p is false and q is true
4   p is false and q is false

So we start with those four possibilities in a table:

The binary operators are ^ and v, "and" and "or". They always have letters
on both sides of them.

The unary operator is ~, "not".  It can only have one letter on the right 
side of it.

^ means "and".  In order to be true it must have true statements on both
sides of it.  It is false any other time.

v means "or".  In order for it to be true it only needs to have just one 
true statement on either side of it.  It is only false when it has false
sentences on BOTH sides.

~ before a letter means that the sentence that follows it is false.

So the truth table for a statement that has two sentences, that is, 
letters, p and q, starts out as this:

        p  q
case 1  T  T
case 2  T  F
case 3  F  T
case 4  F  F

To build a truth table for ~(p^q) v (pvq)

we have to make headings for each of these: 
A.  p
B.  q
C.  (p^q)
D.  ~(p^q) 
E.  (pvq)
F.  ~(p^q) v (pvq)

We start with the four cases for letters p, q, steps A and B 
we can build C from A and B
We can build D from C
We can build E from A and B
We can build F from D and E

So we start with this:

        p  q | (p^q) | ~(p^q) | (pvq) | ~(p^q) v (pvq) |
case 1  T  T |       |        |       |                |
case 2  T  F |       |        |       |                |
case 3  F  T |       |        |       |                |
case 4  F  F |       |        |       |                |

We fill the (p^q) column using the rule for "and".

Case 1 has a T under p and a T under q.
That's both T's, so the (p^q) column gets a T, for case 1.

Case 2 has a T under p and a F under q,
That's not both T's, so the (p^q) column gets a F for case 2.
"And" needs 2 T's to be true.

Case 3 has a F under p and a T under q,
That's not both T's, so the (p^q) column gets a F for case 3.
"And" needs 2 T's to be true.

Case 4 has a F under p and a F under q,
That's certainly not both T's, so the (p^q) column gets a F for case 4.
"And" needs 2 T's to be true.

So the cases under (p^q) go "TFFF"


        p  q | (p^q) | ~(p^q) | (pvq) | ~(p^q) v (pvq) |
case 1  T  T |   T   |        |       |                |
case 2  T  F |   F   |        |       |                |
case 3  F  T |   F   |        |       |                |
case 4  F  F |   F   |        |       |                |  

Now to fill in the next column, we notice it has "~" or "not" before
(p^q), so we put the exact opposite of what is in the (p^q) column.  The
(p^q) column has "TFFF", so the ~(p^q) column has "FTTT":

        p  q | (p^q) | ~(p^q) | (pvq) | ~(p^q) v (pvq) |
case 1  T  T |   T   |    F   |       |                |
case 2  T  F |   F   |    T   |       |                |
case 3  F  T |   F   |    T   |       |                |
case 4  F  F |   F   |    T   |       |                |  

Next we fill in the (pvq) column.

Case 1 has a T under p and a T under q.
All "v" needs is at least 1 T on either or both sides of it. p has a T
and Q has a T, so that's at least one T on at least one side of "v", so
case 1 gets a T.

Case 2 has a T under p and an F under q.
All "v" needs is at least 1 T on either or both sides of it. p has a T
and Q has a F, so that's at least one T on at least one side of "v", so
case 2 gets a T.

Case 3 has a F under p and a T under q.
All "v" needs is at least 1 T on either or both sides of it. p has a F
and Q has a T, so that's at least one T on at least one side of "v", so
case 3 gets a T.

Case 4 has a F under p and a F under q.
But "v" needs at least 1 T on either or both sides of it to be true. It does
NOT have T on either side of it in case 4, so case 4 gets an F.

So the (pvq) column goes TTTF


        p  q | (p^q) | ~(p^q) | (pvq) | ~(p^q) v (pvq) |
case 1  T  T |   T   |    F   |   T   |                |
case 2  T  F |   F   |    T   |   T   |                |
case 3  F  T |   F   |    T   |   T   |                |
case 4  F  F |   F   |    T   |   F   |                |  

Now we just have one more column to fill in.  It has "v" or "OR" between
what's in the previous two columns, ~(p^q), (pvq).  All we need in order to
put a true in that column is for just at least ONE of the preceding two
columns to have a T in it. 

Case 1 has a F under ~(p^q) and a T under (pvq),
That's at least one T, so the last column gets a T, for case 2.

Case 2 has a T under ~(p^q) and a T under (pvq),
That's at least one T, so the last column gets a T for case 2.

Case 3 has a T under ~(p^q) and a T under (pvq),
That's at least one T, so the last column gets a T for case 3.

Case 4 has a T under ~(p^q) and an F under (pvq),
That's at least one T, so the last column gets a T for case 4. 

So the final column gets filled in TTTT.

That means the statement is ALWAYS true in EVERY case.  Since that happened
we say ~(p^q) v (pvq) is ALWAYS true regardless of whether p or q is true or
false.  That's called a "tautology".

By the way, the fancy word for "and" ("^") is "conjunction.
The fancy word for "or" ("v") is "disjunction".
The fancy word for "not" ("~") is "negation".

Edwin</pre>