# Understanding multiple values in a single cell of a truth table (P v ~Q) as (1 1 0 1) when P and Q are 1

This might seem basic to most here but I am struggling with a truth table for a disjunct. As I am looking at it further, I actually think the issue I am struggling with how to interpret truth values of negations.

The proposition is as follows: P v ~Q

The truth table goes

P Q --------P V ~ Q

1 1--------- 1 1 0 1

1 0--------- 1 1 1 0

0 1--------- 0 0 0 1

0 0--------- 0 1 1 0

I'm trying to see if I understand this correctly. In row 1 when Q is said to be true, does that mean 'Q' in isolation is true and so in the phrase '~ Q' the negation is now false? Which would mean '~ Q' is in effect just 'Q'? And the negation gets a false truth value?

And so the reason the (inclusive) disjunct holds in row 1 is because the proposition equals "P (true) or Q (true)" and since a disjunct states that one or both components of its proposition are true, and in this case both are, the disjunct holds?

Is that how row 3 is to be explained? P is not true, Q is. But (true) Q is negated, making it an untrue statement. And so the proposition is saying "(untrue) P or (untrue) Q". So in effect it is neither, and thus the disjunct doesn't hold (as it has to be one or both).

Another one I am struggling with is ~E ^ D

E D -------- ~ E ^ D

1 1---------- 0 1 0 1

1 0---------- 0 1 0 0

0 1---------- 1 0 1 1

0 0---------- 1 0 0 0

If E is true, that means its negation is not? And so in ~E ^ D, we have both E and D as true, and so the conjunct operator should have a positive truth value.. no?

• What are the 4-tuples on the right? If P and Q are both 1, then P v Q is 1. Not "1101". I don't follow your notation. – user4894 Aug 9 '18 at 17:03
• The textbook I am using says, in the case of P v ~Q, if both P and Q are true, then P gets a positive truth value, Q gets a positive truth value, its negation gets a negative one, and the disjunct gets a positive one. P = 1. Q = 1. ~ = 0. v = 1 – freigz Aug 9 '18 at 17:39
• If a sentence is true (or ‘1’), then the sentence’s negation is false (or ‘0’). A disjunction is 1 as long as one disjunct is 1. In P v ~Q, the disjuncts are P and ~Q, rather than P and Q! Also, when Q is 0, ~Q is 1. P v ~Q is then also 1. By contrast, if Q is 1, ~Q is 0. P v ~Q is then also 0, unless P is 1. That’s what happens in the first row: ~Q is 0 (because Q is 1), but P is 1; so one disjunct is 1; so the disjunction is 1. In the third row, both P and ~Q are 0, whence the disjunction is also 0. – MarkOxford Aug 9 '18 at 17:51
• For conjunction: “If E is true, that means its negation is not?” Yes. “And so in ~E ^ D, we have both E and D as true” If both E and D are 1, then ~E is 0, as you said. Yet if ~E is 0, then ~E^D is also 0 because conjunction is 1 only if both conjuncts are 1. (The conjuncts are ~E and D, rather than E and D.) – MarkOxford Aug 9 '18 at 17:59
• Remember that truth tables aren’t built left to right, but from the inside outwards, as it were. We have a disjunction, of the form φ∨ψ. Before we can assign a truth value to ‘∨’, we must first assign values to φ and ψ. As φ=P, we just copy over the value for P. However, ψ is itself complex: ψ=¬Q, so ψ is of the form ¬χ. Once again, before we can assign a truth value to ‘¬’, we must assign a value to χ. Since χ=Q, we copy over the value for Q. Your logic book should discuss the concept of the main connective. The main connective is always the last item to be assigned a truth value. – MarkOxford Aug 16 '18 at 9:12

You begin with the literals.

`````` (P v (~ Q))     ((~ E) ^ D)
(1   (  1))     ((  1)   1)
(1   (  0))     ((  1)   0)
(0   (  1))     ((  0)   1)
(0   (  0))     ((  0)   1)
``````

Then the negation of Q, E

``````       ~ *         ~ *
(P v (~ Q))     ((~ E) ^ D)
(1   (0 1))     ((0 1)   1)
(1   (1 0))     ((0 1)   0)
(0   (0 1))     ((1 0)   0)
(0   (1 0))     ((1 0)   0)
``````

Finally the disjunction of P and the negation, the conjunction of D and the negation

``````  * v  *           *    ^ *
(P v (~ Q))     ((~ E) ^ D)
(1 1 (0 1))     ((0 1) 0 1)
(1 1 (1 0))     ((0 1) 0 0)
(0 0 (0 1))     ((1 0) 1 1)
(0 1 (1 0))     ((1 0) 0 0)
``````

Alternatively

`````` P : Q | ~Q : Pv~Q
1 : 1 | 0  :  1
1 : 0 | 1  :  1
0 : 1 | 0  :  0
0 : 0 | 1  :  1

E : D | ~E : ~E^D
1 : 1 | 0  :   0
1 : 0 | 0  :   0
0 : 1 | 1  :   1
0 : 0 | 1  :   0
``````

There's basically two notations for doing truth-tables. The one you're using is the harder one, so it might be best to look at how it would look with the easier one and then transition to the harder one.

The easier one uses the following format:

1. Columns for each variable
2. Columns for helper rows
3. Columns for the premises
4. A column for the conclusion

(In the case of what you're doing we haven't gotten so far as a conclusion.

For P v ~Q, we have:

1. The variables P and Q
2. The helper row ~Q
3. The premise row P v ~ Q

``````P |  Q  | ~Q | P v ~Q
1 |  1  |  0 |   1
1 |  0  |  1 |   1
0 |  1  |  0 |   0
0 |  0  |  1 |   1
``````

The more advanced version does the same thing but puts the helper rows in situ with spacing (~Q's value is placed directly under ~Q inside the premise).

1. The variables E and D
2. The helper row ~E
3. The output ~E ^ D

``````E |  D  | ~E | ~E ^ D
1 |  1  |  0 |    0
1 |  0  |  0 |    0
0 |  1  |  1 |    1
0 |  0  |  1 |    0
``````

Done the other way:

pedantic first intermediate step:

``````    E |  D  |  ~E ^ D
1 |  1  |  _1 _ 1
1 |  0  |  _1 _ 0
0 |  1  |  _0 _ 1
0 |  0  |  _0 _ 0
``````

(after this I will forgo writing the value E itself there just to make it less painful on the eyes).

intermediate step looks like this:

``````    E |  D  |  ~E ^ D
1 |  1  |  0  _ 1
1 |  0  |  0  _ 0
0 |  1  |  1  _ 1
0 |  0  |  1  _ 0
``````

final product looks like this:

``````    E |  D  | ~E ^ D
1 |  1  |  0 0 1
1 |  0  |  0 0 0
0 |  1  |  1 1 1
0 |  0  |  1 0 0
``````

Simply put, once you get used to it, the condensed format is easier to do, but until you understand the method, the multiple values in what looks like a single column thing is really confusing.

I am unfamiliar with your notation. Here is the standard presentation for the truth table for P v ~Q : http://www.wolframalpha.com/input/?i=truth+table+p+or+not+q

The first line tells us that if P is true and Q is true, then P v ~Q is true.

The second line tells us that if P is true and Q is false, then P v ~Q is true.

The third line tells us that if P is false and Q is true, then P v ~Q is false.

The fourth line tells us that if P is false and Q is false, then P v ~Q is true.