I'm having some difficulties in understanding why there are values under the symbol $\lor$ and $\neg$ in the truth table. Could someone explain me why and/or how you give a value to a symbol or is there something I'm misunderstanding?

truth table

Thanks alot!


I'll be analyzing the second line.

On the third column you should first read the connectives whose scope is the smallest and progress to the connectives with larger scopes.

The 'first level' of these scopes is the second negation in the statement $\neg (P\lor\color{red}{\neg}Q)\lor Q$. This negation affects only the first $Q$ and nothing more.
The 'second level' of these scopes is the first disjunction which affects only the first $P$ and $\neg Q$.
The 'third level' is the first negation which affects only $P\lor \neg Q$.
Finally the 'fourth level' is the second disjunction which affects $\neg(P\lor \neg Q)$ and the second $Q$.

Having established this, I proceed with analyzing the second line.

In the second line $P$'s truth value is $1$ and $Q$'s is $0$.

Now what happens on the first level scope connectives? Since $Q$ is $0$, then $\neg Q$ is $1$, thus a $0$ appears right below all $Q$'s in the statement and a $1$ appears below the second $\neg$ in the second line.

Since $P$ is $1$, a $1$ appears below every $P$ in the statement. The second level is the first disjunction, in which the first $P$ in being disjuncted with $\neg Q$.Since $P$ is $1$, a $1$ appears below $\lor$.

The third level is the negation and it will be flipping the truth value of the disjunction of the truth value just mentioned above, hence a $0$ appears right below the first $\neg$.

In the fourth level, there's a disjunction between the first $\neg$ (and all its scope) which is evaluated as $0$ and the last $P$ which is $1$. Therefore a $1$ appears below the last disjunction.

For a more detailed explanation regarding the concept of 'scope' you can check the section Ambiguity and parentheses on the book Language, Proof and Logic.You can find that section on page 79 of the book (89 of the PDF).

| cite | improve this answer | |

I think that you have the light gray being the values of $P$ and $Q$ repeated.

The values (black) under the second $\neg$ are the values for $\neg Q$. And the valued under the first $\lor$ are the values for $P\lor \neg Q$. So the values under the first $\neg$ are the values for the statement $\neg(P\lor \neg Q)$ and the final $\lor$ are the values for the whole statement.

I would make the table like this: $$ \begin{array}{cc|c|c|c|c|} P&Q&\neg Q & P\lor \neg Q & \neg(P\lor \neg Q) & \neg(P\lor \neg Q) \lor P\\ 0 & 0 & 1& 1& ...& ...\\ 0 & 1 & 0& 0& ...& ...\\ 1 & 0 & 1& 1& ...& ...\\ 1 & 1 & 0& 1& ...& ... \end{array} $$ You can probably fill in the rest.

| cite | improve this answer | |
  • $\begingroup$ is there any other way to set up this formula in a truth table so it becomes easier to read? I get confused by setting up truth table like this. Would appreciate if you could give an example. $\endgroup$ – Dabbish Sep 10 '13 at 15:22
  • $\begingroup$ There's lots of worked examples in standard textbooks, like the one I mention in my answer (which I'm told that students find particularly clear). $\endgroup$ – Peter Smith Jul 10 '14 at 13:41

Truth tables of complicated expressions are sometimes written by making separate columns for each of the separate operations that occur in the expression. So, for your expression, we may first make a column that tabulates the values of $P\lor \neg Q$, then make a separate column for $\neg(P\lor \neg Q)$, then another for $P$ and then a final column for $\neg(P\lor\neg Q)\lor P$. This makes it easy for evaluating the final expression for different values of $P$ and $Q$.

However, a more compact method demonstrated in your table is the following:

Under each variable that occur in your expression ($P,\,Q$), we write their respective values that they take on in their respective columns. Then, if a variable is negated, we write the negated values under the said variable's negation sign. Then the values of expressions like $P\land Q$ or $P\lor \neg Q$, etc. are evaluated and are written below the $\lor$ or $\land$ sign. We repeat this until we have evaluated the final expression.

In your case, the final values are written under the $\lor$ sign connecting $\neg(P\lor\neg Q)$ and $P$.

| cite | improve this answer | |

A basic assumption of propositional logic goes something like that all of the connectives qualify as truth-functional, which in more conventional mathematical terms might get more specifically stated as that all connectives qualify as truth-operational. This means that if we assign truth variables to any variables, there will exist some value V(a, ..., z) where V indicates a function of propositions. For your example we have two truth-operations $\lor$ such that

$\lor$(0, 0)=0, and

$\lor$(0, 1)=$\lor$(1, 0)=$\lor$(1, 1)=1.

$\lnot$(0)=1, $\lnot$(1)=0.

You give a value to a symbol $\lor$ or $\lnot$ by using those functions or something similar.

By no means do you have to set-up truth tables in infix notation. In infix notation the principle connective jumps all over the place, in the sense that it's position varies from well-formed formula to well-formed formula. You can set up truth tables by writing all binary operations (those like $\lor$ and $\land$ which qualify as truth-operations of two truth values) after their arguments. Or you could set up truth tables by writing all binary operations before their arguments. With your example you could write:

  p  q  ¬  ∨ ¬   p ∨
  1  1  0  1  0  1  1
  1  0  1  1  0  1  1
  0  1  0  0  1  0  1
  0  0  1  1  0  0  0


  p|  q| q ¬| p q ¬ ∨| p q ¬ ∨ ¬| p q ¬ ∨ ¬   p ∨
  1|  1|   0|        1|          0|                1
  1|  0|   1|        1|          0|                1
  0|  1|   0|        0|          1|                1
  0|  0|   1|        1|          0|                0
| cite | improve this answer | |

The convention in play is that in the main part of the table (after recording the truth-values of atoms -- if we bother to that again, since that's just copying across the values assigned at the beginning of the line) we write the truth-value under the main connective of the (sub)formula we are evaluating at that step. So it is indeed wffs that are being evaluated -- we just place the resulting valuation under the connective whose impact is relevant at that point in the overall calculation.

You'll find a clear explanation of this convention in most textbooks.

But I'd perhaps particularly recommend looking at Ch. 9, and especially at §9.7 (called "Short working") of P*t*r Sm*th's An Introduction to Formal Logic (CUP).

Well, I would, wouldn't I ...!

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.