MHB Create a Truth Table: (p ^ q) ->(p ▼ ~q)

  • Thread starter Thread starter rymatson406
  • Start date Start date
  • Tags Tags
    Table Truth table
rymatson406
Messages
3
Reaction score
0
(p ^ q) ->(p ▼ ~q)

Need help creating a truth table (6 columns) for the above.Thanks
 
Physics news on Phys.org
I assume the formula is $(p\land q)\to(p\lor \neg q)$.

Have you seen examples of creating a truth table? Can you write a truth table for $p\land q$, which is a part of the required truth table? What exactly is your difficulty?
 
Presumably your first two columns are for $p$ and $q$, which have two possible values each, making the table of necessity four rows "deep". I would go with:

T T
T F
F T
F F

for the first two columns (but other row-orders are possible).

It's hard to say what the next four columns are supposed to be, my guess is:

$\neg q,\ p \wedge q,\ p \vee \neg q$ and $(p \wedge q) \rightarrow (p \vee \neg q)$.

Really, only the last one is "necessary", but it's easier on the ol' noggin to include the other 3.
 
Hello, rymatson406!

Create a truth table for: .(p \wedge q)\;\to\;(p \,\vee \sim q)
. . . . \begin{array}{|c|c|c|c|c|c|c|c|c|} <br /> p &amp; q &amp; (p &amp; \wedge &amp; q) &amp; \to &amp; (p &amp; \vee &amp; \sim q) \\ \hline<br /> T&amp;T &amp; T&amp;T&amp;T &amp;T&amp; T&amp;T&amp;F \\<br /> T&amp;F &amp;T&amp;F&amp;F &amp;T&amp; T&amp;T&amp;T \\<br /> F&amp;T &amp;F&amp;F&amp;T &amp;T&amp; F&amp;F&amp;F \\<br /> F&amp;F &amp; F&amp;F&amp;F &amp;T&amp; F&amp;T&amp;T \\ \hline<br /> &amp;&amp; 1&amp;2&amp;1&amp;3&amp;1&amp;2&amp;1 \\ \hline \end{array}**
 
Hi all, I've been a roulette player for more than 10 years (although I took time off here and there) and it's only now that I'm trying to understand the physics of the game. Basically my strategy in roulette is to divide the wheel roughly into two halves (let's call them A and B). My theory is that in roulette there will invariably be variance. In other words, if A comes up 5 times in a row, B will be due to come up soon. However I have been proven wrong many times, and I have seen some...
Thread 'Detail of Diagonalization Lemma'
The following is more or less taken from page 6 of C. Smorynski's "Self-Reference and Modal Logic". (Springer, 1985) (I couldn't get raised brackets to indicate codification (Gödel numbering), so I use a box. The overline is assigning a name. The detail I would like clarification on is in the second step in the last line, where we have an m-overlined, and we substitute the expression for m. Are we saying that the name of a coded term is the same as the coded term? Thanks in advance.
Back
Top