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

[rymatson406]

(p ^ q) ->(p ▼ ~q)

Need help creating a truth table (6 columns) for the above.Thanks

 

[Evgeny.Makarov]

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?

 

[Deveno]

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.

 

[soroban]

Hello, rymatson406!

Create a truth table for: .[tex](p \wedge q)\;\to\;(p \,\vee \sim q)[/tex]

. . . . [tex]\begin{array}{|c|c|c|c|c|c|c|c|c|}
p & q & (p & \wedge & q) & \to & (p & \vee & \sim q) \\ \hline
T&T & T&T&T &T& T&T&F \\
T&F &T&F&F &T& T&T&T \\
F&T &F&F&T &T& F&F&F \\
F&F & F&F&F &T& F&T&T \\ \hline
&& 1&2&1&3&1&2&1 \\ \hline \end{array}[/tex]**

 

[blue_raver22]

for reaching out! I would be happy to assist you in creating a truth table for the given statement.

Before we begin, let's first define the symbols used in the statement:

^ : represents the logical operator "AND"
-> : represents the logical operator "IF...THEN"
▼ : represents the logical operator "OR"
~ : represents the logical operator "NOT"

Now, let's create the truth table with 6 columns as requested:

| p | q | ~q | p ^ q | p ▼ ~q | (p ^ q) ->(p ▼ ~q) |
|---|---|----|-------|--------|---------------------|
| T | T | F | T | T | T |
| T | F | T | F | T | T |
| F | T | F | F | F | T |
| F | F | T | F | T | T |

In the above truth table, we have considered all possible combinations of the truth values for p and q. As you can see, the first two rows satisfy the given statement, while the last two rows do not.

In summary, the statement (p ^ q) ->(p ▼ ~q) is true in three out of four possible cases, making it a valid statement. I hope this helps in understanding the logic behind the given statement. Let me know if you have any further questions. Keep exploring the world of logic and reasoning!