Fractions and Decimals

Discrete Mathematics

Differential Equations

Solved on Nov 17, 2023

Find the truth table for the logical statement: $(\sim c \vee t) \wedge p$

STEP 1

Assumptions1. The statement is $(\sim c \vee t) \wedge p$ . The symbols represent the following logical operations $\sim$ is NOT, $\vee$ is OR, and $\wedge$ is AND3. The variables $c$ , $t$ , and $p$ can each take on the values of true () or false ()

STEP 2

We will first create a truth table for the variables $c$ , $t$ , and $p$ . Since there are three variables, there will be $2^ =8$ possible combinations of truth values.
$\begin{array}{ccc} c & t & p \\ \hline & & \\ & & F \\ & F & \\ & F & F \\ & & \\ & & F \\ & F & \\ & F & F \\ \end{array}$

STEP 3

Next, we will find the truth values for $\sim c$ . This is the negation of $c$ , so if $c$ is true, $\sim c$ is false, and vice versa.
$\begin{array}{cc} c & \sim c \\ \hline & F \\ & F \\ & F \\ & F \\ & \\ & \\ & \\ & \\ \end{array}$

STEP 4

Now, we will find the truth values for $\sim c \vee t$ . This is the disjunction (OR) of $\sim c$ and $t$ , so it is true if either $\sim c$ or $t$ is true.
$\begin{array}{ccc} \sim c & t & \sim c \vee t \\ \hline & & \\ & & \\ & F & F \\ & F & F \\ & & \\ & & \\ & F & \\ & F & \\ \end{array}$

STEP 5

Finally, we will find the truth values for $(\sim c \vee t) \wedge p$ . This is the conjunction (AND) of $(\sim c \vee t)$ and $p$ , so it is true only if both $(\sim c \vee t)$ and $p$ are true.
$\begin{array}{ccc} \sim c \vee t & p & (\sim c \vee t) \wedge p \\ \hline & & \\ & F & F \\ & & F \\ & F & F \\ & & \\ & F & F \\ & & \\ & F & F \\ \end{array}$
So, the truth table for the statement $(\sim c \vee t) \wedge p$ is as follows $\begin{array}{cccc} c & t & p & (\sim c \vee t) \wedge p \\ \hline & & & \\ & & F & F \\ & F & & F \\ & F & F & F \\ & & & \\ & & F & F \\ & F & & \\ & F & F & F \\ \end{array}$

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