Given [latex]p: 4+7=11[/latex], [latex]q: 11-3=7[/latex], and [latex]r: 7 \times 11=77[/latex], construct a truth table to determine the truth value of each compound statement.

a) [latex]\sim{p} \land q \lor r[/latex]

Step 1: The statement “[latex]\sim{p} \land q \lor r[/latex]” contains three basic logical statements, [latex]p[/latex], [latex]q[/latex], and [latex]r[/latex], and three connectives. When we place parentheses in the statement to indicate the dominance of connectives, the statement becomes [latex](((\sim{p}) \land q) \lor r)[/latex].

Step 2: After we have applied the dominance of connectives, we create a two-row table that includes a column for each basic statement that makes up the compound statement and an additional column for the contents of each set of parentheses. Because we have three sets of parentheses, we include a column for [latex]\sim{p}[/latex], the innermost parentheses; a column for [latex](\sim{p}) \land q[/latex], the next set of parentheses; and [latex]((\sim{p}) \land q) \lor r[/latex] in the last column for the third set of parentheses.

Step 3: Once the table is created, we determine the truth value of each statement starting from left to right. The truth values of [latex]p[/latex], [latex]q[/latex], and [latex]r[/latex] are true, false, and true, respectively, so we place T, F, and T in the second row of the table. Because [latex]p[/latex] is true, [latex]\sim{p}[/latex] is false.

Step 4: Next, evaluate [latex]\sim{p} \land q[/latex] from the table: [latex]\sim{p}[/latex] is false, and [latex]q[/latex] is also false, so [latex]\sim{p} \land q[/latex] is false, because a conjunction is only true if both of the statements that make it are true. Place an F in the table below its heading.

Step 5: Finally, using the table, we understand that [latex](\sim{p} \land q)[/latex] is false and [latex]r[/latex] is true, so the complete statement [latex](\sim{p} \land q) \lor r[/latex] is false or true, which is true (because a disjunction is true whenever at least one of the statements that make it is true). Place a T in the last column of the table. The complete statement [latex]\sim{p} \land q \lor r[/latex] is true.

[latex]\begin{array} {|c|c|c|c|c|c|} \hline p & q & r & \sim{p} & (\sim{p}) \land q & ((\sim{p}) \land q) \lor r \\ \hline \text{T} & \text{F} & \text{T} & \text{F} & \text{F}& \textbf{T} \\ \hline \end{array}[/latex]

b) [latex]\sim{p} \lor q \land r[/latex]

Step 1: Applying the dominance of connectives to the original compound statement [latex]\sim{p} \lor q \land r[/latex], we get [latex](((\sim{p}) \lor q) \land r)[/latex].

Step 2: The table needs columns for [latex]p[/latex], [latex]q[/latex], [latex]r[/latex], [latex]\sim{p}[/latex], [latex](\sim{p}) \lor q[/latex], and [latex]((\sim{p}) \lor q) \land r[/latex].

Step 3: The truth values of [latex]p[/latex], [latex]q[/latex], [latex]r[/latex], and [latex]\sim{p}[/latex] are the same as in Question (a).

Step 4: Next, [latex]\sim{p} \lor q[/latex] is false or false, which is false, so we place an F below this statement in the table. This is the only time that a disjunction is false.

Step 5: Finally, [latex]\sim{p} \lor q[/latex] and [latex]r[/latex] are the conjunction of the statements [latex]\sim{p} \lor q[/latex] and [latex]r[/latex], and so the expressions evaluate to false and true, which is false. Recall that the only time an “and” statement is true is when both statements that form it are also true. The complete statement [latex]\sim{p} \lor q \land r[/latex] is false.

[latex]\begin{array} {|c|c|c|c|c|c|} \hline p & q & r & \sim{p} & (\sim{p}) \lor q & ((\sim{p}) \lor q) \land r \\ \hline \text{T} & \text{F} & \text{T} & \text{F} & \text{F}& \textbf{F} \\ \hline \end{array}[/latex]

c) [latex]\sim{(p \land r)} \lor q[/latex]

Step 1: Applying the dominance of connectives to the original statement, we have [latex]((\sim{(p \land r)}) \lor q)[/latex].

Step 2: So the table needs the following columns: [latex]p[/latex], [latex]q[/latex], [latex]r[/latex], [latex]\sim{(p \land r)}[/latex], and [latex]\sim{(p \land r)} \lor q[/latex].

Step 3: The truth values of [latex]p[/latex], [latex]q[/latex], and [latex]r[/latex] are the same as in Questions 1 and 2.

Step 4: From the table it can be seen that [latex]p \land r[/latex] is true and true, which is true. So the negation of [latex]p[/latex] and [latex]r[/latex] is false, because the negation of a statement always has the opposite truth value of the original statement.

Step 5: Finally, [latex]\sim{(p \land r)} \lor q[/latex] is the disjunction of [latex]\sim{(p \land r)}[/latex] with [latex]q[/latex], and so we have false or false, which makes the complete statement false.

[latex]\begin{array} {|c|c|c|c|c|c|} \hline p & q & r & p \land r & \sim{(p \land r)} & \sim{(p \land r)} \lor q \\ \hline \text{T} & \text{F} & \text{T} & \text{T} & \text{F}& \textbf{F} \\ \hline \end{array}[/latex]

Construct a truth table to analyze all possible outcomes for each of the following arguments.

a) [latex]p \lor \sim{q}[/latex]

Step 1: Because there are two basic statements, [latex]p[/latex] and [latex]q[/latex], and each of these has two possible outcomes, we will have [latex]2(2)=4[/latex] rows in our table to represent all possible outcomes: TT, TF, FT, and FF. The columns will include [latex]p[/latex], [latex]q[/latex], [latex]\sim{q}[/latex], and [latex]p \lor q[/latex].

Step 2: Every value in column [latex]\sim{q}[/latex] will have the opposite truth value of the corresponding value in column [latex]q[/latex]: F, T, F, and T.

Step 3: To complete the last column, evaluate each element in column [latex]p[/latex] with the corresponding element in column [latex]\sim{q}[/latex] using the connective or.

[latex]\begin{array} {|c|c|c|c|} \hline p & q & \sim{q} & p \lor \sim{q} \\ \hline \text{T} & \text{T} & \text{F} & \textbf{T} \\ \hline \text{T} & \text{F} & \text{T} & \textbf{T} \\ \hline \text{F} & \text{T} & \text{F} & \textbf{F} \\ \hline \text{F} & \text{F} & \text{T} & \textbf{T} \\ \hline \end{array}[/latex]

b) [latex]\sim{(p \land q)}[/latex]

Step 1: The columns will include [latex]p[/latex], [latex]q[/latex], [latex]p \land q[/latex], and [latex]\sim{(p \land q)}[/latex]. Because there are two basic statements, [latex]p[/latex] and [latex]q[/latex], the table will have four rows to account for all possible outcomes.

Step 2: The [latex]p \land q[/latex] column will be completed by evaluating the corresponding elements in columns [latex]p[/latex] and [latex]q[/latex], respectively, with the and connective.

Step 3: The final column, [latex]\sim{(p \land q)}[/latex], will be the negation of the [latex]p \land q[/latex] column.

[latex]\begin{array} {|c|c|c|c|} \hline p & q & p \land q & \sim{p \land q} \\ \hline \text{T} & \text{T} & \text{T} & \textbf{F} \\ \hline \text{T} & \text{F} & \text{F} & \textbf{T} \\ \hline \text{F} & \text{T} & \text{F} & \textbf{T} \\ \hline \text{F} & \text{F} & \text{F} & \textbf{T} \\ \hline \end{array}[/latex]

c) [latex](p \lor q) \land r[/latex]

Step 1: This statement has three basic statements, [latex]p[/latex], [latex]q[/latex], and [latex]r[/latex]. Because each basic statement has two possible truth values, true or false, the multiplication principle indicates there are [latex]2(2)(2)=8[/latex] possible outcomes. So eight rows of outcomes are needed in the truth table to account for each possibility. Half of the eight possibilities must be true for the first statement, and half must be false.

Step 2: So the first column for statement [latex]p[/latex] will have four Ts followed by four Fs. In the second column for statement [latex]q[/latex], when [latex]p[/latex] is true, half the outcomes for [latex]q[/latex] must be true, and the other half must be false, and the same pattern will repeat for when [latex]p[/latex] is false. So column [latex]q[/latex] will have TT, FF, TT, FF.

Step 3: The column for the third statement, [latex]r[/latex], must alternate between T and F. Once the three basic propositions are listed, you will need a column for [latex]\sim{q}[/latex], [latex]p \lor \sim{q}[/latex], and [latex](p \lor \sim{q}) \land r[/latex].

Step 4: The column for the negation of [latex]q[/latex], [latex]\sim{q}[/latex], will have the opposite truth value of each value in column [latex]q[/latex].

Step 5: Next, fill in the truth values for the column containing the statement [latex]p \lor \sim{q} .[/latex] The or statement is true if at least one of [latex]p[/latex] or [latex]\sim{q}[/latex] is true; otherwise, it is false.

Step 6: Finally, fill in the column containing the conjunction [latex](p \lor \sim{q}) \land r[/latex]. To evaluate this statement, combine column [latex]p \lor \sim{q}[/latex] and column [latex]r[/latex] with the and connective. Recall that the only time “and” is true is when both values are true; otherwise, the statement is false. The complete truth table is

[latex]\begin{array} {|c|c|c|c|c|c|} \hline p & q & r & \sim{q} & p \lor \sim{q} & (p \lor \sim{q}) \land r \\ \hline \text{T} & \text{T} & \text{T} & \text{F} & \text{T} & \textbf{T} \\ \hline \text{T} & \text{T} & \text{F} & \text{F} & \text{T} & \textbf{F} \\ \hline \text{T} & \text{F} & \text{T} & \text{T} & \text{T} & \textbf{T} \\ \hline \text{T} & \text{F} & \text{F} & \text{T} & \text{T} & \textbf{F} \\ \hline \text{F} & \text{T} & \text{T} & \text{F} & \text{F} & \textbf{F} \\ \hline \text{F} & \text{T} & \text{F} & \text{F} & \text{F} & \textbf{F} \\ \hline \text{F} & \text{F} & \text{T} & \text{T} & \text{T} & \textbf{T} \\ \hline \text{F} & \text{F} & \text{F} & \text{T} & \text{T} & \textbf{F} \\ \hline \end{array}[/latex]

Construct a truth table to determine the validity of each of the following statements.

a) [latex]\sim{p} \land q[/latex]

Step 1: Because there are two statements, [latex]p[/latex] and [latex]q[/latex], and each of these has two possible outcomes, there will be [latex]2(2)=4[/latex] rows in our table to represent all possible outcomes: TT, TF, FT, and FF.

Step 2: The columns will include [latex]p[/latex], [latex]q[/latex], [latex]\sim{p}[/latex], and [latex]\sim{p} \land q[/latex]. Every value in column [latex]\sim{p}[/latex] will have the opposite truth value of the corresponding value in column [latex]p[/latex].

Step 3: To complete the last column, evaluate each element in column [latex]\sim{p}[/latex] with the corresponding element in column [latex]q[/latex] using the connective and . The last column contains at least one false; therefore, the statement [latex]\sim{p} \land q[/latex] is not valid.

[latex]\begin{array} {|c|c|c|c|} \hline p & q & \sim{p} & \sim{p} \land q \\ \hline \text{T} & \text{T} & \text{F} & \textbf{F} \\ \hline \text{T} & \text{F} & \text{F} & \textbf{F} \\ \hline \text{F} & \text{T} & \text{T} & \textbf{T} \\ \hline \text{F} & \text{F} & \text{T} & \textbf{F} \\ \hline \end{array}[/latex]

b) [latex]\sim{(p \land \sim{p})}[/latex]

Step 1: Because the statement [latex]\sim{(p \land \sim{p})}[/latex] only contains one basic proposition, the truth table will only contain two rows. Statement [latex]p[/latex] may be either true or false.

Step 2: The columns will include [latex]p[/latex], [latex]\sim{p}[/latex], [latex]p \land \sim{p}[/latex], and [latex]\sim{(p \land \sim{p})}[/latex]. Evaluate column [latex]p \land \sim{p}[/latex] with the and connective, because the symbol [latex]\land[/latex] represents a conjunction or logical and statement. True and false is false, and false and true is also false.

Step 3: The final column is the negation of each entry in the third column, both of which are false, so the negation of false is true. Because all the truth values in the final column are true, the statement [latex]\sim{(p \land \sim{p})}[/latex] is valid.

[latex]\begin{array} {|c|c|c|c|} \hline p & \sim{p} & p \land \sim{p} & \sim{(p \land \sim{p})} \\ \hline \text{T} & \text{F} & \text{F} & \textbf{T} \\ \hline \text{F} & \text{T} & \text{F} & \textbf{T} \\ \hline \end{array}[/latex]