Chapter 1.2, Problem 16E

(a)

To determine

To find: Whether $\left[(P\Rightarrow Q)\Rightarrow P\right]\Rightarrow P$ is a tautology, a contradiction, or neither.

Answer to Problem 16E

Tautology

Explanation of Solution

Concept Used:

If the propositions $P$ is false, or $Q$ is true if and only if the sentence $P\Rightarrow Q$ is true.

In a truth table, if the given expression holds true for all values then it is a tautology, if it holds false for all values then it is a contradiction, else it is neither.

Calculation:

Below is the truth table for the given expression.

  $\boxed{\begin{array}{lllll}P\hfill & Q\hfill & P\Rightarrow Q\hfill & (P\Rightarrow Q)\Rightarrow P\hfill & \left[(P\Rightarrow Q)\Rightarrow P\right]\Rightarrow P\hfill \\ \text{T}\hfill & \text{T}\hfill & \text{T}\hfill & \text{T}\hfill & \text{T}\hfill \\ \text{T}\hfill & \text{F}\hfill & \text{F}\hfill & \text{T}\hfill & \text{T}\hfill \\ \text{F}\hfill & \text{T}\hfill & \text{T}\hfill & \text{F}\hfill & \text{T}\hfill \\ \text{F}\hfill & \text{F}\hfill & \text{T}\hfill & \text{F}\hfill & \text{T}\hfill \end{array}}$

The given expression holds true for all values, therefore $\left[(P\Rightarrow Q)\Rightarrow P\right]\Rightarrow P$ is a tautology.

(b)

To determine

To find: Whether $P\iff P\wedge (P\vee Q)$ is a tautology, a contradiction, or neither.

Answer to Problem 16E

Tautology

Explanation of Solution

Concept Used:

If the propositions $P$ and $Q$ are true then $P\wedge Q$ is true.

If at least one of $P$ or $Q$ is true then $P\vee Q$ is true.

If the propositions $P$ and $Q$ have the same truth values then $P\iff Q$ is true.

Calculation:

Below is the truth table for the given expression.

  $\boxed{\begin{array}{lllll}P\hfill & Q\hfill & P\vee Q\hfill & P\wedge (P\vee Q)\hfill & P\iff P\wedge (P\vee Q)\hfill \\ \text{T}\hfill & \text{T}\hfill & \text{T}\hfill & \text{T}\hfill & \text{T}\hfill \\ \text{T}\hfill & \text{F}\hfill & \text{T}\hfill & \text{T}\hfill & \text{T}\hfill \\ \text{F}\hfill & \text{T}\hfill & \text{T}\hfill & \text{F}\hfill & \text{T}\hfill \\ \text{F}\hfill & \text{F}\hfill & \text{F}\hfill & \text{F}\hfill & \text{T}\hfill \end{array}}$

The given expression holds true for all values, therefore $P\iff P\wedge (P\vee Q)$ is a tautology.

(c)

To determine

To find: Whether $P\Rightarrow Q\iff P\wedge \sim Q$ is a tautology, a contradiction, or neither.

Answer to Problem 16E

Contradiction

Explanation of Solution

Concept Used:

If the propositions $P$ is false, then $𑨀P$ if true.

Calculation:

Below is the truth table for the given expression.

  $\boxed{\begin{array}{lllll}P\hfill & Q\hfill & P\Rightarrow Q\hfill & P\wedge \sim Q\hfill & P\Rightarrow Q\iff P\wedge \sim Q\hfill \\ \text{T}\hfill & \text{T}\hfill & \text{T}\hfill & \text{F}\hfill & \text{F}\hfill \\ \text{T}\hfill & \text{F}\hfill & \text{F}\hfill & \text{T}\hfill & \text{F}\hfill \\ \text{F}\hfill & \text{T}\hfill & \text{T}\hfill & \text{F}\hfill & \text{F}\hfill \\ \text{F}\hfill & \text{F}\hfill & \text{T}\hfill & \text{F}\hfill & \text{F}\hfill \end{array}}$

The given expression holds true for all values, therefore $P\Rightarrow Q\iff P\wedge \sim Q$ is a contradiction.

(d)

To determine

To find: Whether $P\Rightarrow \left[P\Rightarrow (P\Rightarrow Q)\right]$ is a tautology, a contradiction, or neither.

Answer to Problem 16E

Nether

Explanation of Solution

Calculation:

Below is the truth table for the given expression.

  $\boxed{\begin{array}{lllll}P\hfill & Q\hfill & P\Rightarrow Q\hfill & P\Rightarrow (P\Rightarrow Q)\hfill & P\Rightarrow \left[P\Rightarrow (P\Rightarrow Q)\right]\hfill \\ \text{T}\hfill & \text{T}\hfill & \text{T}\hfill & \text{T}\hfill & \text{T}\hfill \\ \text{T}\hfill & \text{F}\hfill & \text{F}\hfill & \text{F}\hfill & \text{F}\hfill \\ \text{F}\hfill & \text{T}\hfill & \text{T}\hfill & \text{T}\hfill & \text{T}\hfill \\ \text{F}\hfill & \text{F}\hfill & \text{T}\hfill & \text{T}\hfill & \text{T}\hfill \end{array}}$

The given expression holds true for all values, therefore $P\Rightarrow \left[P\Rightarrow (P\Rightarrow Q)\right]$ is neither a tautology, nor a contradiction.

(e)

To determine

To find: Whether $P\wedge (Q\vee \sim Q)\iff P$ is a tautology, a contradiction, or neither.

Answer to Problem 16E

Tautology

Explanation of Solution

Calculation:

Below is the truth table for the given expression.

  $\boxed{\begin{array}{llllll}P\hfill & Q\hfill & \sim Q\hfill & Q\vee \sim Q\hfill & P\wedge (Q\vee \sim Q)\hfill & P\wedge (Q\vee \sim Q)\iff P\hfill \\ \text{T}\hfill & \text{T}\hfill & \text{F}\hfill & \text{T}\hfill & \text{T}\hfill & \text{T}\hfill \\ \text{F}\hfill & \text{T}\hfill & \text{F}\hfill & \text{T}\hfill & \text{F}\hfill & \text{T}\hfill \\ \text{T}\hfill & \text{F}\hfill & \text{T}\hfill & \text{T}\hfill & \text{T}\hfill & \text{T}\hfill \\ \text{F}\hfill & \text{F}\hfill & \text{T}\hfill & \text{T}\hfill & \text{F}\hfill & \text{T}\hfill \end{array}}$

The given expression holds true for all values, therefore $P\wedge (Q\vee \sim Q)\iff P$ is a tautology.

(f)

To determine

To find: Whether $\left[Q\wedge (P\Rightarrow Q)\right]\Rightarrow P$ is a tautology, a contradiction, or neither.

Answer to Problem 16E

Neither

Explanation of Solution

Calculation:

Below is the truth table for the given expression.

  $\boxed{\begin{array}{lllll}P\hfill & Q\hfill & P\Rightarrow Q\hfill & Q\wedge (P\Rightarrow Q)\hfill & \left[Q\wedge (P\Rightarrow Q)\right]\Rightarrow P\hfill \\ \text{T}\hfill & \text{T}\hfill & \text{T}\hfill & \text{T}\hfill & \text{T}\hfill \\ \text{T}\hfill & \text{F}\hfill & \text{F}\hfill & \text{F}\hfill & \text{F}\hfill \\ \text{F}\hfill & \text{T}\hfill & \text{T}\hfill & \text{T}\hfill & \text{F}\hfill \\ \text{F}\hfill & \text{F}\hfill & \text{T}\hfill & \text{F}\hfill & \text{T}\hfill \end{array}}$

The given expression holds true for all values, therefore $\left[Q\wedge (P\Rightarrow Q)\right]\Rightarrow P$ is neither a tautology, nor a contradiction.

(g)

To determine

To find: Whether $(P\iff Q)\iff 𑨀(𑨀P\vee Q)\vee (𑨀P\wedge Q)$ is a tautology, a contradiction, or neither.

Answer to Problem 16E

Contradiction

Explanation of Solution

Calculation:

Below is the truth table for the given expression.

  $\boxed{\begin{array}{lllllllll}P\hfill & Q\hfill & \sim P\hfill & \sim P\vee Q\hfill & \sim (\sim P\vee Q)\hfill & \sim P\wedge Q\hfill & \sim (\sim P\vee Q)\vee (\sim P\wedge Q)\hfill & P\iff Q\hfill & (P\iff Q)\iff \sim (\sim P\vee Q)\vee (\sim P\wedge Q)\hfill \\ T\hfill & T\hfill & F\hfill & T\hfill & F\hfill & F\hfill & F\hfill & T\hfill & F\hfill \\ F\hfill & T\hfill & T\hfill & T\hfill & F\hfill & T\hfill & T\hfill & F\hfill & F\hfill \\ T\hfill & F\hfill & F\hfill & F\hfill & T\hfill & F\hfill & T\hfill & F\hfill & F\hfill \\ F\hfill & F\hfill & T\hfill & T\hfill & F\hfill & F\hfill & F\hfill & T\hfill & F\hfill \end{array}}$

The given expression holds true for all values, therefore $(P\iff Q)\iff \sim (\sim P\vee Q)\vee (\sim P\wedge Q)$ is a contradiction.

(h)

To determine

To find: Whether $\left[P\Rightarrow (Q\vee R)\right]\Rightarrow \left[(Q\Rightarrow R)\vee (R\Rightarrow P)\right]$ is a tautology, a contradiction, or neither.

Answer to Problem 16E

Tautology

Explanation of Solution

Calculation:

Below is the truth table for the given expression.

  $\boxed{\begin{array}{lllllllll}P\hfill & Q\hfill & R\hfill & Q\vee R\hfill & Q\Rightarrow R\hfill & R\Rightarrow P\hfill & (Q\Rightarrow R)\vee (R\Rightarrow P)\hfill & P\Rightarrow (Q\vee R)\hfill & \left[P\Rightarrow (Q\vee R)\right]\Rightarrow \left[(Q\Rightarrow R)\vee (R\Rightarrow P)\right]\hfill \\ \text{T}\hfill & \text{T}\hfill & \text{T}\hfill & \text{T}\hfill & \text{T}\hfill & \text{T}\hfill & \text{T}\hfill & \text{T}\hfill & \text{T}\hfill \\ \text{F}\hfill & \text{T}\hfill & \text{T}\hfill & \text{T}\hfill & \text{T}\hfill & \text{F}\hfill & \text{T}\hfill & \text{T}\hfill & \text{T}\hfill \\ \text{T}\hfill & \text{F}\hfill & \text{T}\hfill & \text{T}\hfill & \text{T}\hfill & \text{T}\hfill & \text{T}\hfill & \text{T}\hfill & \text{T}\hfill \\ \text{F}\hfill & \text{F}\hfill & \text{T}\hfill & \text{T}\hfill & \text{T}\hfill & \text{F}\hfill & \text{T}\hfill & \text{T}\hfill & \text{T}\hfill \\ \text{T}\hfill & \text{T}\hfill & \text{F}\hfill & \text{T}\hfill & \text{F}\hfill & \text{T}\hfill & \text{T}\hfill & \text{T}\hfill & \text{T}\hfill \\ \text{F}\hfill & \text{T}\hfill & \text{F}\hfill & \text{T}\hfill & \text{F}\hfill & \text{T}\hfill & \text{T}\hfill & \text{T}\hfill & \text{T}\hfill \\ \text{T}\hfill & \text{F}\hfill & \text{F}\hfill & \text{F}\hfill & \text{T}\hfill & \text{T}\hfill & \text{T}\hfill & \text{F}\hfill & \text{T}\hfill \\ \text{F}\hfill & \text{F}\hfill & \text{F}\hfill & \text{F}\hfill & \text{T}\hfill & \text{T}\hfill & \text{T}\hfill & \text{T}\hfill & \text{T}\hfill \end{array}}$

The given expression holds true for all values, therefore $\left[P\Rightarrow (Q\vee R)\right]\Rightarrow \left[(Q\Rightarrow R)\vee (R\Rightarrow P)\right]$ is a tautology.

(i)

To determine

To find: Whether $P\wedge (P\iff Q)\wedge 𑨀Q$ is a tautology, a contradiction, or neither.

Answer to Problem 16E

Contradiction

Explanation of Solution

Calculation:

Below is the truth table for the given expression.

  $\boxed{\begin{array}{llllll}P\hfill & Q\hfill & \sim Q\hfill & P\iff Q\hfill & (P\iff Q)\wedge \sim Q\hfill & P\wedge (P\iff Q)\wedge \sim Q\hfill \\ \text{T}\hfill & \text{T}\hfill & \text{F}\hfill & \text{T}\hfill & \text{F}\hfill & \text{F}\hfill \\ \text{F}\hfill & \text{T}\hfill & \text{F}\hfill & \text{F}\hfill & \text{F}\hfill & \text{F}\hfill \\ \text{T}\hfill & \text{F}\hfill & \text{T}\hfill & \text{F}\hfill & \text{F}\hfill & \text{F}\hfill \\ \text{F}\hfill & \text{F}\hfill & \text{T}\hfill & \text{T}\hfill & \text{T}\hfill & \text{F}\hfill \end{array}}$

The given expression holds true for all values, therefore $P\wedge (P\iff Q)\wedge \sim Q$ is a contradiction.

(j)

To determine

To find: Whether $(P\vee Q)\Rightarrow Q\Rightarrow P$ is a tautology, a contradiction, or neither.

Answer to Problem 16E

Neither

Explanation of Solution

Calculation:

Below is the truth table for the given expression.

  $\boxed{\begin{array}{lllll}P\hfill & Q\hfill & P\vee Q\hfill & (P\vee Q)\Rightarrow Q\hfill & (P\vee Q)\Rightarrow Q\Rightarrow P\hfill \\ \text{T}\hfill & \text{T}\hfill & \text{T}\hfill & \text{T}\hfill & \text{T}\hfill \\ \text{T}\hfill & \text{F}\hfill & \text{T}\hfill & \text{F}\hfill & \text{T}\hfill \\ \text{F}\hfill & \text{T}\hfill & \text{T}\hfill & \text{T}\hfill & \text{F}\hfill \\ \text{F}\hfill & \text{F}\hfill & \text{F}\hfill & \text{T}\hfill & \text{F}\hfill \end{array}}$

The given expression holds true for all values, therefore $(P\vee Q)\Rightarrow Q\Rightarrow P$ is neither a tautology, nor a contradiction.

(k)

To determine

To find: Whether $\left[P\Rightarrow (Q\wedge R)\right]\Rightarrow \left[R\Rightarrow (P\Rightarrow Q)\right]$ is a tautology, a contradiction, or neither.

Answer to Problem 16E

Tautology

Explanation of Solution

Calculation:

Below is the truth table for the given expression.

  $\boxed{\begin{array}{llllllll}P\hfill & Q\hfill & R\hfill & Q\wedge R\hfill & P\Rightarrow (Q\wedge R)\hfill & P\Rightarrow Q\hfill & R\Rightarrow (P\Rightarrow Q)\hfill & \left[P\Rightarrow (Q\wedge R)\right]\Rightarrow \left[R\Rightarrow (P\Rightarrow Q)\right]\hfill \\ \text{T}\hfill & \text{T}\hfill & \text{T}\hfill & \text{T}\hfill & \text{T}\hfill & \text{T}\hfill & \text{T}\hfill & \text{T}\hfill \\ \text{T}\hfill & \text{T}\hfill & \text{F}\hfill & \text{F}\hfill & \text{F}\hfill & \text{T}\hfill & \text{T}\hfill & \text{T}\hfill \\ \text{T}\hfill & \text{F}\hfill & \text{T}\hfill & \text{F}\hfill & \text{F}\hfill & \text{F}\hfill & \text{F}\hfill & \text{T}\hfill \\ \text{T}\hfill & \text{F}\hfill & \text{F}\hfill & \text{F}\hfill & \text{F}\hfill & \text{F}\hfill & \text{T}\hfill & \text{T}\hfill \\ \text{F}\hfill & \text{T}\hfill & \text{T}\hfill & \text{T}\hfill & \text{T}\hfill & \text{T}\hfill & \text{T}\hfill & \text{T}\hfill \\ \text{F}\hfill & \text{T}\hfill & \text{F}\hfill & \text{F}\hfill & \text{T}\hfill & \text{T}\hfill & \text{T}\hfill & \text{T}\hfill \\ \text{F}\hfill & \text{F}\hfill & \text{T}\hfill & \text{F}\hfill & \text{T}\hfill & \text{T}\hfill & \text{T}\hfill & \text{T}\hfill \\ \text{F}\hfill & \text{F}\hfill & \text{F}\hfill & \text{F}\hfill & \text{T}\hfill & \text{T}\hfill & \text{T}\hfill & \text{T}\hfill \end{array}}$

The given expression holds true for all values, therefore $\left[P\Rightarrow (Q\wedge R)\right]\Rightarrow \left[R\Rightarrow (P\Rightarrow Q)\right]$ is a tautology.

(l)

To determine

To find: Whether $\left[P\Rightarrow (Q\wedge R)\right]\Rightarrow R\Rightarrow (P\Rightarrow Q)$ is a tautology, a contradiction, or neither.

Answer to Problem 16E

Neither

Explanation of Solution

Calculation:

Below is the truth table for the given expression.

  $\boxed{\begin{array}{llllll}P\hfill & Q\hfill & R\hfill & P\Rightarrow (Q\wedge R)\hfill & \left[P\Rightarrow (Q\wedge R)\right]\Rightarrow R\hfill & \left[P\Rightarrow (Q\wedge R)\right]\Rightarrow R\Rightarrow (P\Rightarrow Q)\hfill \\ \text{T}\hfill & \text{T}\hfill & \text{T}\hfill & \text{T}\hfill & \text{T}\hfill & \text{T}\hfill \\ \text{T}\hfill & \text{T}\hfill & \text{F}\hfill & \text{F}\hfill & \text{T}\hfill & \text{T}\hfill \\ \text{T}\hfill & \text{F}\hfill & \text{T}\hfill & \text{F}\hfill & \text{T}\hfill & \text{F}\hfill \\ \text{T}\hfill & \text{F}\hfill & \text{F}\hfill & \text{F}\hfill & \text{T}\hfill & \text{F}\hfill \\ \text{F}\hfill & \text{T}\hfill & \text{T}\hfill & \text{T}\hfill & \text{T}\hfill & \text{T}\hfill \\ \text{F}\hfill & \text{T}\hfill & \text{F}\hfill & \text{T}\hfill & \text{T}\hfill & \text{T}\hfill \\ \text{F}\hfill & \text{F}\hfill & \text{T}\hfill & \text{T}\hfill & \text{T}\hfill & \text{T}\hfill \\ \text{F}\hfill & \text{F}\hfill & \text{F}\hfill & \text{T}\hfill & \text{T}\hfill & \text{T}\hfill \end{array}}$

The given expression holds true for all values, therefore $\left[P\Rightarrow (Q\wedge R)\right]\Rightarrow R\Rightarrow (P\Rightarrow Q)$ is neither a tautology, nor a contradiction.