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## Here's the Solution to this Question

Let us show that each of these conditional statements is a tautology by using truth tables.

a) (∧ q) → p

$\begin{array}{||c|c||c|c||} \hline\hline p & q &p ∧ q & (p ∧ q) → p \\ \hline\hline 0 & 0 & 0 & 1\\ \hline 0 & 1 & 0 & 1 \\ \hline 1 & 0 & 0 & 1\\ \hline 1 & 1 & 1 & 1\\ \hline\hline \end{array}$

Since the last column contains only 1, we conclude that this formula is a tautology.

b) → (∨ q)

$\begin{array}{||c|c||c|c||} \hline\hline p & q &p \lor q &p → (p ∨ q) \\ \hline\hline 0 & 0 & 0 & 1\\ \hline 0 & 1 & 1 & 1 \\ \hline 1 & 0 & 1 & 1\\ \hline 1 & 1 & 1 & 1\\ \hline\hline \end{array}$

Since the last column contains only 1, we conclude that this formula is a tautology.

c) → (→ q)

$\begin{array}{||c|c||c|c|c||} \hline\hline p & q &\neg p & p \to q &\neg p → (p \to q) \\ \hline\hline 0 & 0 & 1 & 1 & 1\\ \hline 0 & 1 & 1 & 1 & 1 \\ \hline 1 & 0 & 0 & 0 & 1\\ \hline 1 & 1 & 0 & 1 &1\\ \hline\hline \end{array}$

Since the last column contains only 1, we conclude that this formula is a tautology.

d) (∧ q) → (→ q)

$\begin{array}{||c|c||c|c|c||} \hline\hline p & q &p ∧ q & p \to q &(p ∧ q) → (p → q)\\ \hline\hline 0 & 0 & 0 & 1 & 1\\ \hline 0 & 1 & 0 & 1 & 1 \\ \hline 1 & 0 & 0 & 0 & 1\\ \hline 1 & 1 & 1 & 1 &1\\ \hline\hline \end{array}$

Since the last column contains only 1, we conclude that this formula is a tautology.

e) ￢(→ q) → p

$\begin{array}{||c|c||c|c|c||} \hline\hline p & q & p \to q& \neg (p → q) & \neg(p → q) → p \\ \hline\hline 0 & 0 & 1 & 0 & 1\\ \hline 0 & 1 & 1 & 0 & 1 \\ \hline 1 & 0 & 0 & 1 & 1\\ \hline 1 & 1 & 1 & 0 &1\\ \hline\hline \end{array}$

Since the last column contains only 1, we conclude that this formula is a tautology.

﻿f) ￢(→ q) → ￢q

$\begin{array}{||c|c||c|c|c|c||} \hline\hline p & q & p \to q& \neg (p → q) & \neg q & \neg (p → q) \to \neg q\\ \hline\hline 0 & 0 & 1 & 0 & 1 & 1\\ \hline 0 & 1 & 1 & 0 & 0 & 1 \\ \hline 1 & 0 & 0 & 1 & 1 & 1\\ \hline 1 & 1 & 1 & 0 & 0 & 1\\ \hline\hline \end{array}$

Since the last column contains only 1, we conclude that this formula is a tautology.