Question #275316

 

Show that each of these conditional statements is a tautology by using truth tables.

 

a) (p q) → p                b) p → (p q)

c) p → (p q)                            d) (p q) → (p q)

e) ¬(p q) → p                        f) ¬(p q) → ¬q



1
Expert's answer
2021-12-06T16:24:00-0500

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

 

a) (∧ q) → p               


pqpq(pq)p0001010110011111\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)


pqpqp(pq)0001011110111111\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)                            


pq¬ppq¬p(pq)00111011111000111011\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)


pqpqpq(pq)(pq)00011010111000111111\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                        


pqpq¬(pq)¬(pq)p00101011011001111101\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


pqpq¬(pq)¬q¬(pq)¬q001011011001100111111001\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.


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