Discrete Mathematics | Propositional Logic | Test 1 | Question: 8

Question

Consider the following truth table for the connective $\rightarrow:$    
      
$$
\begin{array}{c|c|c}
p & q & p \rightarrow q \\\hline
T & T & T \\
T & F & F \\
F & T & T \\
F & F & T
\end{array}
$$
(T stands for True and F stands for False)    
     
 Now, consider the following statements:    
      
 i. For the last two rows, consider that if both were to result in $F,$ then $p \rightarrow q$ would be evaluated the same as $p \wedge q$      
       
 ii.  If we were to take the result of third row as $T$ and the fourth row as $F,$ then $p \rightarrow q$ would be the same as $q$   
               
 iii. If it were to take the result of third row as $F$ and the last row as
$T,$ then $p \rightarrow q$ would be the same as $q \rightarrow p$     
      
Now, which one of the following statements is correct?     
    

• A. Only (iii) is correct     
       
• B. (i) and (iii) are correct    
       
• C. (i) and (ii) are correct    
       
• D. (i), (ii) and (iii) are correct

Comments

how statement (iii) is right? Since we have altered only the last two rows, the second row is unchanged. when q is F and p is true then p→q would give false and q→p would give true.

---

(iii) should be correct as if we make the changes then p→q is evaluated as p↔q, so p→q should be same as q→p

Answer 1

1.  For the last two rows, consider that if both were to result in $F$, then $p → q$ would be evaluated the same as $p \wedge q$.
   Let’s assume, the table is this according to this statement. 

$p$	$q$	$p → q$	$p\wedge q$
$T$	$T$	$T$	$T$
$T$	$F$	$F$	$F$
$F$	$T$	$F$	$F$
$F$	$F$	$F$	$F$

Yes, then $p → q \equiv p \wedge q$. So, this statement is $True$.

2.  If we were to take the result of third row as $T$ and the fourth row as $F$, then $p → q$ would be the same as $q$. 

$p$	$q$	$p → q$
$T$	$T$	$T$
$T$	$F$	$F$
$F$	$T$	$T$
$F$	$F$	$F$

 Yes, then $p → q \equiv q$. So, this statement is $True$.

3. If it were to take the result of third row as $F$ and the last row as $T$, then $p → q$ would be the same as $q → p$.     

$p$	$q$	$p → q$	$q → p$
$T$	$T$	$T$	$T$
$T$	$F$	$F$	$T$
$F$	$T$	$F$	$F$
$F$	$F$	$T$	$T$

 NO, $p → q \neq q → p$; the truth values are different for both. So, this statement is $False$.

Ans: C. (i) and (ii) are correct.