Discrete Mathematics | Propositional Logic | Test 1 | Question: 10

Question

Consider the following two statements:    
      
i. Sentence $\textit{Neither A nor B}$ can be represented by $A \downarrow B$ where $\downarrow$ is used in Boolean circuits for $\textit{nor}$ function.   
    
ii. Sentence $\textit{not at once A and B}$ can be represented by $A \uparrow B$ where $\uparrow$ is used in Boolean circuits for $\textit{nand}$ function.  
Which one of the following is correct?   
    

• A. Only $(i)$ is correct    
       
• B. Only $(ii)$ is correct    
      
• C. Both $(i)$ and $(ii)$ are correct    
       
• D. None of the above

Answer 1

1. $Neither$ $A$ $nor$ $B$ $\equiv$ $\neg (A \vee B) \equiv A \downarrow B$

$A$	$B$	$A \vee B$	$\neg (A \vee B)\equiv A \downarrow B$
$F$	$F$	$F$	$T$
$F$	$T$	$T$	$F$
$T$	$F$	$T$	$F$
$T$	$T$	$T$	$F$

2. $not$ $at$ $once$ $A$ $and$ $B$ $\equiv \neg (A \wedge B) \equiv A \uparrow B$

$A$	$B$	$A\wedge B$	$\neg (A \wedge B) \equiv A \uparrow B$
$F$	$F$	$F$	$T$
$F$	$T$	$F$	$T$
$T$	$F$	$F$	$T$
$T$	$T$	$T$	$F$

 

Both are correct.

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

Some points about “neither .. nor to logic translation”: 

https://math.stackexchange.com/questions/25091/translating-neither-nor-into-a-mathematical-logical-expression

Similar logic you can use also for “not at once .… and to logic translation”.

NOTE: Truth tables to show how they are equivalent.