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|}
 
|}
  
<b>(A) </b>&not;Q&#9633&not;P
+
(A) $&not;Q&#9633&not;P$
&nbsp;
+
 
<b>(B) </b>P&#9633&not;Q
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(B) '''$P&#9633&not;Q$'''
&nbsp;
+
 
<b>(C) </b>&not;P&#9633Q
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(C) $&not;P&#9633Q$
&nbsp;
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<b>(D) </b>&not;P&#9633&not;Q
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(D) $&not;P&#9633&not;Q$
 
==={{Template:Author|Happy Mittal|{{mittalweb}} }}===
 
==={{Template:Author|Happy Mittal|{{mittalweb}} }}===
  
If we compare column of $P&#9633;Q$ in table with $P &or; Q$, we need both $F$ in $3^{rd}$ row of table, and for that we need  
+
If we compare column of $P&#9633;Q$ in table with $P &or; Q$, we need both F in $3^{rd}$ row of table, and for that we need  
 
$&not;Q$ instead of $Q$. So $P &or; Q$ is equivalent to $P&#9633&not;Q$, and therefore, option <b>(B)</b> is correct.
 
$&not;Q$ instead of $Q$. So $P &or; Q$ is equivalent to $P&#9633&not;Q$, and therefore, option <b>(B)</b> is correct.
  

Revision as of 19:43, 14 July 2014

The binary operation □ is defined as follows

P Q P□Q
T T T
T F T
F T F
F F T

(A) $¬Q&#9633¬P$

(B) $P&#9633¬Q$

(C) $¬P&#9633Q$

(D) $¬P&#9633¬Q$

Solution by Happy Mittal

If we compare column of $P□Q$ in table with $P ∨ Q$, we need both F in $3^{rd}$ row of table, and for that we need $¬Q$ instead of $Q$. So $P ∨ Q$ is equivalent to $P&#9633¬Q$, and therefore, option (B) is correct.




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The binary operation □ is defined as follows

P Q P□Q
T T T
T F T
F T F
F F T

(A) ¬Q&#9633¬P   (B) P&#9633¬Q   (C) ¬P&#9633Q   (D) ¬P&#9633¬Q

Solution by Happy Mittal[edit]

If we compare column of $P□Q$ in table with $P ∨ Q$, we need both $F$ in $3^{rd}$ row of table, and for that we need $¬Q$ instead of $Q$. So $P ∨ Q$ is equivalent to $P&#9633¬Q$, and therefore, option (B) is correct.




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