(p; wedge sim q) wedge (sim p vee q) is | vedclass

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Question

(a) Clearly, $(p\; \wedge \sim q) \wedge (p\; \vee \sim q)$ is a contradiction.


	
		
			
			$p$
			
			
			$q$
			
			
			$~p$

Vedclass · Accepted Answer

A contradiction

Vedclass · Answer

(a) Clearly, $(p\; \wedge \sim q) \wedge (p\; \vee \sim q)$ is a contradiction.


	
		
			
			$p$
			
			
			$q$
			
			
			$~p$
			
			
			$~q$
			
			
			$p \wedge ~q$
			
			
			$~p \vee q$
			
			
			$(p \wedge ~q) \wedge (~p \vee q)$
			
		
		
			
			$T$
			
			
			$T$
			
			
			$F$
			
			
			$F$
			
			
			$F$
			
			
			$T$
			
			
			$F$
			
		
		
			
			$T$
			
			
			$F$
			
			
			$F$
			
			
			$T$
			
			
			$T$
			
			
			$F$
			
			
			$F$
			
		
		
			
			$F$
			
			
			$T$
			
			
			$T$
			
			
			$F$
			
			
			$F$
			
			
			$T$
			
			
			$F$
			
		
		
			
			$F$
			
			
			$F$
			
			
			$T$
			
			
			$T$
			
			
			$F$
			
			
			$T$
			
			
			$F$

$(p\; \wedge \sim q) \wedge (\sim p \vee q)$ is

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