Question

Discrete Math

I'm confused with the questions listed below. Can you please solve and explain in detail? how it transforms one to the other to get the answer?

Using propositional logic properties and other logical equivalences (not truth tables), prove the following statements: 1. (p Vq) V (p V -q) is a tautology 2. ((p-+ r) Л (q r) Л (pv q)) _+ r is a tautology 3. (pVq) Л (-р Л q) is a contradiction 4. (1-p) Λ (p Λ qAs . r) Λ p is a contradiction

Answer 1

1) (pVq) V (p V ¬q)

(pVq) V (¬q V p)

pV (q V ¬q) V p { Commutative law A V B = B V A }

pV (T) V p { we know that A V ¬A = T }

pV (T V p )

pV (T) { we know that A V T = T }

(p V T )

T { we know that A V T = T }

2) [(p  $\rightarrow$ r ) $\wedge$ ( q $\rightarrow$ r) $\wedge$ (p V q ) ] $\rightarrow$ r

[(¬p V r ) $\wedge$ ( ¬q V r) $\wedge$ (p V q )] $\rightarrow$ r { Law of Implies (A$\rightarrow$B) = ¬AVB }

[(¬p V r ) $\wedge$ (p V q) $\wedge$ ( ¬q V r)] $\rightarrow$ r { Law of Implies (A$\rightarrow$B) = ¬AVB }

[(¬p V r ) $\wedge$ (p V q) $\wedge$ (r V ¬q)] $\rightarrow$ r   { Commutative law A V B = B V A }

[(¬p V r ) $\wedge$ (p V r) $\wedge$ (q V ¬q)] $\rightarrow$ r  { Commutative law A V B = B V A }

[(¬p V r ) $\wedge$ (p V r) $\wedge$ (T)] $\rightarrow$ r  { we know that A V ¬A = T }

[(¬p V p ) $\wedge$ (r V r) $\wedge$ (T)] $\rightarrow$ r

[(T) $\wedge$ (r) $\wedge$ (T)] $\rightarrow$ r  { we know that A V A = A }

[(T $\wedge$ r) $\wedge$ (T)] $\rightarrow$ r  { we know that T $\wedge$ A = A }

[(r) $\wedge$ (T)] $\rightarrow$ r

[(r $\wedge$T)] $\rightarrow$ r

[(r)] $\rightarrow$ r { we know that T $\wedge$ A = A }

r $\rightarrow$ r

¬r V r { Law of Implies (A$\rightarrow$B) = ¬AVB }

T   { we know that A V ¬A = T }

3) (p V q) $\wedge$ (¬p $\wedge$ ¬q )

(p V q) $\wedge$ (¬q $\wedge$ ¬p )  { Commutative law A V B = B V A }

p V (q $\wedge$ ¬q) $\wedge$ ¬p  { Associative law (A V B) V C  = A V ( B V C ) }

p V (F) $\wedge$ ¬p  { we know that ¬A $\wedge$ A = F }

(p V F) $\wedge$ ¬p

(p) $\wedge$ ¬p  { we know that A V F = A }

(​​​​​​​p $\wedge$ ¬p )

F  { we know that ¬A $\wedge$ A = F }

4) ¬(q $\rightarrow$ p) $\wedge$ ( p $\wedge$ q $\wedge$ s $\rightarrow$ r) $\wedge$ p

¬(​​​​​​​¬q V p) $\wedge$ ( ​​​​​​​¬(p $\wedge$ q $\wedge$ s) V r) $\wedge$ p

(¬(​​​​​​​¬q) $\wedge$ ¬(​​​​​​​p)) $\wedge$ ( ​​​​​​​(¬p V ¬q V ¬s) V r) $\wedge$ p

(q $\wedge$ ¬​​​​​​​p)) $\wedge$ ( ​​​​​​​(¬p V ¬q V ¬s) V r) $\wedge$ p

(q $\wedge$ (¬​​​​​​​p $\wedge$¬p) V ¬q V ¬s V r) $\wedge$ p

(q $\wedge$¬q V (¬​​​​​​​p $\wedge$¬p) V ¬s V r) $\wedge$ p

((q $\wedge$¬q) V (¬​​​​​​​p $\wedge$¬p) V ¬s V r) $\wedge$ p

((F) V (¬​​​​​​​p $\wedge$¬p) V ¬s V r) $\wedge$ p

(F) V (¬​​​​​​​p $\wedge$¬p) V p $\wedge$ ¬s V r

(F) V ¬​​​​​​​p $\wedge$(¬p V p) $\wedge$​​​​​​​ ¬s V r

(F) V ¬​​​​​​​p $\wedge$(T) $\wedge$​​​​​​​ ¬s V r

(F) V ¬​​​​​​​p $\wedge$ ¬s $\wedge$​​​​​​​ T V r

(F) V ¬​​​​​​​p $\wedge$ ¬s ​​​​​​​$\wedge$​​​​​​​ (T V r)

​​​​​​​(F) V ¬​​​​​​​p $\wedge$ ¬s ​​​​​​​$\wedge$​​​​​​​ (T)

​​​​​​​((F) V ¬​​​​​​​p $\wedge$ ¬s ​​​​​​​$\wedge$​​​​​​​ (T)

​​​​​​​​​​​​​​F V ¬​​​​​​​p $\wedge$ ¬s ​​​​​​​$\wedge$​​​​​​​ (T)

¬​​​​​​​p V ​​​​​​​​​​​​​​F $\wedge$ ¬s ​​​​​​​$\wedge$​​​​​​​ (T)

¬​​​​​​​p V (​​​​​​​​​​​​​​F $\wedge$ ¬s) ​​​​​​​$\wedge$​​​​​​​ (T)

¬​​​​​​​p V (​​​​​​​​​​​​​​F) ​​​​​​​$\wedge$​​​​​​​ (T)

¬​​​​​​​p V (​​​​​​​​​​​​​​T) ​​​​​​​$\wedge$​​​​​​​ (F)

(¬​​​​​​​p V ​​​​​​​​​​​​​​T) ​​​​​​​$\wedge$​​​​​​​ (F)

(​​​​​​​​​​​​​​T) ​​​​​​​$\wedge$​​​​​​​ (F)

F

Answer 2

1) (pVq) V (p V ¬q)

(pVq) V (¬q V p)

pV (q V ¬q) V p { Commutative law A V B = B V A }

pV (T) V p { we know that A V ¬A = T }

pV (T V p )

pV (T) { we know that A V T = T }

(p V T )

T { we know that A V T = T }

2) [(p  $\rightarrow$ r ) $\wedge$ ( q $\rightarrow$ r) $\wedge$ (p V q ) ] $\rightarrow$ r

[(¬p V r ) $\wedge$ ( ¬q V r) $\wedge$ (p V q )] $\rightarrow$ r { Law of Implies (A$\rightarrow$B) = ¬AVB }

[(¬p V r ) $\wedge$ (p V q) $\wedge$ ( ¬q V r)] $\rightarrow$ r { Law of Implies (A$\rightarrow$B) = ¬AVB }

[(¬p V r ) $\wedge$ (p V q) $\wedge$ (r V ¬q)] $\rightarrow$ r   { Commutative law A V B = B V A }

[(¬p V r ) $\wedge$ (p V r) $\wedge$ (q V ¬q)] $\rightarrow$ r  { Commutative law A V B = B V A }

[(¬p V r ) $\wedge$ (p V r) $\wedge$ (T)] $\rightarrow$ r  { we know that A V ¬A = T }

[(¬p V p ) $\wedge$ (r V r) $\wedge$ (T)] $\rightarrow$ r

[(T) $\wedge$ (r) $\wedge$ (T)] $\rightarrow$ r  { we know that A V A = A }

[(T $\wedge$ r) $\wedge$ (T)] $\rightarrow$ r  { we know that T $\wedge$ A = A }

[(r) $\wedge$ (T)] $\rightarrow$ r

[(r $\wedge$T)] $\rightarrow$ r

[(r)] $\rightarrow$ r { we know that T $\wedge$ A = A }

r $\rightarrow$ r

¬r V r { Law of Implies (A$\rightarrow$B) = ¬AVB }

T   { we know that A V ¬A = T }

3) (p V q) $\wedge$ (¬p $\wedge$ ¬q )

(p V q) $\wedge$ (¬q $\wedge$ ¬p )  { Commutative law A V B = B V A }

p V (q $\wedge$ ¬q) $\wedge$ ¬p  { Associative law (A V B) V C  = A V ( B V C ) }

p V (F) $\wedge$ ¬p  { we know that ¬A $\wedge$ A = F }

(p V F) $\wedge$ ¬p

(p) $\wedge$ ¬p  { we know that A V F = A }

(​​​​​​​p $\wedge$ ¬p )

F  { we know that ¬A $\wedge$ A = F }

4) ¬(q $\rightarrow$ p) $\wedge$ ( p $\wedge$ q $\wedge$ s $\rightarrow$ r) $\wedge$ p

¬(​​​​​​​¬q V p) $\wedge$ ( ​​​​​​​¬(p $\wedge$ q $\wedge$ s) V r) $\wedge$ p

(¬(​​​​​​​¬q) $\wedge$ ¬(​​​​​​​p)) $\wedge$ ( ​​​​​​​(¬p V ¬q V ¬s) V r) $\wedge$ p

(q $\wedge$ ¬​​​​​​​p)) $\wedge$ ( ​​​​​​​(¬p V ¬q V ¬s) V r) $\wedge$ p

(q $\wedge$ (¬​​​​​​​p $\wedge$¬p) V ¬q V ¬s V r) $\wedge$ p

(q $\wedge$¬q V (¬​​​​​​​p $\wedge$¬p) V ¬s V r) $\wedge$ p

((q $\wedge$¬q) V (¬​​​​​​​p $\wedge$¬p) V ¬s V r) $\wedge$ p

((F) V (¬​​​​​​​p $\wedge$¬p) V ¬s V r) $\wedge$ p

(F) V (¬​​​​​​​p $\wedge$¬p) V p $\wedge$ ¬s V r

(F) V ¬​​​​​​​p $\wedge$(¬p V p) $\wedge$​​​​​​​ ¬s V r

(F) V ¬​​​​​​​p $\wedge$(T) $\wedge$​​​​​​​ ¬s V r

(F) V ¬​​​​​​​p $\wedge$ ¬s $\wedge$​​​​​​​ T V r

(F) V ¬​​​​​​​p $\wedge$ ¬s ​​​​​​​$\wedge$​​​​​​​ (T V r)

​​​​​​​(F) V ¬​​​​​​​p $\wedge$ ¬s ​​​​​​​$\wedge$​​​​​​​ (T)

​​​​​​​((F) V ¬​​​​​​​p $\wedge$ ¬s ​​​​​​​$\wedge$​​​​​​​ (T)

​​​​​​​​​​​​​​F V ¬​​​​​​​p $\wedge$ ¬s ​​​​​​​$\wedge$​​​​​​​ (T)

¬​​​​​​​p V ​​​​​​​​​​​​​​F $\wedge$ ¬s ​​​​​​​$\wedge$​​​​​​​ (T)

¬​​​​​​​p V (​​​​​​​​​​​​​​F $\wedge$ ¬s) ​​​​​​​$\wedge$​​​​​​​ (T)

¬​​​​​​​p V (​​​​​​​​​​​​​​F) ​​​​​​​$\wedge$​​​​​​​ (T)

¬​​​​​​​p V (​​​​​​​​​​​​​​T) ​​​​​​​$\wedge$​​​​​​​ (F)

(¬​​​​​​​p V ​​​​​​​​​​​​​​T) ​​​​​​​$\wedge$​​​​​​​ (F)

(​​​​​​​​​​​​​​T) ​​​​​​​$\wedge$​​​​​​​ (F)

F