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Consider the following propositional statements:
P1 : ((A ∧ B) → C)) ≡ ((A → C) ∧ (B → C))
P2 : ((A ∨ B) → C)) ≡ ((A → C) ∨ (B → C))

Which one of the following is true?

  1.    P1 is a tautology, but not P2
  2.    P2 is a tautology, but not P1
  3.    P1 and P2 are both tautologies
  4.    Both P1 and P2 are not tautologies
asked in Mathematical Logic by Active (3.7k points)
edited by | 1.1k views
0
sir what does $\equiv$ denotes in this question ??

When i simplify $P1$ ,it comes out to be

$\left ( \left ( A\wedge B \right )\rightarrow C \right )={\left ( AB \right )}'+C$

$A{}'+B{}'+C$ in LHS,

while  In RHS

$\left ( \left ( A\rightarrow C \right ) \wedge \left ( B\rightarrow C \right )\right )$

$\left ( A{}'+C \right ) \wedge \left ( B{}'+C \right )$

=$A{}'B{}'+C$

 

1.Here LHS$\neq$ RHS ,what is this $\equiv$ here?

2.also i made Kmap of LHS ,i am not getting true(all 1)

please help me out arjun sir

while making hte K-Map of LHS ,i am not getting true(all 1) ...
+4
$A\equiv B$ means $\overline{A}$  $\overline{B} + AB$ would be 1.

$\odot$ (EX-NOR)operator is actually 'equivalence operator'
0
A<->B could also be treated as equivalence operator?

2 Answers

+19 votes
Best answer
(D) Both P1 and P2 are not tautologies.

P1: If A is true and B is false, LHS of P1 is true but RHS becomes false. Hence not tautology.

P2: Forward side is true. But reverse side is not true. When A is false and B is true and C is false, RHS is true but LHS is false.

LHS of P2 can be simplified as follows:

((A∨B) → C) = (~(A∨B) ∨ C) = (~A ∧~B) ∨C) = (~A ∨C) ∧ (~B ∨C) = (A→C) ∧ (B→C)
answered by Veteran (342k points)
selected by
0

Can you please covert ((A ∧ B) → C)) to ((A → C) V (B → C))

I can convert into A->(B->C) only

 

 

 

0
they are not equivalent, so how can we convert? You didn't see the last statement- on RHS we get AND in stead of OR, which is a more stricter condition. This makes the forward implication true but reverse implication false.
0
Sorry..I can't get it.Truth table says they are equivalent.In text book they ask to
d) Transform pq -> r into (p -> r) + (q -> r).
+4
Sorry, I misread the expression. I thought it was P2 in question.
 ((A ∧ B) → C))  = ~A V ~B V C = ~A V C V ~B V  C = (A → C) V (B → C)
+1
It's ok :) Thank you for the answer
0
if equivalence hold in above, then will it be tautology??
0
sir what does $\equiv$ denotes in this question ??

When i simplify $P1$ ,it comes out to be

$\left ( \left ( A\wedge B \right )\rightarrow C \right )={\left ( AB \right )}'+C$

$A{}'+B{}'+C$ in LHS,

while  In RHS

$\left ( \left ( A\rightarrow C \right ) \wedge \left ( B\rightarrow C \right )\right )$

$\left ( A{}'+C \right ) \wedge \left ( B{}'+C \right )$

=$A{}'B{}'+C$

 

1.Here LHS$\neq$ RHS ,what is this $\equiv$ here?

2.also i made Kmap of LHS ,i am not getting true(all 1)

please help me out arjun sir

while making hte K-Map of LHS ,i am not getting true(all 1) ...
0
I Understand this concept
+2 votes

Answer is "D" for this question

but what if we exchange in right side of P1 & P2 let's see what happen

P1 : ((A ∧ B) → C)) ≡ ((A → C) ∨ (B → C))       
P2 : ((A ∨ B) → C)) ≡ ((A → C) ∧ (B → C)) 

P1: ~A +~B +C  ≡ ~A +~B+C   

P2:  ~A~B + C ≡(~A+C)(~B+C)     // Here we know that C+~A~B = (C+~A)(C+~B)

 P1 and P2 are both tautologies here

answered by Loyal (7.7k points)


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