1.8k views

Consider the following propositional statements:

• $P_1: ((A ∧ B) → C)) ≡ ((A → C) ∧ (B → C))$
• $P_2: ((A ∨ B) → C)) ≡ ((A → C) ∨ (B → C))$

Which one of the following is true?

1. $P_1$ is a tautology, but not $P_2$
2. $P_2$ is a tautology, but not $P_1$
3. $P_1$ and $P_2$ are both tautologies
4. Both $P_1$ and $P_2$ are not tautologies
edited | 1.8k views
+1
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)

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?
0
<-> and ≡ are same I guess.

Also A<->B is same as (A->B)^(B->A).So we need to check whether (A->B)^(B->A) is tautology for P1 and P2.

(D) Both $P_1$ and $P_2$ are not tautologies.

$P_1:$ If $A$ is true and $B$ is false, LHS of $P_1$ is true but RHS becomes false. Hence not tautology.

$P_2:$ 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 $P_2$ can be simplified as follows:

$((A∨B) → C) \equiv (~(A∨B) ∨ C)$
$\quad \quad \equiv (~A ∧~B) ∨C)$
$\quad \quad \equiv (~A ∨C) ∧ (~B ∨C)$
$\quad \quad \equiv (A→C) ∧ (B→C)$
edited 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.
+1
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)

while making hte K-Map of LHS ,i am not getting true(all 1) ...
0
I Understand this concept
+1
For proving whether it is a tautology or not...we have to check LHS=RHS or not?? or what else?
0
I can't understand the question,what are they want in the question???

please anyone help me to understand this
0
They want to ask, whether, these expressions are tautologies or not?

$\equiv$ treated as Equivalence
0
Yes.

For equivalence if we prove that LHS=RHS then we can say that proposition is tautology.

Since we consider equicalence as $\odot$ (Ex-nor) we can prove either LHS=RHS=TRUE/FALSE.

$P\leftrightarrow Q$ is same as $P\equiv Q.$ if P=Q=TRUE/FALSE. Then they are equivalent.

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