1.4k 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$are not tautologies asked edited | 1.4k views 0 sir what does$\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 )}'+CA{}'+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$(EX-NOR)operator is actually 'equivalence operator' 0 A<->B could also be treated as equivalence operator? 0 <-> and ≡ are same I guess. Also A<->B is same as (A->B)^(B->A).So we need to check whether (A->B)^(B->A) is tautology for P1 and P2. ## 2 Answers +22 votes Best answer (D) Both$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)$answered by Veteran (359k points) edited by 0 Can you please covert ((A ∧ B) → C)) to ((A → C) V (B → C)) I can convert into A->(B->C) only 0 they are not equivalent, so how can we convert? You didn't see the last statement- on RHS we get AND in stead of OR, which is a more stricter condition. This makes the forward implication true but reverse implication false. +1 Sorry..I can't get it.Truth table says they are equivalent.In text book they ask to d) Transform pq -> r into (p -> r) + (q -> r). +4 Sorry, I misread the expression. I thought it was P2 in question. ((A ∧ B) → C)) = ~A V ~B V C = ~A V C V ~B V C = (A → C) V (B → C) +1 It's ok :) Thank you for the answer 0 if equivalence hold in above, then will it be tautology?? 0 sir what does$\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 )}'+CA{}'+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
0
For proving whether it is a tautology or not...we have to check LHS=RHS or not?? or what else?

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