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37 votes
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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
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4 Comments

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) ...
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$A\equiv B$ means $\overline{A}$  $\overline{B} + AB$ would be 1.

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

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39 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:

$((A∨B) → C) \equiv (~(A∨B) ∨ C)$
$\quad \quad \equiv (~A ∧~B) ∨C)$
$\quad \quad \equiv (~A ∨C) ∧ (~B ∨C)$
$\quad \quad \equiv (A→C) ∧ (B→C)$
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They want to ask, whether, these expressions are tautologies or not?

$\equiv$ treated as Equivalence
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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.
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Arjun sir.

For P1 When A is True.
LHS is B --> C. And RHS is C(-B+C). Is it enough to conclude that P1 isn't a tautology?
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4 votes
4 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 not tautologies here

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2 Comments

@Prateek kumar option (D) says they are not tautologies, correct the final statement. 

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“but what if we exchange in right side of P1 & P2”

Prateek has given proof after changing RHS. For that both will be tautologies.

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0 votes
0 votes

Answer D is correct.

Answer:

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