Discrete Mathematics | Propositional Logic | Test 1 | Question: 14

Question

A compound sentence is a $\textit{tautology}$ if it is true independently of the truth values of its component atomic sentences. A sentence is $\textit{atomic}$ if it contains no sentential connectives.     
         
Let $P,Q$ and $R$ be any three distinct atomic sentences. Now consider the following statements:      
       

1. $P \rightarrow [Q \rightarrow (Q \rightarrow P)]$ is a tautology    
2. $[P \vee (\neg P \wedge Q)] \vee (\neg P \wedge \neg Q)$ is not a tautology             
3. $(P \rightarrow Q) \rightarrow (Q \rightarrow P)$ is not a tautology
4. $[(P \rightarrow Q) \leftrightarrow Q] \rightarrow P$ is a tautology
5. $(P \leftrightarrow P) \leftrightarrow P$ is a tautology       

        
Number of correct statements are ______

Comments

Last statement is false when P is false.

Answer 1

Answer). 2 statements are correct: Statement 1 and Statement 3.

Explanation:-

It is given that P, Q, and R are atomic sentences which means they are atomic propositions i.e. either hold true or false values (but not both).

Let’s discuss each statement:

Statement 1:- P→ [Q→ (Q→ P)]  is a tautology.

let's take α as P  and β  as  Q→ (Q → P)    so our statement became α → β 

now assume α as true  so that became the truth value of P: True

AND try to make  simultaneously β  False 

 β: Q→ (Q→ P ) ≡ Q→ (Q→ T)  [ Since P is true in α ]

                            ≡ Q→ T [ anything implies True ≡ True]

                            ≡ T 

so the value of β  is True, and we can’t make it false when the value of α  is True.

so α → β  ≡ T→ T =  T          [ Hence it is Tautology]

hence this statement is correct.

Statement 2:- [P v (~P ∧ Q)] v (~P ∧ ~Q) is not a tautology.

                     ≡ [(P v ~P)  ∧ (P v Q)]  v (~P ∧ ~Q)   [DISTRIBUTION LAW ]

                     ≡ [ T ∧ (P v Q)]  v (~P ∧ ~Q)        [INVERSE LAW]

                    ≡ [(P v Q)]  v (~P ∧ ~Q)     [ IDENTITY LAW]

                    ≡ (P v Q) v ~(P v Q)            [DE-MORGAN’s LAW]

 Assume (P v Q )as logical expressions A 

                  ≡ A v  ~A

                  ≡ T       [INVERSE LAW]

 Hence it is a Tautology, but it is saying that it is not a tautology, so it is an incorrect statement.

Statement 3:- (P→ Q) → (Q→ P) is not a tautology.

Let’s take α  as P→ Q and β  as Q→ P  which makes α → β 

 

now Assume β  is False which means Q→ P is false which directly says Q = True and P = False.

Now, try to make simultaneously α True

  α   = P→ Q    

      = F→ T [ since we already have values of P and Q from β ]

      = T

since we can make α True when β is false.  

so α → β  = T→ F which means not a Tautology.

Hence this statement is a correct statement.

Statement 4:- [(P → Q)↔ Q]→ P is a tautology.

Let’s take α  as  [(P → Q)↔ Q]  and β  as P  which makes α → β 

now Assume β  is False which means P is false.

Now, try to make simultaneously α True

  α   =[(P → Q)↔ Q] 

       =[(F→ Q)↔ Q]    [ since we already have values of P from β ]

      =[T↔ Q]      [False → anything is True always]

     =Q

so α is contingency it can be True or false means it depends on the truth value of Q.

 so α → β  = Q→ F = ~Q 

 Hence it is not a Tautology so that’s the reason it is an incorrect statement.

Statement 5:-(P↔ P)↔P is a tautology

Let’s take α  as P↔ P and β  as P  which makes α ↔ β 

Now solve it “BY- CASE” method,

Case 1: P= True

α  = P↔ P = T↔ T = T

 β  = P = T 

here  α = β and  α ↔ β  = T

 

Case 2: P= False

α  = P↔ P = F↔ F = T

 β  = P = F 

but here α ≠ β  and α ↔ β = F

so α ↔ β  is not a Tautology 

hence it is incorrect Statement.

Comments

Any other way which does not involve making cases to solve this question ?

---

@ankitgupta.1729Yes we can solve it by simplification.