## Propositions-

Before you go through this article, make sure that you have gone through the previous article on Propositions.

We have discussed-

• Propositions are declarative statements that are either true or false but not both.
• Connectives are used to combine the propositions.

## Determining Nature Of Proposition-

Here,

• We will be given a compound proposition.
• We will be asked to determine the nature of the given proposition.

Let us discuss all these terms one by one.

## Tautology-

• A compound proposition is called tautology if and only if it is true for all possible truth values of its propositional variables.
• It contains only T (Truth) in last column of its truth table.

• A compound proposition is called contradiction if and only if it is false for all possible truth values of its propositional variables.
• It contains only F (False) in last column of its truth table.

## Contingency-

• A compound proposition is called contingency if and only if it is neither a tautology nor a contradiction.
• It contains both T (True) and F (False) in last column of its truth table.

## Valid-

• A compound proposition is called valid if and only if it is a tautology.
• It contains only T (Truth) in last column of its truth table.

## Invalid-

• A compound proposition is called invalidÂ if and only if it is not a tautology.
• It contains either only F (False) or both T (Truth) and F (False) in last column of its truth table.

## Falsifiable-

• A compound proposition is called falsifiable if and only if it can be made false for some value of its propositional variables.
• It contains either only F (False) or both T (Truth) and F (False) in last column of its truth table.

## Unfalsifiable-

• A compound proposition is called unfalsifiable if and only if it can never be made false for any value of its propositional variables.
• It contains only T (Truth) in last column of its truth table.

## Satisfiable-

• A compound proposition is called satisfiable if and only if it can be made true for some value of its propositional variables.
• It contains either only T (Truth) or both T (True) and F (False) in last column of its truth table.

## Unsatisfiable-

• A compound proposition is called unsatisfiable if and only if it can not be made true for any value of its propositional variables.
• It contains only F (False) in last column of its truth table.

## Important Points-

It is important to take a note of the the following points-

• All contradictions are invalid and falsifiable but not vice-versa.
• All contingencies are invalid and falsifiable but not vice-versa.
• All tautologies are valid and unfalsifiable and vice-versa.
• All tautologies are satisfiable but not vice-versa.
• All contingencies are satisfiable but not vice-versa.
• All contradictions are unsatisfiable and vice-versa.

Also Read- Converse, Inverse and Contrapositive

## Problem-01:

Determine the nature of following propositions-

1. pÂ âˆ§Â âˆ¼p
2. (pÂ âˆ§ (pÂ â†’ q))Â â†’Â âˆ¼q
3. [ (pÂ â†’ q)Â âˆ§ (qÂ â†’ r) ]Â âˆ§ ( pÂ âˆ§Â âˆ¼r)
4. âˆ¼(pÂ â†’ q)Â âˆ¨ (âˆ¼pÂ âˆ¨ (p âˆ§ q))
5. (p ↔ r)Â â†’ (âˆ¼q â†’ (p âˆ§ r))

## Solution-

Let us solve all the parts one by one-

## Part-01:

### Method-01: Using Truth Table-

 p âˆ¼p pÂ âˆ§ âˆ¼p F T F T F F

Clearly, last column of the truth table contains only F.

Therefore, given proposition is-

• Invalid
• Falsifiable
• Unsatisfiable

### Method-02: Using Algebra Of Proposition-

• The given proposition is pÂ âˆ§Â âˆ¼p
• By complement law,Â pÂ âˆ§Â âˆ¼p = F
• So, given proposition is contradiction, invalid, falsifiable and unsatisfiable.

### Method-03: Using Digital Electronics-

In terms of digital electronics,

• The given proposition can be written as p.p’
• Clearly, p.p’ = 0
• So, given proposition is contradiction, invalid, falsifiable and unsatisfiable.

## Part-02:

### Method-01: Using Truth Table-

 p q pÂ â†’ q pÂ âˆ§ (pÂ â†’ q) âˆ¼qÂ (pÂ âˆ§ (pÂ â†’ q))Â â†’âˆ¼qÂ F F T F T T F T T F F T T F F F T T T T T T F F

Clearly, last column of the truth table contains both T and F.

Therefore, given proposition is-

• Contingency
• Invalid
• Falsifiable
• Satisfiable

### Method-02: Using Algebra Of Proposition-

We have-

(pÂ âˆ§ (pÂ â†’ q))Â â†’Â âˆ¼q

= (pÂ âˆ§ (âˆ¼pÂ âˆ¨ q))Â â†’Â âˆ¼qÂ  Â  Â  Â  Â  Â  Â {Â âˆµ pÂ â†’ q =Â âˆ¼pÂ âˆ¨ q }

= âˆ¼(pÂ âˆ§ (âˆ¼pÂ âˆ¨ q))Â âˆ¨ âˆ¼qÂ  Â  Â  Â  Â  Â Â {Â âˆµ pÂ â†’ q =Â âˆ¼pÂ âˆ¨ q }

=Â âˆ¼((pÂ âˆ§Â âˆ¼p)Â âˆ¨ (p âˆ§ q)) âˆ¨Â âˆ¼qÂ  Â { Using Distributive law }

=Â âˆ¼(FÂ âˆ¨ (p âˆ§ q)) âˆ¨Â âˆ¼qÂ  Â  Â  Â  Â  Â  Â  { Using Complement law }

= âˆ¼(p âˆ§ q) âˆ¨Â âˆ¼qÂ  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â { Using Identity law }

=Â âˆ¼pÂ âˆ¨Â âˆ¼qÂ âˆ¨Â âˆ¼qÂ  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â { Using De Morgans law }

=Â âˆ¼pÂ âˆ¨Â âˆ¼q

• Clearly, the result is neither T nor F.
• So, given proposition is a contingency, invalid, falsifiable and satisfiable.

### Method-03: Using Digital Electronics-

We have-

(pÂ âˆ§ (pÂ â†’ q))Â â†’Â âˆ¼q

= (pÂ âˆ§ (âˆ¼pÂ âˆ¨ q))Â â†’Â âˆ¼qÂ  Â  Â  Â  Â  {Â âˆµ pÂ â†’ q =Â âˆ¼pÂ âˆ¨ q }

= âˆ¼(pÂ âˆ§ (âˆ¼pÂ âˆ¨ q))Â âˆ¨ âˆ¼qÂ  Â  Â  Â  Â {Â âˆµ pÂ â†’ q =Â âˆ¼pÂ âˆ¨ q }

Now in terms of digital electronics, we have-

= (p.(p’ + q))’ + q’

= (p.p’ + p.q)’ + q’

= (p.q)’ + q’Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  { âˆµ p.p’ = 0 }

= p’ + q’ + q’Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â { Using De Morgan’s law }

= p’ + q’

• Clearly, the result is neither 0 nor 1.
• So, given proposition is a contingency, invalid, falsifiable and satisfiable.

## Part-03:

### Method-01: Using Truth Table-

Let [ (pÂ â†’ q)Â âˆ§ (qÂ â†’ r) ]Â âˆ§ ( pÂ âˆ§Â âˆ¼r) = R (say)

 p q r pÂ â†’ q qÂ â†’Â r (pÂ â†’ q)Â âˆ§ (qÂ â†’ r) pÂ âˆ§ âˆ¼r R F F F T T T F F F F T T T T F F F T F T F F F F F T T T T T F F T F F F T F T F T F T F T F F F T T F T F F T F T T T T T T F F

Clearly, last column of the truth table contains only F.

Therefore, given proposition is-

• Invalid
• Falsifiable
• Unsatisfiable

### Method-02: Using Algebra Of Proposition-

We have-

[ (pÂ â†’ q)Â âˆ§ (qÂ â†’ r) ]Â âˆ§ ( pÂ âˆ§Â âˆ¼r)

= [ (âˆ¼pÂ âˆ¨ q)Â âˆ§ (âˆ¼qÂ âˆ¨ r) ]Â âˆ§ ( pÂ âˆ§Â âˆ¼r)Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â {Â âˆµ pÂ â†’ q =Â âˆ¼pÂ âˆ¨ q }

= [ ((âˆ¼pÂ âˆ¨ q)Â âˆ§ âˆ¼q)Â âˆ¨ ((âˆ¼pÂ âˆ¨ q)Â âˆ§ r) ] âˆ§ ( pÂ âˆ§Â âˆ¼r)Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  { Using Distributive law }

= [ ((âˆ¼pÂ âˆ§ âˆ¼q)Â âˆ¨ (q âˆ§ âˆ¼q))Â âˆ¨ ((âˆ¼p âˆ§ r)Â âˆ¨ (qÂ âˆ§ r)) ] âˆ§ ( pÂ âˆ§Â âˆ¼r)Â  Â  Â  Â  Â { Using Distributive law }

= [ ((âˆ¼pÂ âˆ§ âˆ¼q)Â âˆ¨ F)Â âˆ¨ ((âˆ¼p âˆ§ r)Â âˆ¨ (qÂ âˆ§ r)) ] âˆ§ ( pÂ âˆ§Â âˆ¼r)Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  { Using Complement law }

= [ (âˆ¼pÂ âˆ§ âˆ¼q) âˆ¨ (âˆ¼p âˆ§ r)Â âˆ¨ (qÂ âˆ§ r) ] âˆ§ ( pÂ âˆ§Â âˆ¼r)Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â { Using Identity law }

= ((âˆ¼pÂ âˆ§ âˆ¼q) âˆ§ ( pÂ âˆ§Â âˆ¼r)) âˆ¨ ((âˆ¼p âˆ§ r)Â âˆ§Â ( pÂ âˆ§Â âˆ¼r)) âˆ¨ ((qÂ âˆ§ r) âˆ§ ( pÂ âˆ§Â âˆ¼r))Â  { Using Distributive law }

= (âˆ¼pÂ âˆ§ âˆ¼q âˆ§ pÂ âˆ§Â âˆ¼r) âˆ¨ (âˆ¼p âˆ§ r âˆ§ pÂ âˆ§Â âˆ¼r) âˆ¨ (qÂ âˆ§ r âˆ§ pÂ âˆ§Â âˆ¼r)

= FÂ âˆ¨ FÂ âˆ¨ FÂ  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  { Using Complement law }

= F

• Clearly, the result is F.
• So, given proposition is a contradiction, invalid, falsifiable and unsatisfiable.

### Method-03: Using Digital Electronics-

We have-

[ (pÂ â†’ q)Â âˆ§ (qÂ â†’ r) ]Â âˆ§ ( pÂ âˆ§Â âˆ¼r)

= [ (âˆ¼pÂ âˆ¨ q)Â âˆ§ (âˆ¼qÂ âˆ¨ r) ]Â âˆ§ ( pÂ âˆ§Â âˆ¼r)Â  Â  Â Â {Â âˆµ pÂ â†’ q =Â âˆ¼pÂ âˆ¨ q }

Now in terms of digital electronics, we have-

= [ (p’ + q).(q’ + r) ] . (p.r’)

= [ p’.q’ + p’.r + q.q’ + q.r ] . (p.r’)

= [ p’.q’ + p’.r + 0 + q.r ] . (p.r’)Â  Â  Â  Â  Â  Â  Â {Â âˆµ q.q’ = 0 }

= [ p’.q’ + p’.r + q.r ] . (p.r’)

= p’.q’.p.r’ + p’.r.p.r’ + q.r.p.r’

= 0 + 0 + 0

= 0

• Clearly, the result is 0.
• So, given proposition is a contradiction, invalid, falsifiable and unsatisfiable.

## Part-04:

### Method-01: Using Truth Table-

Let âˆ¼(pÂ â†’ q)Â âˆ¨ (âˆ¼pÂ âˆ¨ (p âˆ§ q)) =Â R (say)

 p q âˆ¼p pÂ â†’ q âˆ¼(pÂ â†’ q) pÂ âˆ§Â q âˆ¼pÂ âˆ¨ (pÂ âˆ§ q) R F F T T F F T T F T T T F F T T T F F F T F F T T T F T F T T T

Clearly, last column of the truth table contains only T.

Therefore, given proposition is-

• Tautology
• Valid
• Unfalsifiable
• Satisfiable

### Method-02: Using Algebra Of Proposition-

We have-

âˆ¼(pÂ â†’ q)Â âˆ¨ (âˆ¼pÂ âˆ¨ (p âˆ§ q))

=Â âˆ¼(âˆ¼pÂ âˆ¨ q)Â âˆ¨ (âˆ¼pÂ âˆ¨ (p âˆ§ q))Â  Â  Â  Â  Â  Â  Â  Â {Â âˆµ pÂ â†’ q =Â âˆ¼pÂ âˆ¨ q }

= (pÂ âˆ§Â âˆ¼q)Â âˆ¨ (âˆ¼pÂ âˆ¨ (p âˆ§ q))Â  Â  Â  Â  Â  Â  Â  Â  Â  { Using De Morgans law }

= (pÂ âˆ§Â âˆ¼q)Â âˆ¨ ((âˆ¼pÂ âˆ¨ p) âˆ§ (âˆ¼p âˆ¨ q))Â  Â  Â  Â { Using Distributive law }

= (pÂ âˆ§Â âˆ¼q)Â âˆ¨ (T âˆ§ (âˆ¼p âˆ¨ q))Â  Â  Â  Â  Â  Â  Â  Â  Â  { Using Complement law }

= (pÂ âˆ§Â âˆ¼q)Â âˆ¨ (âˆ¼p âˆ¨ q)Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â { Using Identity law }

= ((pÂ âˆ§Â âˆ¼q)Â âˆ¨ âˆ¼p) âˆ¨ qÂ  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â { Using Associative law }

= ((pÂ âˆ¨ âˆ¼p) âˆ§ (âˆ¼q âˆ¨ âˆ¼p)) âˆ¨ qÂ  Â  Â  Â  Â  Â  Â  Â  { Using Distributive law }

= (T âˆ§ (âˆ¼q âˆ¨ âˆ¼p)) âˆ¨ qÂ  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â { Using Complement law }

= (âˆ¼q âˆ¨ âˆ¼p) âˆ¨ qÂ  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  { Using Identity law }

=Â âˆ¼pÂ âˆ¨ (qÂ âˆ¨Â âˆ¼q)

=Â âˆ¼pÂ âˆ¨ TÂ  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â { Using Complement law }

= TÂ  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  { Using Identity law }

• Clearly, the result is T.
• So, given proposition is a tautology, valid, unfalsifiable and satisfiable.

### Method-03: Using Digital Electronics-

We have-

âˆ¼(pÂ â†’ q)Â âˆ¨ (âˆ¼pÂ âˆ¨ (p âˆ§ q))

=Â âˆ¼(âˆ¼pÂ âˆ¨ q)Â âˆ¨ (âˆ¼pÂ âˆ¨ (p âˆ§ q))Â  Â  Â  Â  Â  {Â âˆµ pÂ â†’ q =Â âˆ¼pÂ âˆ¨ q }

Now in terms of digital electronics, we have-

= (p’ + q)’ + (p’ + p.q)

= (p’ + q)’ + (p’ + p).(p’ + q)Â  Â  Â  Â  Â  Â  { Using Transposition theorem }

=Â (p’ + q)’ + 1.(p’ + q)

=Â (p’ + q)’ + (p’ + q)

= p.q’ + p’ + qÂ  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â { Using De Morgans law }

= (p + p’)(p’ + q’) + qÂ  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  { Using Transposition theorem }

= 1.(p’ + q’) + q

=Â p’ + (q’ + q)

=Â p’ + 1

= 1

• Clearly, the result is 1.
• So, given proposition is a tautology, valid, unfalsifiable and satisfiable.

## Part-05:

### Method-01: Using Truth Table-

Let (p ↔ r)Â â†’ (âˆ¼q â†’ (p âˆ§ r)) = R (say)

 p q r âˆ¼q pÂ â†’Â r rÂ â†’Â p pÂ ↔Â r pÂ âˆ§ r âˆ¼q â†’ (p âˆ§ r) R F F F T T T T F F F F F T T T F F F F T F T F F T T T F T T F T T F T F F F T T T F F T F T F F F T T F T T T T T T T T T T F F F T F F T T T T T F T T T T T T

Clearly, last column of the truth table contains both T and F.

Therefore, given proposition is-

• Contingency
• Invalid
• Falsifiable
• Satisfiable

### Method-02: Using Algebra Of Proposition-

We have-

(p ↔ r)Â â†’ (âˆ¼q â†’ (p âˆ§ r))

= (p ↔ r)Â â†’ (q âˆ¨Â (p âˆ§ r))Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â {Â âˆµ pÂ â†’ q =Â âˆ¼pÂ âˆ¨ q }

= âˆ¼(p ↔ r)Â âˆ¨ q âˆ¨Â (p âˆ§ r)

=Â âˆ¼((pÂ â†’ r)Â âˆ§ (rÂ â†’ p))Â âˆ¨ q âˆ¨Â (p âˆ§ r)Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â {Â âˆµ pÂ ↔ q = (pÂ â†’ q)Â âˆ§ qÂ â†’ p) }

=Â âˆ¼((âˆ¼pÂ âˆ¨ r)Â âˆ§ (âˆ¼rÂ âˆ¨ p))Â âˆ¨ q âˆ¨Â (p âˆ§ r)Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â {Â âˆµ pÂ â†’ q =Â âˆ¼pÂ âˆ¨ q }

= âˆ¼[ ((âˆ¼pÂ âˆ¨ r)Â âˆ§ âˆ¼r) âˆ¨ ((âˆ¼pÂ âˆ¨ r) âˆ§ p) ] âˆ¨ q âˆ¨Â (p âˆ§ r)Â  Â  Â  Â { Using Distributive law }

=Â âˆ¼[ ((âˆ¼pÂ âˆ§ âˆ¼r) âˆ¨ (r âˆ§ âˆ¼r)) âˆ¨ ((âˆ¼pÂ âˆ§ p) âˆ¨ (r âˆ§ p)) ] âˆ¨ q âˆ¨Â (p âˆ§ r)Â  Â { Using Distributive law }

=Â âˆ¼[ ((âˆ¼pÂ âˆ§ âˆ¼r)Â âˆ¨ F) âˆ¨ (F âˆ¨Â (r âˆ§ p)) ] âˆ¨ q âˆ¨Â (p âˆ§ r)Â  Â  Â  Â  Â { Using Complement law }

=Â âˆ¼[ (âˆ¼pÂ âˆ§ âˆ¼r) âˆ¨ (r âˆ§ p) ] âˆ¨ q âˆ¨Â (p âˆ§ r)Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â { Using Identity law }

= [âˆ¼(âˆ¼pÂ âˆ§ âˆ¼r) âˆ§ âˆ¼(r âˆ§ p) ] âˆ¨ q âˆ¨Â (p âˆ§ r)Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â { Using De Morgans law }

= [ (p âˆ¨ r) âˆ§ (âˆ¼r âˆ¨ âˆ¼p) ] âˆ¨ q âˆ¨Â (p âˆ§ r)Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â { Using De Morgans law }

= ((p âˆ¨ r) âˆ§ âˆ¼r) âˆ¨ ((p âˆ¨ r) âˆ§ âˆ¼p) âˆ¨ q âˆ¨Â (p âˆ§ r)Â  Â  Â  Â  Â  Â  Â  Â  Â { Using Distributive law }

= ((p âˆ§ âˆ¼r) âˆ¨ (r âˆ§ âˆ¼r)) âˆ¨ ((p âˆ§ âˆ¼p) âˆ¨ (r âˆ§ âˆ¼p)) âˆ¨ q âˆ¨Â (p âˆ§ r)Â  Â { Using Distributive law }

= ((p âˆ§ âˆ¼r)Â âˆ¨ F) âˆ¨ (F âˆ¨ (r âˆ§ âˆ¼p)) âˆ¨ q âˆ¨Â (p âˆ§ r)Â  Â  Â  Â  Â  Â  Â  Â  Â { Using Complement law }

= (p âˆ§ âˆ¼r) âˆ¨ (r âˆ§ âˆ¼p) âˆ¨ q âˆ¨Â (p âˆ§ r)Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â { Using Identity law }

= (p âˆ§ âˆ¼r) âˆ¨ q âˆ¨Â (âˆ¼p âˆ§ r) âˆ¨ (p âˆ§ r)

= (p âˆ§ âˆ¼r) âˆ¨ q âˆ¨ ((âˆ¼p âˆ¨ p) âˆ§ r)Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â { Using Distributive law }

= (p âˆ§ âˆ¼r) âˆ¨ q âˆ¨ (T âˆ§ r)Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  { Using Complement law }

= (p âˆ§ âˆ¼r) âˆ¨ q âˆ¨ rÂ  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â { Using Identity law }

=Â Â r âˆ¨ (p âˆ§ âˆ¼r) âˆ¨ q

= ((rÂ âˆ¨ p)Â âˆ§ (rÂ âˆ¨Â âˆ¼r)) âˆ¨ qÂ  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â { Using Distributive law }

= ((rÂ âˆ¨ p)Â âˆ§ T) âˆ¨ qÂ  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  { Using Complement law }

= pÂ âˆ¨ qÂ âˆ¨ rÂ  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â { Using Identity law }

• Clearly, the result is neither T nor F.
• So, given proposition is a contingency, invalid, falsifiable and satisfiable.

### Method-03: Using Digital Electronics-

We have-

(p ↔ r)Â â†’ (âˆ¼q â†’ (p âˆ§ r))

= (p ↔ r)Â â†’ (q âˆ¨Â (p âˆ§ r))Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â Â {Â âˆµ pÂ â†’ q =Â âˆ¼pÂ âˆ¨ q }

= âˆ¼(p ↔ r)Â âˆ¨ (q âˆ¨Â (p âˆ§ r))Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â {Â âˆµ pÂ â†’ q =Â âˆ¼pÂ âˆ¨ q }

Now in terms of digital electronics, we have-

= (p.r + p’.r’)’ + (q + p.r)

= (p.r)’ . (p’.r’)’ + (q + p.r)Â  Â  Â  Â  Â  Â  Â  Â  Â  (Using De Morgans Theorem)

= (p’ + r’) . (p + r) + (q + p.r)Â  Â  Â  Â  Â  Â  Â (Using De Morgans Theorem)

= p’.p + p’.r + r’.p + r’.r + q + p.r

= 0 +Â p’.r + r’.p + 0 + q + p.r

= p’.r + r’.p + q + p.r

= (p’ + p).r + r’.p + q

=Â r + r’.p + q

= (r + r’).(r + p) + qÂ  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â (Using Transposition Theorem)

= p + q + r

• Clearly, the result is neither 0 nor 1.
• So, given proposition is a contingency, invalid, falsifiable and satisfiable.