## 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 invalid 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. p ∧ ∼p
2. (p ∧ (p → q)) → ∼q
3. [ (p → q) ∧ (q → r) ] ∧ ( p ∧ ∼r)
4. ∼(p → q) ∨ (∼p ∨ (p ∧ q))
5. (p ↔ r) → (∼q → (p ∧ r))

## Solution-

Let us solve all the parts one by one-

## Part-01:

### Method-01: Using Truth Table-

 p ∼p p ∧ ∼p 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 p ∧ ∼p
• By complement law, p ∧ ∼p = 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 p → q p ∧ (p → q) ∼q (p ∧ (p → q)) →∼q 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-

(p ∧ (p → q)) → ∼q

= (p ∧ (∼p ∨ q)) → ∼q             { ∵ p → q = ∼p ∨ q }

= ∼(p ∧ (∼p ∨ q)) ∨ ∼q            { ∵ p → q = ∼p ∨ q }

= ∼((p ∧ ∼p) ∨ (p ∧ q)) ∨ ∼q   { Using Distributive law }

= ∼(F ∨ (p ∧ q)) ∨ ∼q              { Using Complement law }

= ∼(p ∧ q) ∨ ∼q                       { Using Identity law }

= ∼p ∨ ∼q ∨ ∼q                       { Using De Morgans law }

= ∼p ∨ ∼q

• 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 ∧ (p → q)) → ∼q

= (p ∧ (∼p ∨ q)) → ∼q          { ∵ p → q = ∼p ∨ q }

= ∼(p ∧ (∼p ∨ q)) ∨ ∼q         { ∵ p → q = ∼p ∨ q }

Now in terms of digital electronics, we have-

= (p.(p’ + q))’ + q’

= (p.p’ + p.q)’ + q’

= (p.q)’ + q’                            { ∵ p.p’ = 0 }

= p’ + q’ + q’                           { 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 [ (p → q) ∧ (q → r) ] ∧ ( p ∧ ∼r) = R (say)

 p q r p → q q → r (p → q) ∧ (q → r) p ∧ ∼r 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-

[ (p → q) ∧ (q → r) ] ∧ ( p ∧ ∼r)

= [ (∼p ∨ q) ∧ (∼q ∨ r) ] ∧ ( p ∧ ∼r)                                               { ∵ p → q = ∼p ∨ q }

= [ ((∼p ∨ q) ∧ ∼q) ∨ ((∼p ∨ q) ∧ r) ] ∧ ( p ∧ ∼r)                            { Using Distributive law }

= [ ((∼p ∧ ∼q) ∨ (q ∧ ∼q)) ∨ ((∼p ∧ r) ∨ (q ∧ r)) ] ∧ ( p ∧ ∼r)         { Using Distributive law }

= [ ((∼p ∧ ∼q) ∨ F) ∨ ((∼p ∧ r) ∨ (q ∧ r)) ] ∧ ( p ∧ ∼r)                    { Using Complement law }

= [ (∼p ∧ ∼q) ∨ (∼p ∧ r) ∨ (q ∧ r) ] ∧ ( p ∧ ∼r)                               { Using Identity law }

= ((∼p ∧ ∼q) ∧ ( p ∧ ∼r)) ∨ ((∼p ∧ r) ∧ ( p ∧ ∼r)) ∨ ((q ∧ r) ∧ ( p ∧ ∼r))  { Using Distributive law }

= (∼p ∧ ∼q ∧ p ∧ ∼r) ∨ (∼p ∧ r ∧ p ∧ ∼r) ∨ (q ∧ r ∧ p ∧ ∼r)

= F ∨ F ∨ F                                                                                    { 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-

[ (p → q) ∧ (q → r) ] ∧ ( p ∧ ∼r)

= [ (∼p ∨ q) ∧ (∼q ∨ r) ] ∧ ( p ∧ ∼r)      { ∵ p → q = ∼p ∨ 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’)             { ∵ 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 ∼(p → q) ∨ (∼p ∨ (p ∧ q)) = R (say)

 p q ∼p p → q ∼(p → q) p ∧ q ∼p ∨ (p ∧ 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-

∼(p → q) ∨ (∼p ∨ (p ∧ q))

= ∼(∼p ∨ q) ∨ (∼p ∨ (p ∧ q))               { ∵ p → q = ∼p ∨ q }

= (p ∧ ∼q) ∨ (∼p ∨ (p ∧ q))                  { Using De Morgans law }

= (p ∧ ∼q) ∨ ((∼p ∨ p) ∧ (∼p ∨ q))       { Using Distributive law }

= (p ∧ ∼q) ∨ (T ∧ (∼p ∨ q))                  { Using Complement law }

= (p ∧ ∼q) ∨ (∼p ∨ q)                           { Using Identity law }

= ((p ∧ ∼q) ∨ ∼p) ∨ q                           { Using Associative law }

= ((p ∨ ∼p) ∧ (∼q ∨ ∼p)) ∨ q                { Using Distributive law }

= (T ∧ (∼q ∨ ∼p)) ∨ q                           { Using Complement law }

= (∼q ∨ ∼p) ∨ q                                    { Using Identity law }

= ∼p ∨ (q ∨ ∼q)

= ∼p ∨ T                                               { Using Complement law }

= T                                                        { 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-

∼(p → q) ∨ (∼p ∨ (p ∧ q))

= ∼(∼p ∨ q) ∨ (∼p ∨ (p ∧ q))          { ∵ p → q = ∼p ∨ q }

Now in terms of digital electronics, we have-

= (p’ + q)’ + (p’ + p.q)

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

= (p’ + q)’ + 1.(p’ + q)

= (p’ + q)’ + (p’ + q)

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

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

= 1.(p’ + q’) + q

= p’ + (q’ + q)

= 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) → (∼q → (p ∧ r)) = R (say)

 p q r ∼q p → r r → p p ↔ r p ∧ r ∼q → (p ∧ 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) → (∼q → (p ∧ r))

= (p ↔ r) → (q ∨ (p ∧ r))                                             { ∵ p → q = ∼p ∨ q }

= ∼(p ↔ r) ∨ q ∨ (p ∧ r)

= ∼((p → r) ∧ (r → p)) ∨ q ∨ (p ∧ r)                             { ∵ p ↔ q = (p → q) ∧ q → p) }

= ∼((∼p ∨ r) ∧ (∼r ∨ p)) ∨ q ∨ (p ∧ r)                           { ∵ p → q = ∼p ∨ q }

= ∼[ ((∼p ∨ r) ∧ ∼r) ∨ ((∼p ∨ r) ∧ p) ] ∨ q ∨ (p ∧ r)       { Using Distributive law }

= ∼[ ((∼p ∧ ∼r) ∨ (r ∧ ∼r)) ∨ ((∼p ∧ p) ∨ (r ∧ p)) ] ∨ q ∨ (p ∧ r)   { Using Distributive law }

= ∼[ ((∼p ∧ ∼r) ∨ F) ∨ (F ∨ (r ∧ p)) ] ∨ q ∨ (p ∧ r)         { Using Complement law }

= ∼[ (∼p ∧ ∼r) ∨ (r ∧ p) ] ∨ q ∨ (p ∧ r)                           { Using Identity law }

= [∼(∼p ∧ ∼r) ∧ ∼(r ∧ p) ] ∨ q ∨ (p ∧ r)                         { Using De Morgans law }

= [ (p ∨ r) ∧ (∼r ∨ ∼p) ] ∨ q ∨ (p ∧ r)                             { Using De Morgans law }

= ((p ∨ r) ∧ ∼r) ∨ ((p ∨ r) ∧ ∼p) ∨ q ∨ (p ∧ r)                 { Using Distributive law }

= ((p ∧ ∼r) ∨ (r ∧ ∼r)) ∨ ((p ∧ ∼p) ∨ (r ∧ ∼p)) ∨ q ∨ (p ∧ r)   { Using Distributive law }

= ((p ∧ ∼r) ∨ F) ∨ (F ∨ (r ∧ ∼p)) ∨ q ∨ (p ∧ r)                 { Using Complement law }

= (p ∧ ∼r) ∨ (r ∧ ∼p) ∨ q ∨ (p ∧ r)                                   { Using Identity law }

= (p ∧ ∼r) ∨ q ∨ (∼p ∧ r) ∨ (p ∧ r)

= (p ∧ ∼r) ∨ q ∨ ((∼p ∨ p) ∧ r)                                         { Using Distributive law }

= (p ∧ ∼r) ∨ q ∨ (T ∧ r)                                                    { Using Complement law }

= (p ∧ ∼r) ∨ q ∨ r                                                             { Using Identity law }

=  r ∨ (p ∧ ∼r) ∨ q

= ((r ∨ p) ∧ (r ∨ ∼r)) ∨ q                                                   { Using Distributive law }

= ((r ∨ p) ∧ T) ∨ q                                                            { Using Complement law }

= p ∨ q ∨ r                                                                       { 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) → (∼q → (p ∧ r))

= (p ↔ r) → (q ∨ (p ∧ r))                      { ∵ p → q = ∼p ∨ q }

= ∼(p ↔ r) ∨ (q ∨ (p ∧ r))                     { ∵ p → q = ∼p ∨ q }

Now in terms of digital electronics, we have-

= (p.r + p’.r’)’ + (q + p.r)

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

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

= p’.p + p’.r + r’.p + r’.r + q + p.r

= 0 + p’.r + r’.p + 0 + q + p.r

= p’.r + r’.p + q + p.r

= (p’ + p).r + r’.p + q

= r + r’.p + q

= (r + r’).(r + p) + q                           (Using Transposition Theorem)

= p + q + r

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