Tautology Contradiction Contingency

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.

 

 

In this article, we will discuss about the nature of proposition.

 

Also Read- Converting English Sentences to Propositional Logic

 

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.

 

Contradiction-

 

  • 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

 

PRACTICE PROBLEMS BASED ON DETERMINING NATURE OF PROPOSITIONS-

 

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∼pp ∧ ∼p
FTF
TFF

 

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

Therefore, given proposition is-

  • Contradiction
  • 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-

 

pq qp ∧ (p  q)∼q (p ∧ (p  q)) ∼q 
FFTFTT
FTTFFT
TFFFTT
TTTTFF

 

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)

 

pqr q r(p  q) ∧ (q  r)p ∧ ∼rR
FFFTTTFF
FFTTTTFF
FTFTFFFF
FTTTTTFF
TFFFTFTF
TFTFTFFF
TTFTFFTF
TTTTTTFF

 

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

Therefore, given proposition is-

  • Contradiction
  • 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)

 

pq∼p
q∼(p  q)p ∧ q∼p ∨ (p ∧ q)
R
FFTTFFTT
FTTTFFTT
TFFFTFFT
TTFTFTTT

 

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)

 

pqr∼q
 r p rp ∧ r∼q (p ∧ r)R
FFFTTTTFFF
FFTTTFFFFT
FTFFTTTFTT
FTTFTFFFTT
TFFTFTFFFT
TFTTTTTTTT
TTFFFTFFTT
TTTFTTTTTT

 

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.

 

To gain better understanding about Tautology, Contradiction and Contingency,

Watch this Video Lecture

 

Get more notes and other study material of Propositional Logic.

Summary
Tautology Contradiction Contingency
Article Name
Tautology Contradiction Contingency
Description
Tautology, Contradiction, Contingency, Valid, Invalid, Falsifiable, Unfalsifiable, Satisfiable, Unsatisfiable with their definition, truth table and examples are discussed.
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Gate Vidyalay
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