Tag: Falsifiable

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 ∼p p ∧ ∼p
F T F
T F F

 

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-

 

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-

  • 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)

 

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.

 

To gain better understanding about Tautology, Contradiction and Contingency,

Watch this Video Lecture

 

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