**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-

- p ∧ ∼p
- (p ∧ (p → q)) → ∼q
- [ (p → q) ∧ (q → r) ] ∧ ( p ∧ ∼r)
- ∼(p → q) ∨ (∼p ∨ (p ∧ q))
- (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-__

__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,

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