# Problems on Tautology

**Pre-requisite:** Basics of Propositions, Laws

**Proposition**– The meaning of proposition in literature is an idea, a plan or an offer, or a suggestion that can be proved True or False. The same goes for mathematical propositions. They are declarative sentences that can be True or False. Propositions are the fundamental building blocks of logic.

Examples:1. The magnetic lines emerge from the North and merge into South pole.

2. 2 + 1 = 3

3. ‘p’ is a vowel.

All of the above three sentences are proper propositions, where the first two are True and the third one is False.

## Tautology

Propositional logic is said to be a tautology if it is always true regardless of the true/false of the atomic formulas. A tautology is always â€œTrue”. To check whether a given logic is a tautology or not, we often use the truth table method. Although the truth table method is not effective when the logic contains a number of atomic formulas.

Example:Odd number = A

Even number = B

1. If we add one odd number and one even number then we get odd number.

Conve

g statement-1 into mathematical logic:

A âˆ§ B â‡’ A

Let’s prove that the above logic is a tautology. In order to build the truth table, we need to convert the logic statements into a clausal form.

The truth table of A âˆ§ B â‡’ A, clausal form is: ï¿¢(A âˆ§ B)âˆ¨A

A | B | (A âˆ§ B) | ï¿¢(A âˆ§ B) | ï¿¢(A âˆ§ B)âˆ¨A |
---|---|---|---|---|

T | T | T | F | T |

T | F | F | F | T |

F | T | F | T | T |

F | F | F | T | T |

All entries are True, irrespective of the True/False values of atomic literals. So, this is a tautology.

Examples of Tautology with logic symbols:

- ï¿¢Aâˆ¨A
- (Pâˆ¨Q)â‡’(Pâˆ¨Q)

A mathematical sentence consisted of logic. A proposition is either True or False. The proposition is made up of mathematical logic. Various propositional logics are given below in their priority order:

- Negation (No)
- Conjunction (and)
- Disjunction (or)
- Implication (â‡’)
- Equivalence (â‡”)

**Tautology**– A proposition that is always true. The truth table is evaluated for the given proposition and if in every case the result is True, then that proposition is called Tautology.

**Truth table**

It is a table that gives the output of the propositional logic against each input component. The result is binary, either True or False for each row of inputs.

**Problems: Find if the given propositional logic is a tautology or not.**

**1) P**

**Truth table:**

P |
---|

T |

F |

The truth table of P contains a False value. *Thus, it can not be a tautology.*

**2) Pâ‡’P**

We shall

the truth table for this proposition.**Implication:**

Pâ‡’Q =ï¿¢Pâˆ¨Q

The simplified expression of the given proposition is: ï¿¢Pâˆ¨P

**Truth Table:**

P | ï¿¢P | ï¿¢Pâˆ¨P |
---|---|---|

T | F | T |

F | T | T |

The truth table of ï¿¢Pâˆ¨P consists only of True values.* Therefore, Pâ‡’P is a tautology.*

**3) (P â‡’ P) â‡’ P**

We shall draw the truth table for this proposition.

**Implication:**

Pâ‡’Q=ï¿¢Pâˆ¨Q

The simplified expression of the given proposition is:

¢Pâˆ¨P) â‡’ P

ï¿¢(ï¿¢Pâˆ¨P)âˆ¨P

(ï¿¢(ï¿¢P) âˆ§ ï¿¢P)âˆ¨P {By Demorgan’s Law}

(P âˆ§ ï¿¢P) âˆ¨P

(P âˆ§ ï¿¢P)= False {Complement laws: â€“ Pâˆ§ï¿¢P=F }

False âˆ¨ P = P {Absorption law}

Thus, (P â‡’ P) â‡’ P is equivalent to P. We have already solved this in problem-1.

Therefore,

this is not a Tautology.

**4) (p â†’ q) â†’ [(p â†’ q) â†’ q]**

Solving: (p â†’ q) = ï¿¢pâˆ¨q {Implication}

Solving: [(p â†’ q) â†’ q]

= [(ï¿¢pâˆ¨q) â†’ q]

sp; = [ï¿¢(ï¿¢pâˆ¨q)âˆ¨q]

= [(ï¿¢(ï¿¢p)âˆ§ï¿¢q)âˆ¨q] {Demorgan’s Law}

= [(pâˆ§ï¿¢q)âˆ¨q] {Involution law}

= [(pâˆ¨q)âˆ§(ï¿¢qâˆ¨q)] {Distributive law}

= [(pâˆ¨q)âˆ§T] {Complement law}

= (pâˆ¨q) {Absorption law}

Solving (p â†’ q) â†’ [(p â†’ q) â†’ q]

ï¿¢(ï¿¢pâˆ¨q)âˆ¨(pâˆ¨q)

[ï¿¢(ï¿¢p)âˆ§(ï¿¢q)]âˆ¨(pâˆ¨q){Demorgan’s Law}

(pâˆ§ï¿¢q)âˆ¨(pâˆ¨q) {Involution law}

Thus,

al expression is: (pâˆ§ï¿¢q)âˆ¨(pâˆ¨q)

**Truth table: **

p | q | ï¿¢q | (pâˆ§ï¿¢q) | (pâˆ¨q) | (pâˆ§ï¿¢q)âˆ¨(pâˆ¨q) |
---|---|---|---|---|---|

T | T | F | F | T | T |

T | F | T | T | T | T |

F | T | F | F | T | T |

F | F | T | F | F | F |

Since there is a False entry in the truth table, **it implies it is not a Tautology**.

**5) ((Pâ‡’Q)âˆ§P)â‡’Q**

Solving (Pâ‡’Q): ï¿¢Pâˆ¨Q

Solving ((Pâ‡’Q)âˆ§P): ((ï¿¢Pâˆ¨Q)âˆ§P)

= (ï¿¢Pâˆ§P)âˆ¨(Qâˆ§P) {Distributive Law}

= (F)âˆ¨(Qâˆ§P) {Complement Law}

= (Qâˆ§P) {Absorption Law}

Solving ((Pâ‡’Q)âˆ§P)â‡’Q: (Qâˆ§P)â‡’Q

= ï¿¢(Qâˆ§P)âˆ¨Q

= (ï¿¢Qâˆ¨ï¿¢P)âˆ¨Q {Demorgan Law}

= (ï¿¢Qâˆ¨Q)âˆ¨ï¿¢P) {Associative Law}

= Tâˆ¨(ï¿¢P) {Complement Law}

= T {Absorption Law}

Final CNF is:

True

Here, no need of finding the Truth Table. Given logic is a tautology.