**Logical reasoning**

**Statements**

A statement in a logical context is a declaration, verbal or written that is either true or false but not both. A true statement is said to have a truth value **T,** while a false statement is said to have a truth value **F.**

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**Example 1**

The following are statements:

(a) Nigeria is an African country.

(b) The earth is conical in shape.

(c) If I run I shall not late.

(d) Japanese are hard working people.

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**Example 2**

The following are not statements in the logical context.

(a) Who is he?

(b) What a lovely man!

(c) Take the pencil away.

(d) If I think of my family.

In general, questions, exclamations, commands and expressions of feelings which cannot be assigned a truth value T or F are not statements in the logical context.

By convention, we shall use letter *P*.*q.r,…* to denote statements.

**Negation**

Given a statement P, the negation ofP, written ̴P is the statement; “it is false that P” or “nor P”.

If P is true, ̴P is false and if P is false ̴P is true. In other words, if P has the truth value T then ̴P has the truth value F and if P has the truth value F then ̴P has the truth value T.

The relationship between P and ̴P can be summarizer in the following table.

** Table 1**

P | ̴P |

T | F |

F | T |

Table 1 is called a truth table.

If P is a statement: “Nigeria is a rich country” then ̴P is the statement: “it is false that Nigeria is a rich country” or in a more reasonable English “Nigeria is not a rich country”.

Let *q* be the statement “some lawyers are honest people” then ̴*q* is the statement” it is false that some lawyers are honest people”. In a more reasonable English, we can also write ̴*q* as some lawyers are not honest people.

Let P be the statement 3 + 4 = 8 then ̴P is the statement 3 + 4 ≠ 8.

Let q be the statement X + 1 ≥ 4 then ̴*q* is the statement X + 1 ≥ 4.

Let P be the statement “The set of numbers 2, 4, 6, 8, … is a set of even numbers” then ̴P is the statement “The set of numbers 2, 4, 6, 8, … is a set of odd numbers.

- State which of the following are statements in the logical context:

(a) Caesar was a great leader.

(b) Stop talking to the boys.

(c) Decide whether you are going to the club’s meeting now.

(d) Oh Mansa Musa, you are wonderful!

(e) The Broking House in Ibadan, is a magnificent building.

- State which of the following are statements in the logical context:

(a) As old as Methuselah.

(b) The set of numbers 3, 5 and 4 is not a Pythagorean triplet.

(c) Is he a serious teacher at all?

(d) If 6 is an odd number, then 3 + 5 = 10.

- Write the negation of each of the following statements:

(a) He is a handsome man.

(b) It is very cold in Siberia.

(c) It is very hot in tropics.

(d) The sky is blue.

- Write the negation of each of the following statements:

(a) The party leader will win the election.

(b) The football captain scored the first goal.

(c) Short cuts are dangerous.

(d) Honest men are very rare to come by.

- Statement with reasons whether the statement
*q*is a negation of the statement*P*in each of the following:

(a) *p*. The line AB is parallel to the line CD,

- The line AB is perpendicular to the line CD.

(b) *p*. He is a good leader.

- He is a bad leader.

(c) *p*. She is a good leader.

- She is a good follower.

(6) Write the negation of each of the following avoiding the word ‘not’ as much as possible.

(a) The car is moving fast.

(b) He was present in school yesterday.

(c)The Equator is a Great circle.

(d) His friend is younger than my brother.

(7) Write the negation of each of the following avoiding the word ‘not’ as much as possible.

(a) He obtained the least mark in the examination.

(b) She is the shortest girl in the class.

(c) He is an ugly man.

(d) The hospital is in a bad state.

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**Conditional Statement**

Let *p *stand for the statement ‘Lagos is a state in Nigeria’ and *q* stand for the statement ‘Lagos is a state in Africa’. One way the two statements can be combined is ‘If Lagos is a city in Nigeria then Lagos is a city in Africa or ‘if *p* and *q*’.

The statement ‘‘if *p* and *q*’ is a combination of two simple statements *p* and *q* it is therefore called a **compound statement.**

Symbolically, we can write the compound statement if *p* then *q*as *p q*.

The statement*p* *q* is read as:

*p* implies q or

If p then q or

*q* is necessary for *p* or

*p* is sufficient for *q* or

*p* only if *q* or

*p* follows from *p* or

*q*if*p.*

The symbol is an operation. In the compound statement*p* is called the **antecedent** while the sub statement*q* is called the **consequent **of *p q.*

The truth or falsity of the implication *p q* is illustrated in Table 2.

**Table 2**

p | Q | P q |

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

The statement *p* *q* is false if the antecedent is true and the consequent is false. The statement *p* *q* is sometimes called a **Conditional Statement.**

Consider the following statements;

(a) If Cairo is in Africa then 8 is an even number.

(b) If Cairo is in Africa then 8 is an odd number.

(c)If Cairo is in Asia then 8 is an even number.

(d) If Cairo is in Asia then 8 is an odd number.

The statements (a) (c) and (d) are all true but the statement (b) is not true for the simple reason that the antecedent is true while the consequent is false. Note that although in statement (d) both the antecedent and consequent are false, yet the whole statement is true.

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**Converse Statement**

Let*p* be the statement ‘a triangle is equilateral’ and *q* the statement ‘a triangle is equiangular’. The statement *p* implies *q.*i.e*p q*. The statement also implies p, i.e*q p*. In other words, if a triangle is equiangular then it is equilateral. The statement *q p* is called the **converse of the statement**

*P q.*

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**Inverse Statement**

If *p* is the statement ‘a triangle is equilateral’ and *q* is the statement ‘a triangle is equiangular’ the statement ̴*p * ̴*q* is the statement ‘if a triangle is not equilateral than it is not equiangular.

The statement ̴*p * ̴*q*is called the **inverse of the statement ***p q.*

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**Contrapositive Statement**

If *P* is the statement ‘a triangle is equilateral’ and *q* is the statement ‘a triangle is equiangular’ the statement ̴*q * ̴*p* is ‘if a triangle is not equiangular then it is not equilateral’. The statement

*q p* is called the **contrapositive statement **of *p q.*

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**Biconditional Statement**

Let *p* be the statement ‘the interior angles of a polygon are equal’ and *q* be the statement ‘a polygon is regular’.

*p q* is the statement’If the interior angles of a polygon are equal then the polygon is regular’

*q p* is the statement ‘If a polygon is regular then the interior angles of the polygon are equal’. We see here that*p q *and *q p.*

The two conditional statements are valid. We say that *p *and *q* imply ‘each other or p is equivalent to *q* and we write *p q*. The statement *p q* is called a **Biconditional Statement** of *p and q* and the

statements*p* and *q* are said to be equivalent to each other.

Other terminologies for *p q* are:

*q* is equivalent to *p*

*p* is necessary and sufficient for *q*

*q*if and only if *p*

*p* if and only if *q*

If *p*thenq and if *q* then*p.*

The truth of falsity of *pq* is completely illustrated by Table 3.

**Table 3**

p | q | P q |

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

So a biconditional statement is true when the two substatements have the same truth value.

Consider the following four statements:

(a) Nyerere is an African name if and only if a, e, i, o, u are vowels;

(b) Nyerere is an African name if and only if a, e, i, o, u are consonants;

(c) Nyerere is a European name if and only if a, e, i, o, u are vowels;

(d) Nyerere is a European name if and only if a, e, i, o, u are are consonants.

The statements (a) and (d) are both true since the substatements of each have the same truth value. The statements (b) and (c) are false since the substatement of each have different truth values.

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**The Chain Rule**

If *p. q *and*r* are three statements such that *p q* and *q r,* then *p r.* This is called the **chain rule**and it may have several links.

Consider the argument:

T_{1}: If a student works very hard, he passes his examination.

T_{2}: If a student passes his examination, he is awarded a certificate.

T_{1}: If a student works very hard, he is awarded a certificate.

Let *P* be the statement ‘a student works very hard.

Let *q* be the statement ‘a student passes his examination.

Let *r* be the statement ‘a student is awarded a certificate.

The argument has the following structural form:

*p p. q r : p r*

This argument follows the chain rule link hence it is said to be valid.

The statement T_{1} and T_{2} are called the **Premises**while the statement T_{3} is called the **Conclusion** of the argument.

The argument is valid not on the basis of the truth or falsity of its premises but on the basis of the structural from which follows the chain rule link.

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**Compound Statement**

A statement may consist of two or more simple statements or substatements. Such a statement is called

a**Compound **or **Composite Statement.** For example, the compound statement: ‘John is an intelligent and courageous boy’, consists of the substatements: ‘John is an intelligent boy’ and ‘John is a courageous boy’.

By convention, we shall use letters *p, q, r,…* to denote statements.

We proceed now to introduce logical symbols called **Connectives.**

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**Disjunction**

Two statements can be combined by the use of the connective ‘or’.

The statement: ‘He is a philosopher’ can be combined with the statement ‘He is a teacher as follow: ‘He is a philosopher or he is a teacher’. In a more refined English, the combined statement is ‘Either he is a philosopher, or he is a teacher’.

If the statement ‘He is a philosopher’ is denoted *p* and the statement ‘He is a teacher is denoted *q,* the combined statement is either *p* or *q*or simply *q or q.* In symbolic logic, *q *or *q *is designated *p* v *q,* where the connective *p *v *q.*Where the connective v means ‘or’.

The word ‘or; in English Language is used in two different sense.

In the statement, ‘the professor will deliver the lecture at Bayero University or remain at the hotel’. The ‘or’ is used in an exclusive sense, in that it is not possible for the professor to deliver lecture at Bayero University, and at the same time remain at the hotel. The ‘or’ is said to be used in the exclusive sense.

In the statement, ‘He is a good politician or a good statement’, the or is used in the inclusive sense, in that it is possible for the good politician to be a good statesman as well. We shall clear this apparent ambiguity, it we settle for one meaning.

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**In this teat, we use ‘or’ in the inclusive sense.** The statement *p v q *can be read; either *p or q *or both or simply *p v q* or both. The truth table for *p v q* is illustrated in Table 4

**Table 4**

p | q | P v q |

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

The statement *p v q* is false when both *p *and *q* are false, otherwise *p v q* is true.

Consider the following statements:

(a) Cairo is in Africa or 8 is an even number.

(b) Cairo is in Africa or 8 is an odd number.

(c) Cairo is in Asia or 8 is an even number.

(d) Cairo is in Asia or 8 is an odd number.

The statements (a) (b) and (c) are all true since at least one of the substatements is true. The statement (d) is false since the two substatements are false.

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**Conjunction**

Another way by which two statements can be combined with the statement ‘she is proud’ using the connective ‘and’ as ‘she is pretty and she is proud’. In a more refined English, the combined statement is: ‘she is pretty and proud’.

If the statement ‘she is pretty’ is denoted *p* and the statement ‘she is proud’ is denoted *q,* the statement ‘she is pretty and proud’ can be written as *p *and *q.* In symbolic logic, *p* and *q* is designated *p ᴧq,* where

the connective *ᴧ* means ‘and’.

The truth table for *p ᴧ q* is illustrated in Table 5

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**Table 5**

p | q | P ᴧ q |

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

*The statementp ᴧ q ***is true when the sub statements p and q are both true, otherwise p ᴧ q is false.**

Consider the following statements:

(a) Cairo is in Africa and 8 is an even number.

(b) Cairo is in Africa and 8 is an odd number.

(c) Cairo is in Asia and 8 is an even number.

(d) Cairo is in Asia and 8 is an odd number.

The statements (a) (b) and (c) are all true since at least one of the substatements is true. The statement (d) is false since the two substatements are false

Of the four given statements, only (a) is true. The statements (b), (c) and (d) are since at least one if their substatement is false.

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**Equivalent Statements**

**Two compound statements are said to be logically equivalent if they have the same truth value.**

Use the truth table to establish that:

̴(*p v q)* = ̴*p* ᴧ ̴*q*

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**Solution**

** Table 6**

p | Q | p ᴧ q | ̴(p v q) |

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

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P | q | ̴p | ̴q | ̴pᴧ ̴q |

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

(a) (b)

We observe that the last columns of Table 6(a) and Table 6(b) have the same truth values.

Hence ̴(*p v q)* = ̴*p ᴧ* ̴*q*

Use the truth table to prove that:

̴(*p ᴧ q)* = ̴*p v* ̴*q*

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**Solution**

** Table 7**

p | q | p ᴧ q | ̴(p v q) |

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

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P | Q | ̴p | ̴q | ̴pᴧ ̴q |

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

(a) (b)

The last columns of the two tables have the same truth values, hence ̴(*p ᴧ q)* = ̴*p v * ̴*q*

Use the truth table to show that:

(a) the connective ᴧ distributes over the connective v;

(b) the connective v distributes over the connective ᴧ.

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**Solution**

Given the statements *p, q *and* r* and the connectives ᴧ and v, we wish specifically to show that:

(a) p ᴧ (q v r) = (p ᴧ q) v (p ᴧ r)

(b)p v (q ᴧ r) = (p v q) ᴧ (p v r)

(a)

**Table 8**

p | Q | R | q v r | pᴧ (q v r) |

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

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**Table 9**

p | q | r | p ᴧ q | p ᴧ r | (p ᴧ q) v (p ᴧ r) |

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

We notices that the last two columns of the two tables above the same truth values,

Hence p ᴧ (q v r) = (p ᴧ q) v (p ᴧ r)

(b)

**Table 10**

p | q | R | q ᴧ r | p v (q ᴧ r) |

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

**Table 11**

p | q | r | p v q | p v r | (p v q) ᴧ (p v r) |

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

We see here also that the last two columns of Table 10 and Table 11 have the same truth value, hence

p v (q ᴧ r) = (p v q) ᴧ (p v r)

Given the statements p, q and r show that:

(a) p ᴧ (q ᴧ r) = (p ᴧ q) ᴧ r

(b) p v (q v r) = (p v q) v r

**Table 12**

(a

p | q | R | q ᴧ r | p v (q ᴧ r) |

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

P | Q | r | p ᴧ q | (p ᴧ q) ᴧ r |

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

(a) (b)

The last two columns of Table 12(a) and Table 12(b) are identical, hence;

p ᴧ (q ᴧ r) = (p ᴧ q) ᴧ r

We see here that the conjunctive connective as an operator, is associative.

(b) ** Table 13**

P | Q | r | p ᴧ q | (p v q) v r |

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

p | q | r | q v r | p v (q v r) |

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

Table last two columns of Table 13a and Table 13b are identical, hence p v (q v r) = (p v q) v r

**The disjunctive connective as an operator is seen as being associative.**

**Tautology and Contradiction**

A compound statement which is always true irrespective of the truth values of the substatements, is called a **Tautology**. A tautology is represented as **T.**

A compound statement which is always false, irrespective of the truth values of the substatements is called a **Contradiction.** A contradiction is usually represented as **F.**

Ues the truth table to show that the statement p v ̴p is a tautology.

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**Solution**

**Table 14**

p | ̴p | p v ̴p |

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

We observe that the last column of the table has the truth value T irrespective of the truth values of the substatements. Hence the statement p v ̴p is a tautology.

Use the truth table to show the statement p ᴧ ̴p is a contradiction.

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**Solution**

**Table 15**

p | ̴p | p ᴧ ̴p |

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

The last column of Table 15 has the truth value F irrespective of the truth values of the substatements hence, p ᴧ ̴p = F.

Show that the statement

p ᴧ (( ̴p ᴧ q) v ( ̴p ᴧq)) is a contradiction.

**Table 16**

p | Q | ̴p | ̴q | ̴p ᴧ q | ̴p ᴧ ̴q | ( ̴p ᴧ q) v ( ̴p ᴧ ̴q) | p ᴧ (( ̴p ᴧ q) v ( ̴p ᴧ ̴q)) |

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

We see that the truth value of expression in the last column of Table 16 is F irrespective of the truth values of the substatements, hence p ᴧ (( ̴p ᴧ q) v( ̴p ᴧ ̴q)) = F

**Laws of the Algebra of Logical Statements**

There is a close relationship between the algebra of sets and algebra of logical statements. The logical connectives as operations obey the laws of algebra.

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**Commutative Laws**

- (a) p ᴧ q = q ᴧ p

I (b) p v q = q v p

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**Associative Laws**

Ii (a) p ᴧ (q ᴧ r) = (p ᴧ q) ᴧ r

Ii (b) p v (q v r) = (p v q ) v r

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**Distributive Laws**

iii. (a) p ᴧ (q ᴧ r) = (p ᴧ q) ᴧ (p ᴧ r)

iii. (b) p v (q v r) = (p v q ) v (p v r)

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**Laws of Absorption**

- (a) p ᴧ (p v q) = p
- (b) p v (p ᴧ q) = p

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**Idempotent Laws**

- (a) p ᴧ p = p
- (b) p v p = p

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**De Morgan’s Laws**

- (a) ̴(p ᴧ q) = ̴p c ̴q
- (b) ̴(p v q) = ̴p ᴧ ̴q

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**Laws of Complementation**

vii. (a) p ᴧ ̴p = F (b) p v ̴p = T

vii. (a) ̴F = T (b) ̴T = F

- ̴( ̴p) = P

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**Laws of Contrapositivity**

- p q = ̴q ̴p

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**Laws of Identity**

- P ᴧ T = P

xii. P ᴧ F = F

xiii. P v T = T

xiv. P v F = P

All these laws can be verified using the truth table technique.

Using the truth table technique, show that if p, q and r are arbitrary statements, then:

(p q) ᴧ (q r) (p r) = T

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**Solution**

** Table 17**

p | q | r | p q | q r
| (p q)r (q r) | p r | (p q) ᴧ (q r) (p r) |

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

The truth value of the expression in the last column is T, hence

(p q) ᴧ (q r) (p r) = T

Use the truth table technique to show that

p q = (p q) ᴧ (q p)

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**Solution**

**Table 18**

(a)

p | Q | p q |

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

(b)

p | Q | p q | q p | (p q) ᴧ (q p) |

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

The truth values of the expression in the last columns of each of the table are identical, hence

p q = (p q) ᴧ (q p)

Use the truth table technique to show that

̴p ̴q = q p

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**Solution**

**Table 19**

p q ̴p ̴q ̴p ̴q

T | T | F | F | T |

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

p q q p

T | T | T |

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

(a) (b)

The last column of the two tables are identical, hence

̴p ̴q = q p

- Let
*P*be the statement ‘He is funny’ and*q*be the statement ‘He is serious’. Write each of the following in a simple English:

(a) p v q (b) p ᴧ q

(c) p ᴧ ̴q (d) ̴p v ̴q

- Let
*p*be the statement ‘she is beautiful’ and*q*be the statement ‘she is soft – spoken’. Write each of the following in symbolic form:

(a) She is beautiful and soft – spoken.

(b) Either she is beautiful or she is soft – spoken.

(c) She is beautifulbut not soft – spoken.

(d) She is ugly but soft – spoken.

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**Evaluation**

- Fin the truth value of these statements (a) If 11 > 8 then -1 < -8(b) If 3 + 4 ≠ 10 then 2 + 3 ≠ 5.

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**General Evaluation**

(1) Let P be the statement: “He is funny” and q be the statement: “He is serious”. Write each of the following in simple English (i) p v q (b) p ᴧ ̴q (c) ̴p v ̴q

(2) If p and q represent two statements ‘he is good in physics” and “he is good in mathematics” respectively. Write the following in symbolic form; “he is good in physics if and only if he is good in mathematrics”.

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**Reading Assignment**

F/Maths Project 1 pages 126 – 130 Exercise 9b Q 2, 3 and 4

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**Weekend Assignment**

**Objective**

P is the statement “Ayo has determination and q is the statement”Ayo will succeed”. Use this information to answer the questions. Which of these symbols represent these statements?

(1) Ayo has no determination (a) p q (b) ̴p q (c) ̴p

(2) If Ayo has no determination then he won’t succeed (a) ̴p ̴q (b) p ̴q (c) p q

(d) p ̴q

(3) If Ayo won’t succeed then he has no determination (a) ̴q p (b) ̴q ̴q (c) ̴q p

(d) q p

(4) If Ayo has determination then he will succeed (a) ̴p q (b) ̴p ̴q (c) ̴q ̴p

(d) p q

(5) If Ayo has no determination then he will succeed (a) ̴p q (b) ̴q ̴p (c) ̴p

(d) ̴p ̴q

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**Theory**

(1) Write down the inverse, converse and contrapositive of each of these statements.

(i) If the bank workers work hard they will be adequately compensated.

(ii) If he is humble and prayerful, he will meet with God’sfavour.

(iii) If he set a good example, he will get a good followership.

(2) Consider the following statements P: some dogs are tame Q: all tame animals are small.

Which of the following is a valid conclusion from the above statements?

(i) All dogs are tame (ii) No dog is small (iii) All small animals are tame (iv) Some dogs are small (v) All tame animals are dogs

See also

Cubic equations and their factorization