DISCRETE MATHEMATICS CHAPTER I

CHAPTER I Propositions Sentences considered in propositional logic are not arbitrary sentences but are the ones that are either true or false, but not both. This kind of sentences are called propositions. If proposition is true, then we say it has a truth value of “true”; if a proposition is false, its truth value is “false”. Examples:1. “Grass is green”, and “2+5=5” are propositions. The first proposition has the truth value of “true “ and the second proposition “false”. 2. “Close the door” , and “Is it hot outside?” are not propositions. 3. “x is greater than 2”, where x is a variable representing a number, is not a proposition, because unless a specific value is given to x we can not say whether it is true or false, nor do we know what x represents. 4. “He is intelligent or studies every night” is a compound statement with sub statements “He is intelligent” and “He studies every night”. 5. “Where are you going?” is not proposition since it is neither true nor false.

Examples 2 is irrational (T) 3 + 4 = 6 (F) 6 + 1 = 4 + 3 (T) Here are some sentences that are not statements: 6) What did you say? 7) X = 5 8) Give me your book.

Implication(Conditional) Propositions will be denoted by the letters p, q, r . The fundamental property of a proposition is that it is either true or false, but not both. Connectives The words and symbols used to form compound propositions are called connectives. Symbolic form Nature of the compound statement formed by the connective Connective word Symbol used ~ p Negation not ~ , ¬ p ˄ q Conjunction and ˄ p ˅ q Disjunction or ˅ P → q Implication(Conditional) If…..then → P ↔ q Equivalence(Bi-conditional) If and only if ↔

Propositions and Truth Tables As mentioned the truth value of a proposition is either true ( denoted by T)or false (denoted by F). A truth table is a table that shows the truth value of a compound propositions for all possible cases. The truth value of the compound statement ( p ˄ q) is given by the following table: p ˄ q q p T F

The truth value of the compound statement ( p ˅ q ) is given by the following table: 3. ( ~p ) , Negation 4. Conditional , p → q p ˄ q q p T F p ˅ q T T ~p p F T p → q q p T F

5. Equivalence (Bi-conditional) , p ↔ q F 1Example. Example.1 Find the truth table of: ~( p ˄~ q ) p˅~p p ˄ ~p Solutions. a) a b c ~( p ˄~ q ) p ˄~ q q~ q p T F

Tautologies and Contradictions b) p˅~p ~p p T F c) p˄~p ~p p F T Tautologies and Contradictions Some propositions P(p, q, r, …) contain only T in the last column of there truth table, such propositions are called tautologies. Similarly, a proposition P(p, q, r, …) is called a contradiction if it contain only F in the last column of there truth table. For ex. ,Example.1 (b) is a tautology, and the proposition p ˄ ~p in Ex.1 (c) is contradiction.

Find the truth table of ~p ˅ q Example.2 Find the truth table of ~p ˅ q ~p ˅ q ~p q p T F Observe that the above truth table is identical to the truth table of p → q . Hence p → q is logically equivalent the proposition ~p ˅ q . p → q ≡ ~p ˅ q ≡

Law of the Algebra of propositions Let p , q and r be statements, and let (1) represent the true statement and (0) the false statement. Note. All the laws of table (*) come in pairs, called dual pairs. For each expression, one finds the dual by replacing all T(1) by F(0) and all F(0) by T(1) and replacing all ˄ by ˅ and all ˅ by ˄ .

Table * (1b) p ˄ p ≡ p (1a) p ˅ p ≡ p (2b) (p˄ q) ˄ r ≡ p ˄ (q˄ r) (2a) (p ˅ q) ˅ r ≡ p ˅ (q˅ r) (3b) p ˄ q ≡ q ˄ p (3a) p ˅ q ≡ q ˅ p (4b) p ˄ (q˅ r) ≡ (p ˄q) ˅(p ˄ r) (4a) p ˅(q˄ r) ≡ (p ˅q) ˄(p ˅ r) (5b) p ˄ 1 ≡p (5a) p ˅ 0 ≡ p (6b) p ˄ 0 ≡ 0 (6a) p˅ 1 ≡ 1 (7b) p ˄ ~p ≡ 0 (7a) p ˅~ p ≡ 1 (8b) ~F ≡ T (8a) ~T ≡ F (9) ~ (~ p) ≡ p (10b) ~( p ˄ q) ≡ ~ p ˅~ q (10a) ~( p ˅ q) ≡ ~ p ˄ ~ q (11) p→q ≡ ~ p ˅ q ≡ (~ p ˅ q ) ˄(~ q ˅ p ) (12) p↔q ≡ (p→q) ˄ (q→p)

Problems. Verify that the proposition p˅~ (p ˄ q) is a tautology. Verify that the proposition (p ˄ q) ˄ ~ (p ˅ q) is contradiction. Prove that: (a) p ˅ q ≡ ~ (~ p ˄ ~ q) , (b) ~ (p ˄ q) ≡ ~ p ˅ ~ q , (c) ~ (p ˅ q) ≡ ~ p ˄ ~ q (d) (p→q) ˄ (q→p) ≡ p↔q Verify that the proposition [(p↔q) ˄ q] →p is a tautology. Determine the truth table of: (a) ( p→q ) ˅ ~ ( p↔ ~ q ) , (b) ( p→q ) → ( p ˄ q ) Use truth tables to prove that : (a) p ˅ (q ˄ r) ≡ (p ˅ q ) ˄ (p ˅ r) , (b) [ ~ (p ˄ q) ˅ r] → ~ p is a tautology. 7. Simplify each proposition by using the law listed in table * (a) p ˅ (p ˄ q) , (b) ~ (p ˅ q) ˅ (~ p ˄ q ) , (c) ~ (p→ ~ q) ˅ (p˅ q) (d) ~ p ˄ ~(~ p ˅ ~ q ) , (e) ~( p ˄ ~q) ˅(p) ˅ (p˅ q) (f) ~ [~( ~p ˅ q)˅ ( p ˄ ~q)] ˅ q .