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Discrete Mathematics University of Jazeera College of Information Technology & Design Khulood Ghazal

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Predicate logic Suppose we have Propositional Logic, with statements like: p is "All people with red hair have fiery tempers“ q is "Joe has red hair“ r is "Joe has a fiery temper", To make a statement about Joe having red hair, and therefore a fiery temper, we can write: P ^ q ⇒ r To make a statement about Brenda having red hair, and therefore a fiery temper, we need further propositions, like this: s is "Brenda has red hair" t is "Brenda has a fiery temper" and so: p ^s ⇒ t Each time we want to make a statement about another person having red hair, and therefore a fiery temper, we need further propositions.

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A much better way of representing these ideas is to use predicates like this: redHair means "... has red hair“ (redHair is not a proposition.) We can use this predicate to form statements about anyone who has red hair; like this: redHair(Joe) means "Joe has red hair“ is a proposition redHair(Brenda) means "‘Brenda has red hair“ is a proposition... and so on. we can define the predicate fieryTemper means "... has a fiery temper" So "If Joe has red hair, then he has a fiery temper" can be represented by: redHair(Joe) ⇒ fieryTemper(Joe) And "If Brenda has red hair, then she has a fiery temper" by: redHair(Brenda) ⇒ fieryTemper(Brenda) Predicate logic

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Write each of the following propositions using predicate notation: 1. Jimmy is a friend of mine. friend(Jimmy) 2. Sue is wealthy and clever. wealthy(Sue) ^ clever(Sue) 3.Jane is wealthy but not clever. wealthy(Jane) ^ ¬clever(Jane) 4. Both Mark and Elaine are friends of mine. friend(Mark) ^ friend(Elaine) 5. If Peter is a friend of mine, then he is not boring. friend(Peter) ⇒ ¬ boring(Peter) 6. If Jimmy is wealthy and not boring, then he is a friend of mine. (wealthy(Jimmy) ^ ¬boring(Jimmy)) ⇒ friend(Jimmy) Example: The following predicates are defined: friend "… is a friend of mine" wealthy "… is wealthy“ clever "… is clever" boring "… is boring“

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A function has the property that it returns a unique value when we know the value(s) of any parameter(s) supplied to it. P (x) is a function since it returns a value which depends upon the value of its parameter, x. P (x) can then be described as a propositional function whose predicate is P. Propositional Functions

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Define suitable propositional functions for : (a)All of my friends are wealthy. (b) Some of my friends are boring. we will making statements about my friends in general, without referring to a particular individual. So we need to define propositional functions as follows: friend(x) is "x is a friend of mine" wealthy(x) is "x is wealthy" boring(x) is "x is boring" We can re_ write the two statements above as: (a)For all x, friend(x) ⇒ wealthy(x) (b)For some x, friend(x) ^ boring(x) Quantifiers

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Notation: ∀ and ∃ The symbol ∀ (called the universal quantifier) stands for the phrase "For all …“ The symbol ∃ (called the existential quantifier) stands for the phrase "For some …" So we can write (a) above as: (a) ∀ x, friend(x) ⇒ wealthy(x) Means "For each value of x, if x is a friend of mine, then x is wealthy". So we can write (b) above as: (b) ∃ x, friend(x) ^ boring(x) Means "For at least one value of x, x is a friend of mine and x is boring". Quantifiers

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(a)Some cats understand French. Re-write in the singular: "At least one cat understands French". So we need to define propositional functions as: cat (x) is "x is a cat" French(x) is "x understands French" So there is at least one x that is a cat and understands French; or, in symbols: ∃ x, cat (x) ^ French(x) (b) At least one lecturer is not boring. This is already in the singular; so: lecturer(x) is "x is a lecturer" boring(x) is "x is boring" So: ∃ x, lecturer(x) ^ ¬ boring(x) Example : Define suitable propositional functions and then express in symbols:

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(c) I go swimming every sunny day. sunny(x) is "x is a sunny day" swimming(x) is "x is a day when I go swimming" we can re-write it as: "For each day, if it is a sunny day then it is a day when I go swimming" So, in symbols: ∀ x, sunny(x) ⇒ swimming(x) (d) No footballers can sing. we might re-write "No footballers can sing" as "For each x, if x is a footballer, then x cannot sing". In symbols, then, this gives the equally valid solution: ∀ x, footballer(x) ⇒ ¬ sing(x) Re-write in the singular: "It is not true that at least one footballer can sing". So: footballer(x) is "x is a footballer" sing(x) is “ x can sing" In symbols, then: ¬ ( ∃ x, footballer(x) ^ sing(x)) So, x P(x) is the same as x P(x).

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Using the above predicates, symbolize each of the following: (a) Some of my friends are clever. ∃ x, friend(x) ^ clever(x) (b) All clever people are boring. ∀ x, clever(x) ⇒ boring(x) (c) None of my friends is wealthy. ∀ x, friend(x) ⇒ ¬wealthy(x) OR: ¬( ∃ x, friend(x) ^ wealthy(x)) (d) Some of my wealthy friends are clever. ∃ x, friend(x) ^ wealthy(x) ^ clever(x) (e) All my clever friends are boring. ∀ x, (clever(x) ^ friend(x)) ⇒ boring(x) (f) All clever people are either boring or wealthy. ∀ x, clever(x) ⇒ (boring(x) ∨ wealthy(x)) Example :The following predicates are defined : friend "… is a friend of mine" wealthy "… is wealthy" clever "… is clever" boring "… is boring“

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