1 / 12

More on Composite of Existential and Universal Quantifiers

More on Composite of Existential and Universal Quantifiers. (x) (y) <=> (y) (x) ? Let’s look at this with an example: Let x be father and y be children (x) (y) : There exists a father for all children (y) (x) : For all the children there is a father

kalia-randall
Download Presentation

More on Composite of Existential and Universal Quantifiers

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. More on Composite of Existential and Universal Quantifiers • (x) (y) <=> (y) (x) ? • Let’s look at this with an example: • Let x be father and y be children • (x) (y) : There exists a father for all children • (y) (x) : For all the children there is a father • These are clearly different sentences in that: • There is one father for all the children --- very unlikely • For all the children, there is a father. (Every child has a father. ---- much more reasonable! )

  2. View the two cases in tabular form Let F( x, y ) be the predicate that states x is the father of y. Note x is the bound variable and y is free; as you change y, the expression still holds true. X1 X2 X3 (x)F( x, y) T T y1 F F T F T y2 F V (y) Ξ(x)F( x, y) F T y3 F T Yes F F T T y4 V (y) F(x,y) F F F Ξ (x) V (y) F(x,y) No y is the bound and x is free; as you Iterate through the x’s, the expression Is never true

  3. Existential and Universal Quantifiers Basics • Note that (x) P(x) is the equivalent to saying: P(a1), P(a2), - - - - , P(an) is true for all the n instances of a’s. or (x) P(x)  P(a1) /\ P(a2)/\ - - - - /\ P(an) • Note that (x) P(x) is the same as saying: at least one of the following P(a1); P(a2); - - - -; P(an) is true or (x) P(x)  P(a1) \/ P(a2) \/ - - - - \/ P(an)

  4. Relationships Between Existential and Universal Quantifiers • From previous slide, we can have: ~ (x) P(x) = ~ ( P(a1) /\ P(a2) /\ - - - /\ P(an) ) = ~P(a1) \/ ~P(a2) \/ - - - \/ ~P(an) = (x) ~P(x) Thus we have the following: ~ (x) P(x)  (x)~P(x) Your try the: ~ (x) P(x)  (x) ~P(x)

  5. More Than One Predicate • Consider the following: (x) ( P(x) /\ Q(x) ) = [ (P(a1)/\Q(a1)) \/ (P(a2)/\Q(a2)) \/ - - - ] ~ (x)(P(x) /\ Q(x)) = ~ [ (P(a1)/\Q(a1)) \/ (P(a2)/\Q(a2)) \/ - - - ] = ~(P(a1)/\Q(a1)) /\ ~(P(a2)/\Q(a2)) /\ - - - - ] = (x) [ ~ ( P(x) /\ Q(x) ) ] = (x) (~P(x) \/ ~Q(x) ) So we have : ~ (x) (P(x)/\Q(x))  (x) (~P(x) \/ ~Q(x) )

  6. Relationship Between Existential and Universal Quantifiers • ~ (x) p(x) <=> (x) ~p(x) • ~ (x) p(x) <=> (x) ~p(x) • ~ (x) (p(x) /\ q(x)) <=> (x) (~p(x) \/ ~q(x) )

  7. More Relationships in Predicate Calculus • (x) P(x) -> (x) P(x) • (x) P(x) <-> ~ (x) ~P(x) • (x) P(x) <-> ~ (x) ~P(x) • (x) (y) P(x,y) <-> (y) (x) P(x,y) • (x) (y) P(x,y) <-> (y) (x) P(x,y) • (x) (y) P(x,y) -> (y) (x) P(x,y) Note: the one way arrow versus the two way arrow

  8. Sample Proof • V(x) P(x) -> Ξ (x) P(x) • P(x1) ------ P(xn) -> P(x1) V ------ V P(xn) • ~ [ P(x1) ----- P(xn)] V P(x1) V ------ V P(xn) • ~ P(x1) V ------V ~P(xn) V P(x1) V ----- VP(xn) • ~P(x1) V P(x1) V ---------------------V ~P(xn) V P(xn) • T V -------------------V T • True • A tautology

  9. Does “Spot” have a tail? • Now that we have gone through the fundamentals of predicate calculus, let’s try the earlier reasoning process: • Let p(x) indicate that x is a dog • Let q(y) indicate that y has a tail • Let spot be the specific dog in question. • (x) (p(x) -> q(x)) : every dog has a tail (premise 1) • p(spot) : spot is a dog (premise 2) • p(spot) -> q(spot) : universal instantiation with “spot”& premise 1 • p(spot), p(spot) -> q(spot) => q(spot) : using modus ponens rule • therefore q(spot) : spot has a tail !

  10. Additional Concepts • Singular existential quantifier • There exists only one object that has the desired characteristic • We represent that by placing a 1 subscript to the existential quantifier (x) P(x) • Counting quantifiers • Represents the number of objects in a group that has the specific characteristic • We will use Ω to represent counting quantifiers. • Example 1: Ω(x): 1,2, - - - ,10 x>6 will yield 4 • Example 2: Ω(x) : programs has_bug(x) stands for the number of programs that are defective 1

  11. Additional Concepts • Pre-condition and Post-condition • These were used when we covered code correctness • How should we articulate and what should we use to articulate the pre-condition and post-conditions of computing systems. • Variables • Global • Local • In the case of a variable, x, we may differentiate the post-condition by using an apostrophe,’, • X for the variable for pre-condition • X’ for the same variable for post-condition

  12. Converting English Phrases to Pre/Post Conditions in Predicate Calculus • Example from page 86 of your text: “ The procedure SunUpPost has two parameters. The first parameter is an integer array, Vals, which has a range of 1, - - -,10. The second parameter is a boolean variable, Outrange. This is set to true if any value of the Vals is less than or greater than 2000.” • Let the counting quantifier, omega, be represented by Ω, • Pre-conditions: • [Ω par: variables param_of(par, SunUpPost) ] = 2 • param_of(Vals, SunUpPost) /\ param_of( OutRange, SunupPost) • i: 1, - - -,10 Integer( Vals(i) ) • boolean(Outrange) • Post-conditions: • i : 1, - - - , 10 Vals’(i) = Vals(i) • ( i: 1, - - -, 10 Vals(i) ~= 2000 ) -> ( Outrange’ = True )

More Related