More on Composite of Existential and Universal Quantifiers

1
More on Composite of Existential and Universal
Quantifiers

• (x) (y) ltgt (y) (x) ?
• Lets 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) ? (x) F( x, y)
F
T
y3
T
F
Yes
F
T
F
T
y4
V (y) F(x,y)
F
F
F
? (x) V (y) F(x,y)
No
y is the bound and x is free as you Iterate
through the xs, 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 as.
• 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) ltgt (x) p(x)
• (x) p(x) ltgt (x) p(x)
• (x) (p(x) /\ q(x)) ltgt (x) (p(x) \/
  q(x) )

7
More Relationships in Predicate Calculus

• (x) P(x) -gt (x) P(x)
• (x) P(x) lt-gt (x) P(x)
• (x) P(x) lt-gt (x) P(x)
• (x) (y) P(x,y) lt-gt (y) (x)
  P(x,y)
• (x) (y) P(x,y) lt-gt (y) (x)
  P(x,y)
• (x) (y) P(x,y) -gt (y) (x)
  P(x,y)

Note the one way arrow versus the two way arrow

8
Sample Proof

• V(x) P(x) -gt ? (x) P(x)
• P(x1) ------ P(xn) -gt 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, lets 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) -gt q(x)) every dog has a
  tail (premise 1)
• p(spot) spot is a dog
  (premise 2)
• p(spot) -gt q(spot) universal
  instantiation with spot premise 1
• p(spot), p(spot) -gt q(spot) gt 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 O to represent counting quantifiers.
• Example 1 O(x) 1,2, - - - ,10 xgt6
  will yield 4
• Example 2 O(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 O,
• Pre-conditions
• O 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
  ) -gt ( Outrange True )