CSE390D Notes for Friday, 9/27/24
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Predicate logic is like methods that take arguments and return a boolean result
For example, given:
odd(x) means "x is odd"
even(x) means "x is even"
We can use predicates with constants:
odd(5) ^ ~even(22)
odd(4) -> even(19)
More often we use variables:
odd(x) -> ~even(x)
Variables need to be bound to make a proposition. So bind them to constants:
odd(x) -> ~even(x) for x = 17
or bind them with quantifiers:
(all x) (odd(x) -> ~even(x))
Is this a true statement? Depends. What is the domain? Does it include
things other than integers? Can people be odd? Might have to say something
like:
(all x) (integer(x) ^ odd(x) -> ~even(x))
Two different quantifiers:
universal: (all x), "for all x"
existential: (exists x), "there exists an x"
What about negation? More DeMorgan's Laws:
~(all x) P(x) <--> (exists x) ~P(x)
~(exists x) P(x) <--> (all x) ~P(x)
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Programming analogy for quantifiers:
Is it true that "All unicorns are President of the United States?"
need domain: all living things on Earth
answer? maybe...not clear
Think of "(all x) P(x)" as boolean method with this pseudocode:
for (d : domain) {
if (!P(d))
return false;
}
return true;
Then unicorn answer is clear: true (vacuously true)
Think of "(exists x) P(x)" as boolean method with this pseudocode:
for (d : domain) {
if (P(d))
return true;
}
return false;
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Aristotelian forms:
All P's are Q's: All men are mortal.
(all x) (P(x) -> Q(x))
~(exists x) (P(x) ^ ~Q(x))
bad: (all x) (P(x) ^ Q(x))
Some P's are Q's: Some cars are green.
(exists x) (P(x) ^ Q(x))
~(all x) (P(x) -> ~Q(x))
bad: (exists x) (P(x) -> Q(x))
No P's are Q's: No strawberries are blue.
(all x) (P(x) -> ~Q(x))
~(exists x) (P(x) ^ Q(x))
Some P's are not Q's: Some cars are not green.
(exists x) (P(x) ^ ~Q(x))
~(all x) (P(x) -> Q(x))
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Practice:
P(x): x is a professor
I(x): x is ignorant
V(x): x is vain
No professors are ignorant (all x) (P(x) -> ~I(x))
All ignorant people are vain (all x) (I(x) -> V(x))
No professors are vain (all x) (P(x) -> ~V(x))
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S(x): x is a student in this class
C(x): x has visited Canada
M(x): x has visited Mexico
Some student in this class has visited Mexico:
(exists x) (S(x) ^ M(x))
Every student in the class has visited Canada or Mexico
(all x) (S(x) -> (C(x) v M(x)))
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Stuart Reges
Last modified: Mon Sep 30 10:58:21 PDT 2024