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During the study of discrete mathematics, I found this course very informative and applicable.The main points in these lecture slides are:Propositional Function, Nested Quantifiers, Rules of Inference, Universe of Discourse, Quantified Predicates, Quantified Variables, Existential Quantifier, Universal Quantifier, Predicate Logic and Propositions
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Nested quantifiers Rules of inference Proofs
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An expression with zero free variables is an actual proposition
Ex. Q(x) : x > 0, R(y): y < 10
∃ x Q(x) ∧ ∃y R(y)
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When dealing with polyadic predicates, each argument may be quantified with its own quantifier. Each nested quantifier occurs in the scope of another quantifier.
Examples: (L=likes, UoD(x)=kids, UoD(y)=cars) ∀x∀y L(x,y) reads ∀x(∀y L(x,y)) ∀x∃y L(x,y) reads ∀x(∃y L(x,y)) ∃x∀y L(x,y) reads ∃x(∀y L(x,y)) ∃x∃y L(x,y) reads ∃x(∃y L(x,y)) Another example ∀x (P(x) ∨ ∃y R(x,y))
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To negate a quantifier, move negation to the right, changing quantifiers as you go. Example:
¬∀x∃y∀z P(x,y,z) ≡∃x^ ∀y^ ∃z^ ¬P(x,y,z).
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Assume that the following statements are true:
I have a total score over 96. If I have a total score over 96, then I get an A in the class.
What can we claim? I get an A in the class.
How do we know the claim is true? Logical Deduction.
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“∴” means “therefore”
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I have a total score over 96.
If I have a total score over 96, then I get an A for the class.
∴ I get an A for this class
p
p → q
∴ q
Tautology:
(p ∧ (p → q)) → q
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∴ I am a student or I am a visitor.
p
∴ p ∨ q
Tautology:
p → (p ∨ q)
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∴ I am a student.
p ∧ q
∴ p
Tautology:
(p ∧ q) → p
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I am a student or I am a soccer player. I am a not soccer player.
∴ I am a student.
p ∨ q ¬q
∴ p
Tautology:
((p ∨ q) ∧ ¬q) → p
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If I get a total score over 96, I will get an A in the course. If I get an A in the course, I will have a 4.0 semester average.
∴ If I get a total score over 96 then
p → q q → r
∴ p → r
Tautology: ((p → q) ∧ (q → r)) → (p → r)
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I have taken CS2610 or I have taken CS1301. If I have taken CS2610 then I can register for CS If I have taken CS1301 then I can register for CS
∴ I can register for CS
p ∨ q p → r q → r
∴ r
Tautology: ((p ∨ q ) ∧ (p → r) ∧ (q → r)) → r
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∴ You must have the flu.
Fallacy:
(q ∧ (p → q)) → p
q
p → q
∴ p
Abductive, rather than deductive reasoning!