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Unformatted text preview: Administration 1. Two Grading schemes. Higher of two. 2. Homework reconciliation day. Administration 1. Two Grading schemes. Higher of two. 2. Homework reconciliation day. Tuesday 123. December 6th. Administration 1. Two Grading schemes. Higher of two. 2. Homework reconciliation day. Tuesday 123. December 6th. Will have all old homeworks. Administration 1. Two Grading schemes. Higher of two. 2. Homework reconciliation day. Tuesday 123. December 6th. Will have all old homeworks. Location: TBA. Administration 1. Two Grading schemes. Higher of two. 2. Homework reconciliation day. Tuesday 123. December 6th. Will have all old homeworks. Location: TBA. waiting on room scheduling. Administration 1. Two Grading schemes. Higher of two. 2. Homework reconciliation day. Tuesday 123. December 6th. Will have all old homeworks. Location: TBA. waiting on room scheduling. 3. Watch Piazza for final review session! Extra office hours. Etc. CS70: Satish Rao: Lecture 38. I Undecidability. Barber paradox. Barber announces: Barber paradox. Barber announces: “The barber shaves every person who does not shave themselves.” Barber paradox. Barber announces: “The barber shaves every person who does not shave themselves.” Who shaves the barber? Barber paradox. Barber announces: “The barber shaves every person who does not shave themselves.” Who shaves the barber? Get around paradox? Barber paradox. Barber announces: “The barber shaves every person who does not shave themselves.” Who shaves the barber? Get around paradox? The barber lies. Russell’s Paradox. Naive Set Theory: Any definable collection is a set. Russell’s Paradox. Naive Set Theory: Any definable collection is a set. ∃ y ∀ x ( x ∈ y ⇐⇒ P ( x )) (1) Russell’s Paradox. Naive Set Theory: Any definable collection is a set. ∃ y ∀ x ( x ∈ y ⇐⇒ P ( x )) (1) y is the set of elements that satifies the proposition P ( x ) . Russell’s Paradox. Naive Set Theory: Any definable collection is a set. ∃ y ∀ x ( x ∈ y ⇐⇒ P ( x )) (1) y is the set of elements that satifies the proposition P ( x ) . P ( x ) = x 6∈ x . Russell’s Paradox. Naive Set Theory: Any definable collection is a set. ∃ y ∀ x ( x ∈ y ⇐⇒ P ( x )) (1) y is the set of elements that satifies the proposition P ( x ) . P ( x ) = x 6∈ x . There exists a y that satisfies statement 1 for P ( · ) . Russell’s Paradox. Naive Set Theory: Any definable collection is a set. ∃ y ∀ x ( x ∈ y ⇐⇒ P ( x )) (1) y is the set of elements that satifies the proposition P ( x ) . P ( x ) = x 6∈ x . There exists a y that satisfies statement 1 for P ( · ) . Take x = y . Russell’s Paradox. Naive Set Theory: Any definable collection is a set....
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This note was uploaded on 02/29/2012 for the course COMPSCI 70 taught by Professor Rau during the Fall '11 term at Berkeley.
 Fall '11
 Rau

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