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Chapter 1, Section 1 TRUE FALSE F T T F T T F T T T T T T F T T T F

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Presentation on theme: "Chapter 1, Section 1 TRUE FALSE F T T F T T F T T T T T T F T T T F"β€” Presentation transcript:

1 Chapter 1, Section 1 TRUE FALSE F T T F T T F T T T T T T F T T T F
F F T F T T negations ops in ( ) final πΆβˆ¨π‘† 𝑆 β€² ∨𝐢 β€² 𝐢 β€² ∧ 𝑆 β€² 𝐢 β€² →𝑆 𝐡→𝑅 ≑ 𝐡 β€² βˆ¨π‘… 𝐡→𝑅 𝐡∧ 𝑅 β€² 𝐡 β€² ∧ 𝑅 β€² π΅β€²βˆ¨π‘…β€² 𝐡→𝑅 β€² ≑ 𝐡 β€² βˆ¨π‘… β€² β‰‘π΅βˆ§π‘…β€² 𝐡→𝐢 𝐡→𝐢 β€² ≑ 𝐡 β€² ∨𝐢 β€² β‰‘π΅βˆ§πΆβ€² 𝐡→𝐢′ π΅β€²βˆ§πΆ π΅βˆ§πΆβ€²

2 note: necessary β‰  sufficient If A is true then B must be true 𝐴→𝐡
𝐴∨𝐡 →𝐢 𝐢∧ 𝐴∨𝐡 β€² πΆβˆ§π΄β€²βˆ§π΅β€² (𝐴→𝐡)∧( 𝐡 β€² →𝐢) 𝐡→𝐴 note: necessary β‰  sufficient If A is true then B must be true 𝐴→𝐡 𝑃→(𝐢∧𝐡) P = project finished C = client happy B = bills paid L = lights out 𝐡′→𝐿 𝑃↔𝐿′ ( 𝐡 β€² ∧𝐿)→𝐢′ (𝐡↔𝑃)′→𝐿 𝐡↔(π‘ƒβˆ¨πΏ)

3 A B B->A A->(B->A)
tautology A B A^B B’ A’ B’v A’ * contradiction A B AvB’ A^B (A^B)’ * A B C AvB (AvB)^C’ A’vC * A B C A’ BvC’ *

4 Chapter 1 Section 3 TRUE TRUE FALSE TRUE A(x) is β€œx=0” B(x,y) is β€œx times y = 0” A(x) is β€œxβ‰ 0” B(x,y) is β€œx times y = 0” if A(x) is always true then B(x) is always true for each x for which A(x) then also B(x) X A(x) B(x) : : : A(x) is β€œx is a surgeon” B(x) is β€œx is a physician” A(x) is β€œx = 42” B(x) is β€œx <> 42” P(x) is β€œx is an integer” Q(x) is β€œx is a real number” P(x) is β€œx is a real number” Q( x) is β€œx is an integer” βˆ€π‘₯[𝐿 π‘₯ →𝑃 π‘₯ ] βˆƒπ‘₯[𝐿 π‘₯ βˆ§π‘… π‘₯ ] βˆ€π‘₯[𝐿 π‘₯ →𝑅 π‘₯ ] βˆƒπ‘₯[𝐿(π‘₯)βˆ§βˆ€π‘¦ 𝑍 𝑦 →𝐸 π‘₯,𝑦 ] βˆ€π‘₯βˆ€π‘¦[ 𝐿 π‘₯ βˆ§π‘ 𝑦 →𝐸 π‘₯,𝑦 ]

5 βˆ€π‘₯(¬𝑃 π‘₯ ) βˆƒπ‘₯(𝑃 π‘₯ ) βˆ€π‘₯ 𝐢 π‘₯ ∧𝐿 π‘₯ βˆƒπ‘₯ 𝐢′ π‘₯ βˆ¨πΏβ€² π‘₯ ((βˆƒπ‘₯ 𝐴 π‘₯ β€² β€² β‰‘βˆ€(π‘₯)[𝐴 π‘₯ ]β€²β€²β‰‘βˆ€(π‘₯)(𝐴 π‘₯ ) Let P(x) be false for all x Let Q(x) be true for some x βˆ€π‘₯(𝑃 π‘₯ →𝑄 π‘₯ ) Will always be true (vacuously) βˆƒπ‘₯(𝑃 π‘₯ βˆ¨π‘„ π‘₯ )β‰‘βˆƒπ‘₯(𝑄 π‘₯ ) Since P(x) is never true But then βˆƒπ‘₯(𝑃 π‘₯ βˆ§π‘„ π‘₯ ) Can never be true Let x be the domain of dogs Let y be the domain of cats Let P(x) be β€œx purrs” Let Q(x) be β€œx barks” In English: If all dogs purr or bark then either all dogs purr or there exists a cat that barks.

6 Are these system specifications consistent?
The router can send packets to the edge system only if it supports the new address space. For the router to support the new address space it is necessary that the latest software release be installed. The router can send packets to the edge system if the latest software release is installed. The router does not support the new address space R -> S S -> I I -> R ~S R = router can send packets to the edge system S = router support new address space I = latest SW release is installed R S I R->S S->I I->R ~S system is consistent

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