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logic和boolean algebra 的题目

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发表于 1-3-2012 11:05 PM | 显示全部楼层 |阅读模式
1 Consider the following argument.
If Abu likes to drive to work or his father’s car is old, then he will buy a new car.
Abu does not buy a new car or he takes a train to work.
Abu did not take a train to work.
Therefore, Abu does not like to drive to work.
(a) Rewrite the argument using statement variables and connectives. [2 marks]
(b) Test the argument for validity. [5 marks]

2 Let set B with binary operations ∨ and ∧ be a Boolean algebra. Show that, for all x, y and z in B,
(X'∧Y'∧Z')∨(X'∧Y'∧Z)∨(X∧Y'∧Z)≡Y' [5 marks]
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发表于 2-3-2012 01:58 AM | 显示全部楼层
1 Consider the following argument.
If Abu likes to drive to work or his father’s car is old, then  ...
oceanheng 发表于 1-3-2012 11:05 PM



    (1)(a) p=Abu likes to drive to work

          q=Abu father’s car is old

           r=Abu will buy a new car

           s=Abu takes a train to work


          (p v q) → r

          ~r v s

          ~s

          ---------------

     ∴ ~p


     (b)用rule of inference, modus tollens, law of syllogism, simplification 就可以prove到了


(2)查一下题目有没有错误。

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发表于 2-3-2012 02:12 AM | 显示全部楼层
2 Let set B with binary operations ∨ and ∧ be a Boolean algebra. Show that, for all x, y and z in B,
(X'∧Y'∧Z')∨(X'∧Y'∧Z)∨(X∧Y'∧Z)≡Y' [5 marks]
oceanheng 发表于 1-3-2012 11:05 PM



    题目是不是少了这个?

(X'∧Y'∧Z')∨(X'∧Y'∧Z)∨(X∧Y'∧Z')∨(X∧Y'∧Z)≡Y'
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 楼主| 发表于 2-3-2012 08:33 PM | 显示全部楼层
回复 3# Allmaths
是是~copy error
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发表于 2-3-2012 11:02 PM | 显示全部楼层
回复  Allmaths
是是~copy error
oceanheng 发表于 2-3-2012 08:33 PM



    (X'∧Y'∧Z')∨(X'∧Y'∧Z)(X∧Y'∧Z') ∨(X∧Y'∧Z)
≡[(X'∧Y')∧(Z'∨Z)](X∧Y'∧Z') ∨(X∧Y'∧Z)
≡[(X'∧Y')∧E](X∧Y'∧Z') ∨(X∧Y'∧Z)
≡(X'∧Y')(X∧Y'∧Z') ∨(X∧Y'∧Z)
≡(X'∧Y')[(X∧Y')∧(Z'∨Z)]
≡(X'∧Y')[(X∧Y')∧E]
≡(X'∧Y')(X∧Y')
≡Y'∧(XX')
≡Y'∧E
≡Y'
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 楼主| 发表于 22-3-2012 09:19 PM | 显示全部楼层
回复 5# Allmaths


   proove 的是这样对吗?~r v s
~s
------------------
∴~r


(p v q) --> r
~ r
---------------------
∴~(p v q)


~p ^ ~q
∴~p
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