我都想知
原帖由 siteelstink 於 10-12-25 04:02 PM 發表
Let P(n):係,除非唔係
P(係,除非唔係)=
P(係)+P(唔係)=
P(係)+P(~係)
Let P(係)=x
P(係)+P(~係)= x+(1-x)= 1
Since this proposition is always true
Therefore P(n) is a tautology

原帖由 siteelstink 於 2010-12-25 16:02 發表
Let P(n):係,除非唔係
P(係,除非唔係)=
P(係)+P(唔係)=
P(係)+P(~係)
Let P(係)=x
P(係)+P(~係)= x+(1-x)= 1
Since this proposition is always true
Therefore P(n) is a tautology

原帖由 siteelstink 於 10-12-25 04:02 PM 發表
Let P(n):係,除非唔係
P(係,除非唔係)=
P(係)+P(唔係)=
P(係)+P(~係)
Let P(係)=x
P(係)+P(~係)= x+(1-x)= 1
Since this proposition is always true
Therefore P(n) is a tautology
原帖由 siteelstink 於 10-12-25 04:02 PM 發表
Let P(n):係,除非唔係
P(係,除非唔係)=
P(係)+P(唔係)=
P(係)+P(~係)
Let P(係)=x
P(係)+P(~係)= x+(1-x)= 1
Since this proposition is always true
Therefore P(n) is a tautology
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