Re: It's a Bianary Operation 00000001 ?

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Re: It's a Bianary Operation 00000001 ?

Posted By: Neurotic <junkherethx@hotmail.com>
Date: 12 June 2001, 01:35

In Response To: Re: It's a Bianary Operation 00000001 ? (Shade)

> given two functions, "foo" and "bar"
> foo [ bar (read "foo is a square subset of bar") if and only if
> foo(x) = bar(x) for all defined values of foo(x).

Heh. Off topic I know...but when I read that I just thought of the term "FUBAR", which reminds me of these puzzles. =P

Messages in This Thread

for more GI info LOOK HERE!!!
Lara -- 11 June 2001, 22:54
- It's a Bianary Operation 00000001 ?
  Lara -- 11 June 2001, 23:01
  - Re: It's a Bianary Operation 00000001 ?
    Angus -- 11 June 2001, 23:25
    - Re: It's a Bianary Operation 00000001 ?
      Lara -- 11 June 2001, 23:42
    - Re: It's a Bianary Operation 00000001 ?
      Angus -- 11 June 2001, 23:45
      - Re: It's a Bianary Operation 00000001 ?
        Shade -- 11 June 2001, 23:57
        
        Re: It's a Bianary Operation 00000001 ?
        Lara -- 12 June 2001, 00:03
        
        Re: It's a Bianary Operation 00000001 ?
        novaHunter -- 12 June 2001, 00:20
        
        More Unicode
        novaHunter -- 12 June 2001, 00:29
        
        Re: It's a Bianary Operation 00000001 ?
        Shade -- 12 June 2001, 07:42
        
        Re: It's a Bianary Operation 00000001 ?
        Shade -- 12 June 2001, 07:45
        
        Re: It's a Bianary Operation 00000001 ?
        Sludge -- 12 June 2001, 00:58
        
        Re: It's a Bianary Operation 00000001 ?
        Neurotic -- 12 June 2001, 01:35
  - Re: It's a Bianary Operation 00000001 ?
    Lara -- 11 June 2001, 23:47

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