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From: Lester Zick on 24 Apr 2007 18:12 On Tue, 24 Apr 2007 09:27:05 -0400, Tony Orlow <tony(a)lightlink.com> wrote: >>> If I say 3x+3=3(x+1) is that true? >>> Yes, it's true for all x. >> >> How about for x=3/0? >> > >Division by pure 0 is proscribed because it produces an unmeasurable oo. >If x is any specific real, or hyperreal, or infinitesimal, then that >statement is true for all x. 3/0 is not a specific number. In other words the statement isn't true for all x. ~v~~
From: Lester Zick on 24 Apr 2007 18:13 On Tue, 24 Apr 2007 09:27:05 -0400, Tony Orlow <tony(a)lightlink.com> wrote: >>> If I say a=not b, is that true? Not if a and b >>> are both true. >> >> How do you arrive at that assumption, Tony? If a and b are both true >> of the same thing they can still be different from each other. > >You are confusing logical "not" with non-identity. "a=not(b)" means that >if b is true, a is false, and vice versa. You are talking about >"not(a=b)", which means simply that there is some difference between a >and b, but not that there is a contradiction between the two. > >If I say, "A woman cannot be an engineer", then >woman(x)=not(engineer(x)) and engineer(x)=not(woman(x)). If there IS a >woman who is an engineer, then woman(x) and engineer(x), and >"engineer(x)=not(woman(x))" is false. > >If I say, "A woman is not an engineer", that does not have any bearing >on whether the same being cannot at both times be a woman and an >engineer. That's not a matter of logical contradiction, but of some set >of distinctions between the two classes of objects. No clue as to what you're talking about here. ~v~~
From: Lester Zick on 24 Apr 2007 18:15 On Tue, 24 Apr 2007 09:27:05 -0400, Tony Orlow <tony(a)lightlink.com> wrote: >>> If I say a or not a, that's true for all a. a and b are >>> variables, which may each assume the value true or false. >> >> Except you don't assign them the value true or false; you assign them >> the value 1 or 0 and don't bother to demonstrate the "truth" of either >> 1 or 0. >> > >1 is true, 0 is false. If a is 0 or 1, then we have "0 or 1", or "1 or >0", respectively. Since or(a,b) is true whenever a is true or b is true, >or both, or(1,0) and or(0,1), the only possible values for the >statement, are both true. So, or(a,not(a)) is always true, in boolean >logic, or probability. > >Intuitively, if a is a subset of the universe, and not(a) is everything >else, then the sum of a and not(a) is very simply the universe, which is >true. Yeah but you still haven't proven that 1 is true and 0 false or what either of these terms has to mean in mechanically exhaustive terms. ~v~~
From: Lester Zick on 24 Apr 2007 18:16 On Tue, 24 Apr 2007 09:27:05 -0400, Tony Orlow <tony(a)lightlink.com> wrote: >>> If you want to talk about the truth values of individual facts used in >>> deduction, by all means, go for it. >> >> I don't; I never did. All I ever asked was how people who assume the >> truth of their assumptions compute the truth value of the assumptions. >> >> ~v~~ > >By measuring the logical implications of their assumptions. Or perhaps by measuring the truth of their assumptions instead. ~v~~
From: Lester Zick on 24 Apr 2007 18:23
On Tue, 24 Apr 2007 22:54:15 +0100, Ben newsam <ben.newsam.remove.this(a)gmail.com> wrote: >On Tue, 24 Apr 2007 11:31:02 -0700, Lester Zick ><dontbother(a)nowhere.net> wrote: > >>On Tue, 24 Apr 2007 00:43:24 +0100, Ben newsam >><ben.newsam.remove.this(a)gmail.com> wrote: >> >>>On Mon, 23 Apr 2007 11:54:05 -0400, Tony Orlow <tony(a)lightlink.com> >>>wrote: >>> >>>>If you want to talk about the truth values of individual facts used in >>>>deduction, by all means, go for it. >>> >>>I would counsel you seriously not to attempt it, Lester will only >>>obfuscate the "discussion", and then will start to hurl personal >>>insults at you. >> >>Or you could bore us to death with the experimental confirmation of >>spatial dimentionality. Is there nothing you can contribute to the >>advancement of science besides gratuitous meretricious ad hominem >>attacks on moi? > >You don't know what "meretricious" means, do you? No but I didn't figure you did either. ~v~~ |