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#71 | |
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Curiosity demands that I examine all possiblities. My apologies for offending anyone, if I did. |
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#72 | |
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Sincerely, Goliath |
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#73 | |
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#74 | |
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So why did you delay for so long? Why did you try to fool me? Why the dishonesty? Sincerely, Goliath |
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#75 | |
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May you live long and prosper ![]() I could also ask you why you are being a "monkey on my back"? with an almost "religious" zeal? ![]() Will this thread also be banished? Metaphorically speaking, of course... |
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#76 | |||
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Sincerely, Goliath |
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#77 | |
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Join Date: Feb 2001
Location: Hiding from Julian ;)
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It's not like a mathematical argument where the numbers we are dealing with have certain intrinsic properties; with god, we can't even decide what his intrinsic properties are! And if the ones you decide to give god end up not working there's nothing stopping you from ditching them and pulling new ones out of your ass! It's like constantly redefining the concept of 'number' as something nobody but you accepts in order to prove that there is a largest integer. God is not logical. |
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#78 |
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Off to ~e~
xorbie EoG Mod |
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#79 |
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Join Date: Jan 2001
Location: Folding@Home in upstate NY
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OK, Chimp, let's break this down real simple like.
First, a lot of people here would reject outright your axiom #1. But for the sake of argument, let's give you your axiom and even #2, since it too, is an assumption. As Soralis pointed out, #3 is a contradiction, so using it anywhere else in the "argument" is meaningless. One of the early things they teach you in a class on logic is that contradictions can then be used to prove anything: If <contradiction> then <anything>. However, again from basic logic, if you have ((a-->b) /\ b), and try to make the conclusion 'b' TRUE, you cannot have 'a' FALSE. Thanks to Soralis for exposing #3 for the contradiction it is. I quickly felt it was definitely wrong, but didn't work it out for myself (at least not symbolically). |
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#80 |
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The Universal Set is the set of all that exists.
A set of all sets that exist can not exist in any self-consistent set theory. This is why: One of the axioms of the set theory is that for every set X there exists a set Y which is composed of all elements of X for which f is true where f is any statement. (Note that in a set theory every element of a set is a set itself.) Suppose that a set U such that for every set x: x is an element of U. (By definition, U is an element of U.) Statement f is "x is not an element of x" (i.e., of itself). This means that a set U' exist which is composed of all sets that are not an element of themselves. Is U' an element of itself? If so, then by definition of U' it is not an element of itself, and if not, then it is an element of itself. Since both options lead to a contradiction, the set U can not exist. (I could rewrite this proof in formal language, but there are not enough characters in the ASCII.) Mike Rosoft |
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