Re: Question: What conditions are sufficient to prove a subset does not exist?
- From: MoeBlee <jazzmobe@xxxxxxxxxxx>
- Date: Thu, 02 Aug 2007 16:02:59 -0700
On Aug 2, 2:32 pm, george <gree...@xxxxxxxxxx> wrote:
On Aug 2, 5:18 pm, MoeBlee <jazzm...@xxxxxxxxxxx> wrote:
On Aug 2, 1:00 pm, Scott <ToaTe...@xxxxxxxxx> wrote:
the following would still be true?
You had:
(1) A <-> f
(2) ~A
(3) ~A (again)
(4) ~f
A -> ~f (2,4)
Right.
Oops, I overlooked the annotation "(2, 4)" which need only be "(2)" or
"(4)".
A -> ~f follows from 2 alone or from 4 alone.
This is NOT right.
If you are going to insist on (2)~A, then
ANY conditional with A in front is true;
(4) is not even relevant. RIGHT would be
A -> ~f (2),
IF EITHER OF A OR f WERE A WFF or sentence.
Either (2) alone or (4) alone justifies inferring A -> ~f.
And I'm supposing, for the sake of the mere question of the sentential
logic, that 'A' and 'f' stand for well formed formulas.
SINCE NEITHER IS (A is a set and f is a function, and
if the goal was to avoid having to repeat "exists" all the time,
well, guess what: this is set theory: EVERYthing is a set
wherefore EVERYthing exists).
I'm taking "A" and "f" to stand for formulas of some kind such as, but
not necessarily, "a set having properties such and such exists" and "a
function having properties so and so exists". Be assured I do not
endorse such things as "Ex x" in ordinary predicate logic; also that I
think begninners invite confusion when they say things like "A does
not exist" (or even "A exists") since 'A', whatever it denotes, does
denote some member of the universe of discourse of whatever
interpretation of the language, and so what should really be said is
"There does not exist an object having such and such properties" or
"There does exist an object having such and such properties".
~f -> ~A (1)
Right.
Again, not really.
RIGHT would be
~f --> ~A (2),
Actually ~f -> ~ A does follow from A <-> f. There the annotation is
okay. It also follows from (2) alone also, but saying it follows from
(1) alone is not incorrect.
since if you have ~A, ANYconditional with ~A on the right is true.
Again (1) is NOT relevant to this conclusion!
The rest of yours was correct but my point is there is no point
in arguing correct reasoning and logic with someone WHO
DOESN'T EVEN UNDERSTAND THE BASIC COMPONENTS
of the paradigm!
I don't disagree that the poster seems to be confused at a very basic
level. I was just commenting on the particular matter of the
sentential logic. I leave the pedagogical wisdom to you; I'm not
claiming anything about how best to educate the poster. I just
commented on the sentential logic, and with the tacit assumption that
the letters stand for formulas.
MoeBlee
.
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