'When Are Relations Neither True Nor False?
in [Math]
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From: Jesse F. Hughes on 2 Mar 2010 09:30 Newberry <newberryxy(a)gmail.com> writes: > On Feb 28, 3:54 pm, Aatu Koskensilta <aatu.koskensi...(a)uta.fi> wrote: >> Newberry <newberr...(a)gmail.com> writes: >> > b) Is this >> >> > (x)((2 > x > 4) -> ~(x < x + 1)) >> >> > "ordinary mathematical reasoning"? >> >> It isn't reasoning at all. It's a formula that doesn't formalize any >> statement we would ordinarily meet in mathematical reasoning. > > So why do you exhort people that it is true? It's not that sentence that we're interested in, but there are similar sentences that *are* used in mathematical reasoning. Suppose, for instance, that we have a proof (Ex)Px -> (Ex)(Px & Qx). For instance, let Px stand for (E a,b,c)(a^x + b^x = c^x) and Qx stand for x is prime. Then we know that the above sentence is true: if there is a counterexample to Fermat's last theorem, then there is a counterexample in which the exponent is prime. Equivalently, ~(Ex)(Px & Qx) -> ~(Ex)Px. (1) Suppose further that I prove (according to the usual notion of mathematical proof) that ~(Ex)(Px & Qx). (2) Then, (1) and (2) allow me to conclude ~(Ex)Px. (3) Ah, but here's the rub! According to your comments, if (3) is true, then (2) is neither true nor false and hence not true. If (2) is not true, then surely there is something wrong with my proof of (3), since my proof of (3) depends on (2). I thought I had a proof of (2), but it turns out that I didn't. At this point, you may say that what I really had was a proof of a second-order statement ~T((Ex)(Px & Qx)), (2') and that I can somehow use (2') and (1) to derive (3), but you are only blowing steam at present, since you haven't given us any second-order logic to conclude that (1) & (2') |- (3). Statements like (2) are useful in mathematical reasoning. We use them to derive statements like (3). -- "I told her that I loved her. She said she loved me too. Neither one was lying, Yet it wasn't true." -- Del McCoury Band
From: Marshall on 2 Mar 2010 10:11 On Mar 2, 6:30 am, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: > Newberry <newberr...(a)gmail.com> writes: > > > At this point, you may say that what I really had was a proof of a > second-order statement > > ~T((Ex)(Px & Qx)), (2') > > and that I can somehow use (2') and (1) to derive (3), but you are > only blowing steam at present, since you haven't given us any > second-order logic to conclude that (1) & (2') |- (3). > > Statements like (2) are useful in mathematical reasoning. We use them > to derive statements like (3). One thing that the cranks and crankophiles never understand is that the systems they come up with add a lot of complexity while actually removing functionality or utility. To do so merely to avoid some counterintuitive but harmless property (such as vacuous truth, in Newberry's case) is a huge waste of time. Less powerful; more work to use: that's a crank theory for you. Marshall PS. To the extent that you can call them "theories." PPS. This post applies equally well in the current Easterly thread.
From: Newberry on 2 Mar 2010 10:31 On Mar 2, 6:30 am, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: > Newberry <newberr...(a)gmail.com> writes: > > On Feb 28, 3:54 pm, Aatu Koskensilta <aatu.koskensi...(a)uta.fi> wrote: > >> Newberry <newberr...(a)gmail.com> writes: > >> > b) Is this > > >> > (x)((2 > x > 4) -> ~(x < x + 1)) > > >> > "ordinary mathematical reasoning"? > > >> It isn't reasoning at all. It's a formula that doesn't formalize any > >> statement we would ordinarily meet in mathematical reasoning. > > > So why do you exhort people that it is true? > > It's not that sentence that we're interested in, but there are similar > sentences that *are* used in mathematical reasoning. > > Suppose, for instance, that we have a proof > > (Ex)Px -> (Ex)(Px & Qx). > > For instance, let Px stand for (E a,b,c)(a^x + b^x = c^x) and Qx stand > for x is prime. Then we know that the above sentence is true: if > there is a counterexample to Fermat's last theorem, then there is a > counterexample in which the exponent is prime. > > Equivalently, > > ~(Ex)(Px & Qx) -> ~(Ex)Px. (1) > > Suppose further that I prove (according to the usual notion of > mathematical proof) that > > ~(Ex)(Px & Qx). (2) > > Then, (1) and (2) allow me to conclude > > ~(Ex)Px. (3) > > Ah, but here's the rub! According to your comments, if (3) is true, > then (2) is neither true nor false and hence not true. If (2) is not > true, then surely there is something wrong with my proof of (3), since > my proof of (3) depends on (2). I thought I had a proof of (2), but > it turns out that I didn't. > > At this point, you may say that what I really had was a proof of a > second-order statement > > ~T((Ex)(Px & Qx)), (2') > > and that I can somehow use (2') and (1) to derive (3), but you are > only blowing steam at present, since you haven't given us any > second-order logic to conclude that (1) & (2') |- (3). > > Statements like (2) are useful in mathematical reasoning. We use them > to derive statements like (3). > > -- > "I told her that I loved her. > She said she loved me too. > Neither one was lying, > Yet it wasn't true." -- Del McCoury Band Was this selected at random?
From: Jesse F. Hughes on 2 Mar 2010 13:10 Newberry <newberryxy(a)gmail.com> writes: > On Mar 2, 6:30 am, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: >> Newberry <newberr...(a)gmail.com> writes: >> > On Feb 28, 3:54 pm, Aatu Koskensilta <aatu.koskensi...(a)uta.fi> wrote: >> >> Newberry <newberr...(a)gmail.com> writes: >> >> > b) Is this >> >> >> > (x)((2 > x > 4) -> ~(x < x + 1)) >> >> >> > "ordinary mathematical reasoning"? >> >> >> It isn't reasoning at all. It's a formula that doesn't formalize any >> >> statement we would ordinarily meet in mathematical reasoning. >> >> > So why do you exhort people that it is true? >> >> It's not that sentence that we're interested in, but there are similar >> sentences that *are* used in mathematical reasoning. >> >> Suppose, for instance, that we have a proof >> >> (Ex)Px -> (Ex)(Px & Qx). >> >> For instance, let Px stand for (E a,b,c)(a^x + b^x = c^x) and Qx stand >> for x is prime. Then we know that the above sentence is true: if >> there is a counterexample to Fermat's last theorem, then there is a >> counterexample in which the exponent is prime. >> >> Equivalently, >> >> ~(Ex)(Px & Qx) -> ~(Ex)Px. (1) >> >> Suppose further that I prove (according to the usual notion of >> mathematical proof) that >> >> ~(Ex)(Px & Qx). (2) >> >> Then, (1) and (2) allow me to conclude >> >> ~(Ex)Px. (3) >> >> Ah, but here's the rub! According to your comments, if (3) is true, >> then (2) is neither true nor false and hence not true. If (2) is not >> true, then surely there is something wrong with my proof of (3), since >> my proof of (3) depends on (2). I thought I had a proof of (2), but >> it turns out that I didn't. >> >> At this point, you may say that what I really had was a proof of a >> second-order statement >> >> ~T((Ex)(Px & Qx)), (2') >> >> and that I can somehow use (2') and (1) to derive (3), but you are >> only blowing steam at present, since you haven't given us any >> second-order logic to conclude that (1) & (2') |- (3). >> >> Statements like (2) are useful in mathematical reasoning. We use them >> to derive statements like (3). >> >> -- >> "I told her that I loved her. >> She said she loved me too. >> Neither one was lying, >> Yet it wasn't true." -- Del McCoury Band > > Was this selected at random? Yes. Just another coincidence from the random number generator. -- Jesse F. Hughes "Students said they wanted to make people feel more comfortable by not having to choose a gender at the bathroom door." -- Boston Globe article on gender-neutral bathrooms at universities.
From: Jesse F. Hughes on 2 Mar 2010 13:15
Newberry <newberryxy(a)gmail.com> writes: > On Mar 2, 6:30 am, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: >> Suppose, for instance, that we have a proof >> >> (Ex)Px -> (Ex)(Px & Qx). >> >> For instance, let Px stand for (E a,b,c)(a^x + b^x = c^x) and Qx stand >> for x is prime. Then we know that the above sentence is true: if >> there is a counterexample to Fermat's last theorem, then there is a >> counterexample in which the exponent is prime. >> >> Equivalently, >> >> ~(Ex)(Px & Qx) -> ~(Ex)Px. (1) >> >> Suppose further that I prove (according to the usual notion of >> mathematical proof) that >> >> ~(Ex)(Px & Qx). (2) >> >> Then, (1) and (2) allow me to conclude >> >> ~(Ex)Px. (3) >> >> Ah, but here's the rub! According to your comments, if (3) is true, >> then (2) is neither true nor false and hence not true. If (2) is not >> true, then surely there is something wrong with my proof of (3), since >> my proof of (3) depends on (2). I thought I had a proof of (2), but >> it turns out that I didn't. >> >> At this point, you may say that what I really had was a proof of a >> second-order statement >> >> ~T((Ex)(Px & Qx)), (2') >> >> and that I can somehow use (2') and (1) to derive (3), but you are >> only blowing steam at present, since you haven't given us any >> second-order logic to conclude that (1) & (2') |- (3). >> >> Statements like (2) are useful in mathematical reasoning. We use them >> to derive statements like (3) > > Obviously if we do not use clasical logic the derivation system will > be different. If we have a semantically complete system and > > ~(Ex)Px. (3) > > is true then we will be able to prove it. Your example is useful when > we use classical logic. If we do not then it will not be useful. It is > that simple. I just don't see any reason to give up a perfectly straightforward and sensible argument (like that above) in order to fit your intuitions that, if (3) is true, then (2) is neither true nor false. Perhaps t-relevance logic is useful, but it does not seem to be satisfactory for mathematical reasoning. In mathematics, the argument I outlined above is a perfectly good argument for (3). Seems to me that you've got quite a job convincing mathematicians otherwise. -- Jesse F. Hughes "I'm a geek hatchling." -- Quincy P. Hughes, age 7 |