Contra Equivalence?
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From: Jan Burse on 13 Aug 2010 16:46 Nam Nguyen wrote: > As far as FOL rules of inference is concerned, if all of > the following conditions hold: > > 1) {F1} |- ~F2 > 2) {F2} |- ~F1 > 3) F1 is NOT logically equivalent (reducible) to ~F2 > (and vice versa) > > then: > > q1) Would we consider F1 as a contradiction to F2, and > vice versa, observing condition 3) above? Well I would say contradiction is weaker. I would say it could expresses that F1 is the *negation* of F2. And this is much stronger than saying F1 contradicts F2. For example the following two formulas contradict each other: p & q (i) p & ~q (ii) The reason that (i) and (ii) contradict each other is that they are not in line in some common aspects. So if you take for example the conjunction of (i) and (ii) you get a contradictory formula, i.e. a formula which is everywhere false. (*) (**) But (ii) is not the negation of (i). I guess a formula that is the negation of another formula will be not in line in all aspects to the other formula. But in natural language you will very seldom find this pure form of negation. For example an antonym is not in general a pure negation. Best Regards (*) There should be a B, such that: (i) & (ii) |- B and (i) & (ii) |- ~B Well simplest is to take p, p follows from identity from (i) & (ii) and ~p follows also from identity from (i) & (ii). (**) Maybe we should start calling such a formula inconsistent.
From: MoeBlee on 13 Aug 2010 17:04 On Aug 13, 3:05 pm, Nam Nguyen <namducngu...(a)shaw.ca> wrote: First, I unnecessarily complicated this by misreading you. Second, I posted something I didn't mean. So it's better for me just to start over (and I'll mention the correction at the end, just for the record): Here's what we have: It is not the case that {GC} |- ~cGC. Rather, {GC} |-_Z ~cGC. It is not the case that {cBC} |- ~GC Rather, {cGC} |-_Z ~Gc. However, we do agree on 3): It is not the case that |- GC <-> ~cGC. We do have |-_Z GC -> ~cGC Whether we have |-_Z ~cGC -> GC, I don't know offhand, would have to think about it more. Now, I'm curious what significance you are seeking from this? Here are my self-corrections: (1) You wrote: > What I had in mind is F1 and F2 aren't variables but just 2 formula-constants. > In my latest post I had: > > >> An example of the situation above is where: > >> > >> F1 = GC (Goldbach Conjecture) > >> F2 = cGC ("There are infinitely many counter examples of GC"). (Aside from the terminology 'formula constants'), for me, what you just said is tantamount to saying 'F1' and 'F2' are meta-variables ranging over formulas and then you're instantiating them to particular formulas. And that's perfectly fine, and also it works perfectly with my point, where I instantiated 'F1' and 'F2' to a couple of other formulas I used to illustrate a particular matter. But I just realized I misread what you wrote, so even though, just for the sake of clarity, I would include my remarks about variables, I see that for what you're concerned with it doesn't matter that much. That is, I missed that you said "if" 1) - 3) hold. So this was a needless complication about meta-variables. I get what you mean anyway. (2) At this point at least, I didn't mean to say |-_Z GC <-> ~cGC. I'd have to think more about it to see whether |- GC -> ~cGC ? That is, in the contrapositive, in Z, if there is an even number that is not the sum of two primes then does it follow that there are infinitely many even numbers that are not the sum of two primes? Anyway, to reiterate, at least we know: It is not the case that {GC} |- ~cGC. Rather, {GC} |-_Z ~cGC. It is not the case that {cBC} |- ~GC Rather, {cGC} |-_Z ~Gc. However, we do agree on 3): It is not the case that |- GC <-> ~cGC. We do have |-_Z GC -> ~cGC MoeBlee
From: MoeBlee on 13 Aug 2010 17:12 On Aug 13, 3:46 pm, Jan Burse <janbu...(a)fastmail.fm> wrote: > For example the following two formulas contradict > each other: > > p & q (i) > p & ~q (ii) > > The reason that (i) and (ii) contradict each other is > that they are not in line in some common aspects. So if > you take for example the conjunction of (i) and (ii) you > get a contradictory formula, i.e. a formula which is > everywhere false. (*) (**) > > But (ii) is not the negation of (i). I agree. Maybe, though, some quibble about referring to 'false' when the matter doesn't have to involves semantics? Otherwise, yes, we agree on this distinction: 'P and Q are inconsistent' (i.e., 'P and Q are contradictory') versus 'P is the negation of Q (also 'Q is the negation of P'). .. MoeBlee
From: MoeBlee on 13 Aug 2010 17:18 On Aug 13, 4:12 pm, MoeBlee <jazzm...(a)hotmail.com> wrote: > we > agree on this distinction: 'P and Q are inconsistent' (i.e., 'P and Q > are contradictory') versus 'P is the negation of Q (also 'Q is the > negation of P'). Just to be clear, and to reiterate: Where 'P' and 'Q' are any formulas: If P is the negation of Q or Q is the negation of P, then P and Q are contradictory. But the converse doesn't always hold. I.e., from the mere fact that P and Q are contradictory we can't infer that either P is the negation of Q or Q is the negation of P. By the way, also, "P is the negation of Q" is not symmetrical. I.e., it is not the case that "P is the negation of Q" implies that "Q is the negation of P". Even stronger, if P is the negation of Q then Q is NOT the negation of P. MoeBlee
From: Nam Nguyen on 13 Aug 2010 18:59
MoeBlee wrote: > On Aug 13, 2:01 pm, Nam Nguyen <namducngu...(a)shaw.ca> wrote: > >> An example of the situation above is where: >> >> F1 = GC (Goldbach Conjecture) >> F2 = cGC ("There are infinitely many counter examples of GC"). > > Well, that's not in the pure predicate calculus (not in any language > using only pure predicate calculus for that language). (If I'm not > mistaken, it's not in first order PA either, since I don't know how > you'd formulate "infinitely many" in the language of PA.) So is this > is in some theory such as a set theory? Let's look at Z set theory. I think L(PA) would be sufficient. Iirc, I had a definition of "infinitely many" in one post of the past which would require the language to have a 2-ary predicate symbol <, but would still be general otherwise. It's something resembling the following: Assuming there's a defined unary predicate P(x), the statement "There are infinitely many examples of P" would be written as the formula (*)P(x): (*)P(x) <-> ( Ex[P(x)] /\ AxEy[P(x) -> (P(y) /\ (x < y))] ) So, cGC is just (*)P(x): cGC <-> (*)P(x), where P(x) <-> ~GC(x), and where: GC(x) <-> (even(x) -> ((x > 2) /\ Ep1p2[prime(p1) /\ prime(p2) /\ (x = p1+p2)] )) [and where even(x) and prime(x) are the familiar expressions defined in L(PA)]. I could overlook at things but I think we'd have: {GC} |- ~cGC {cGC} |- ~GC Also, the only 2 relevant formal systems are {GC} and {cGC}, NOT PA. *** On the issue of 2 FOL formulas being "logically equivalent (or reducible)", the context I intend is the following. If A1 and A2 are atomic formulas each with a different non-logical symbol, then though in an inconsistent T, from A1 we could infer A2 and vice versa, but we can not logically reduce say A1 to A2, the way to reduce say ~~A to A. Hope this has clarified what I'd like to say. (I'll be back later with more responses to you, JB, or perhaps others). > > 'F1' is, say, some forumulation of "for every n, if n is even then n > is the sum of two primes" > > 'F2' is some formulation of "{n | n is even & n is not the sum of two > primes} is infinite". > > Notice, I write '|-_Z' to indicate we're talking about the theory Z > not just the pure predicate calculus. > > 1) {F1} |-_Z ~F2 > > Yes. > > 2) {F2} |-_Z ~F1 > > Yes. > > 3) F1 is NOT logically equivalent (reducible) to ~F2 > > Yes. (For example, 'infinite' might not necessarily be interpreted in > the ordinary way, via 'e' itself not being interpreted as membership, > or all kinds of ways we could cook up.) > > So maybe what you're driving at this this?: > > F1 and ~F2 are not logically equivalent. Yes. > > But F1 and ~F2 are equivalent in Z. Yes. I.e., from some finite set of > axioms of Z along with ordinary definitions of the various non- > primitive terminology, we derive F1 <-> ~F2. > > This is a quite ordinary thing. I don't know what you're trying to > fathom about it. > > It is quite ordinary that we may have two formulas P and Q such that P > and Q are not logically equivalent but for certain theories T we have > |-_T P <-> Q. > > MoeBlee > -- ----------------------------------------------------------- I'm not a crank; I'm only just as difficult to reason with. NN ----------------------------------------------------------- |