Contra Equivalence?
in [Logic]
|
Prev: EINSTEINIANS AS MARAUDERS
Next: More Common Error? Confounding 2 Fundamentally Different Things? OrNot Lumping Together What Is Only Superficially Different?
From: Nam Nguyen on 13 Aug 2010 13:55 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? q2) If the answer for q1 is "No", would the term "contra- equivalence" be adequate to describe the relationship between F1 and F2 (as adequate as "contradiction" between F1 and ~F1, or between F2 and ~F2)? q3) What would be the significance about FOL proofs, via rules of inference, of a situation where conditions 1), 2), 3) hold, and where consistency of the underlying T is assumed? Thanks. -- ----------------------------------------------------------- Normally, we do not so much look at things as overlook them. Zen Quotes by Alan Watt -----------------------------------------------------------
From: Nam Nguyen on 13 Aug 2010 15:01 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? > > q2) If the answer for q1 is "No", would the term "contra- > equivalence" be adequate to describe the relationship > between F1 and F2 (as adequate as "contradiction" > between F1 and ~F1, or between F2 and ~F2)? > > q3) What would be the significance about FOL proofs, via > rules of inference, of a situation where conditions 1), > 2), 3) hold, and where consistency of the underlying T > is assumed? An example of the situation above is where: F1 = GC (Goldbach Conjecture) F2 = cGC ("There are infinitely many counter examples of GC"). -- ----------------------------------------------------------- I'm not a crank; I'm only just as difficult to reason with. NN -----------------------------------------------------------
From: MoeBlee on 13 Aug 2010 15:47 On Aug 13, 12:55 pm, Nam Nguyen <namducngu...(a)shaw.ca> 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) It depends on what specific formulas F1 and F2 are. (Unless you tell me otherwise, I'm taking that 'F1' and 'F2' are meta-variables ranging over formulas.) For example let F1 be the formula ~P and F2 be the formula P. Then F1 and ~F2 are logically equivalent. True, we cannot prove that IN GENERAL F1 and ~F2 are logically equivalent. But that doesn't preclude that there are certain INSTANCES in which F1 and ~F2 are logically equivalent. > then: > > q1) Would we consider F1 as a contradiction to F2, and > vice versa, observing condition 3) above? We can't conclude 3). But F1 and F2 are contradictory anyway. Unless you stipulate otherwise, I'm taking the definition of contradictory as: P and Q are contradictory iff there is formula R such that {P Q} |- R and {P Q} |- ~R. I.e., P and Q are contradictory iff {P Q} is inconsistent. If by 'contradictory' you mean something different, then you'd need to say what it is. > q2) If the answer for q1 is "No", would the term "contra- > equivalence" be adequate to describe the relationship > between F1 and F2 I've haven't happened to have seen the term 'contra-equivalence' before; so I don't see any reason you can't define it however you like. But I don't know what purpose the added term serves. > (as adequate as "contradiction" > between F1 and ~F1, or between F2 and ~F2)? If 'contra-equivalent' means the same as 'contradictory' then I suppose it's just as adequate. But when you mention F1 and ~F1, those are not JUST contradictory, but ALSO one is the NEGATION of the other. A formula P is the negation of a formula P iff Q is the actual formula ~P. Two formulas can be contradictory though one is not the negation of the other. For example: ~~~Q and Q. Another example: P and ~(P & P). Another example: P and Q & ~Q. > q3) What would be the significance about FOL proofs, via > rules of inference, of a situation where conditions 1), > 2), 3) hold, and where consistency of the underlying T > is assumed? I take it that 1), 2) , and 3) are in the pure predicate calculus. I.e., that '|-' refers to derivability with no other non-logical axioms other than those specifically mentioned on the left side of '|-'. If some other theory T is involved, then the notation might be: |-_T to indicate that derivability allows also any finite set of theorems of T to be on the left side of '|-'. Beyond that, I don't know what you're asking really. What kind of "significance" do you have in mind? Of course, there are formulas F1 and F2 such that 1) - 3) hold. Meanwhile, some theory or another might or might not be consistent. I don't see what connection you're driving at or what significance you're wondering about. MoeBlee
From: Nam Nguyen on 13 Aug 2010 16:05 MoeBlee wrote: > On Aug 13, 12:55 pm, Nam Nguyen <namducngu...(a)shaw.ca> 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) > > It depends on what specific formulas F1 and F2 are. (Unless you tell > me otherwise, I'm taking that 'F1' and 'F2' are meta-variables ranging > over formulas.) 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"). [Any rate, I have to be away from keyboard but will make a response later.] > > For example let F1 be the formula ~P and F2 be the formula P. Then F1 > and ~F2 are logically equivalent. > > True, we cannot prove that IN GENERAL F1 and ~F2 are logically > equivalent. But that doesn't preclude that there are certain INSTANCES > in which F1 and ~F2 are logically equivalent. > >> then: >> >> q1) Would we consider F1 as a contradiction to F2, and >> vice versa, observing condition 3) above? > > We can't conclude 3). But F1 and F2 are contradictory anyway. > > Unless you stipulate otherwise, I'm taking the definition of > contradictory as: > > P and Q are contradictory iff there is formula R such that {P Q} |- R > and {P Q} |- ~R. > > I.e., P and Q are contradictory iff {P Q} is inconsistent. > > If by 'contradictory' you mean something different, then you'd need to > say what it is. > >> q2) If the answer for q1 is "No", would the term "contra- >> equivalence" be adequate to describe the relationship >> between F1 and F2 > > I've haven't happened to have seen the term 'contra-equivalence' > before; so I don't see any reason you can't define it however you > like. But I don't know what purpose the added term serves. > >> (as adequate as "contradiction" >> between F1 and ~F1, or between F2 and ~F2)? > > If 'contra-equivalent' means the same as 'contradictory' then I > suppose it's just as adequate. > > But when you mention F1 and ~F1, those are not JUST contradictory, but > ALSO one is the NEGATION of the other. > > A formula P is the negation of a formula P iff Q is the actual formula > ~P. > > Two formulas can be contradictory though one is not the negation of > the other. For example: ~~~Q and Q. Another example: P and ~(P & P). > Another example: P and Q & ~Q. > >> q3) What would be the significance about FOL proofs, via >> rules of inference, of a situation where conditions 1), >> 2), 3) hold, and where consistency of the underlying T >> is assumed? > > I take it that 1), 2) , and 3) are in the pure predicate calculus. > I.e., that '|-' refers to derivability with no other non-logical > axioms other than those specifically mentioned on the left side of > '|-'. > > If some other theory T is involved, then the notation might be: > > |-_T > > to indicate that derivability allows also any finite set of theorems > of T to be on the left side of > '|-'. > > Beyond that, I don't know what you're asking really. What kind of > "significance" do you have in mind? > > Of course, there are formulas F1 and F2 such that 1) - 3) hold. > Meanwhile, some theory or another might or might not be consistent. I > don't see what connection you're driving at or what significance > you're wondering about. > > MoeBlee -- ----------------------------------------------------------- I'm not a crank; I'm only just as difficult to reason with. NN -----------------------------------------------------------
From: MoeBlee on 13 Aug 2010 16:07
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. '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 |