A question on XOR
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From: William Elliot on 9 Aug 2010 01:57 On Sat, 7 Aug 2010, Nam Nguyen wrote: > I have a question: Let A be an atomic formula of an L. > As far as FOL proof is concerned, if (A \/ ~A) signifies > a tautology and (A /\ ~A) a contradiction, what would > (A xor ~A) signify? A tautology.
From: Nam Nguyen on 9 Aug 2010 02:04 William Elliot wrote: > On Sat, 7 Aug 2010, Nam Nguyen wrote: > >> I have a question: Let A be an atomic formula of an L. >> As far as FOL proof is concerned, if (A \/ ~A) signifies >> a tautology and (A /\ ~A) a contradiction, what would >> (A xor ~A) signify? > > A tautology. Right. That's what JB said and I now realize it so. I've also moved the topic toward the revised (*) and a possible replacement of the familiar CON(T). -- ----------------------------------------------------------- Normally, we do not so much look at things as overlook them. Zen Quotes by Alan Watt -----------------------------------------------------------
From: Jan Burse on 11 Aug 2010 12:03 Nam Nguyen wrote: > Be that as it may, a contraction could also be defined > as a formula of the form (F /\ ~F) for some formula F. Well you may find the definition that a theory T is contradictory iff there is a formula B, such that B and ~B are derivable from it. But this is not the same as to say that a contradictory formula A has the form B /\ ~B. For example the following formulae are all also contradictory in itself, but do not have this form: f (i) p /\ (p -> ~p) (ii) ~(p \/ ~p) (iii) p /\ ~~~p (iv) Etc... They are contradictory in the sense that their negation can be derived, and in the sense that they are always false and in the sense when taken as a theory, i.e. a theory consisting of this single formula A, then we can find formula B and ~B that are derivable from this theory. In fact it happens that we can pick A itself as a witness B for a contradiction. Because when A is contradictory its negation can be derived: |- ~A And by weakening we have trivially: A |- ~A Further by identity we have: A |- A So indeed A and ~A follow from the theory A, when A is contradictory. More meta reasoning would show that we can pick any formula C as a witness, since classical logic explodes upon inconsistency. Best Regards
From: Jan Burse on 11 Aug 2010 12:08 Nam Nguyen wrote: > William Elliot wrote: >> On Sat, 7 Aug 2010, Nam Nguyen wrote: >> >>> I have a question: Let A be an atomic formula of an L. >>> As far as FOL proof is concerned, if (A \/ ~A) signifies >>> a tautology and (A /\ ~A) a contradiction, what would >>> (A xor ~A) signify? >> >> A tautology. > > Right. That's what JB said and I now realize it so. I've also > moved the topic toward the revised (*) and a possible replacement > of the familiar CON(T). > Oops, only reading this now after I have already posted something about "A contradiction". Bye
From: Nam Nguyen on 11 Aug 2010 12:26
Jan Burse wrote: > Nam Nguyen wrote: >> Be that as it may, a contraction could also be defined >> as a formula of the form (F /\ ~F) for some formula F. Despite what you said below, (F /\ ~F) can be defined as a contraction purely on the semantics of the logical symbols /\, and ~ (and no theory needs to be involved). > > Well you may find the definition that a theory T is > contradictory iff there is a formula B, such that > B and ~B are derivable from it. > > But this is not the same as to say that a contradictory > formula A has the form B /\ ~B. For example the following > formulae are all also contradictory in itself, but do > not have this form: > > f (i) > p /\ (p -> ~p) (ii) > ~(p \/ ~p) (iii) > p /\ ~~~p (iv) > Etc... > > They are contradictory in the sense that their negation > can be derived, and in the sense that they are always > false and in the sense when taken as a theory, i.e. > a theory consisting of this single formula A, then we > can find formula B and ~B that are derivable from this > theory. > > In fact it happens that we can pick A itself as a witness > B for a contradiction. Because when A is contradictory > its negation can be derived: > > |- ~A > > And by weakening we have trivially: > > A |- ~A > > Further by identity we have: > > A |- A > > So indeed A and ~A follow from the theory A, when A is > contradictory. More meta reasoning would show that we > can pick any formula C as a witness, since classical > logic explodes upon inconsistency. > > Best Regards > -- ----------------------------------------------------------- Normally, we do not so much look at things as overlook them. Zen Quotes by Alan Watt ----------------------------------------------------------- |