Is the Empty Set Bounded? Proof and Contradiction

[evagelos]

Depending on the axioms that you use for logic, that A => B is true whenever A is false can be either an axiom or a theorem.

Vacuous Truth is just what it's called when A is false because it doesn't at all matter what B is.

up to now i was informed of three possibilities:

morphism said a definition with a question mark, you said a theorem or an axiom depending on the set of axioms. You succeeded in making me to open the books of logic to find out
myself.
so wait i get back to you

 

[Dragonfall]

Your logic book will only tell you the formalism preferred by its author.

 

[evagelos]

The laws and the facts about logic are eternal and no author and his preferred formalism
can change that

 

[CRGreathouse]

The laws and the facts about logic are eternal and no author and his preferred formalism can change that

Some logic books define $\phi \vee \psi$ as $\neg\phi\rightarrow\psi$ while others define it as its own privative with 3-4 axioms (for example, in paraconsistent logics). What would you say to that?

 

[Dragonfall]

The laws and the facts about logic are eternal and no author and his preferred formalism
can change that

It's as if you're talking about a religion.

 

[evagelos]

It's as if you're talking about a religion.

If you subtract religion from our life we will live in a better world,but if you take away

logic can you live in para consistent logic as GR put it

 

[Dragonfall]

You sound just like a postmodernist mathematician! You wouldn't happen to be a disciple of Jacques Lacan, would you?

 

[evagelos]

pardon me but i like to postmortem mathematics.

Who the hell is that Lucas guy anyway

 

[LukeD]

The laws and the facts about logic are eternal and no author and his preferred formalism
can change that

Why do you think that the laws of mathematics are such? Why are they "true" and "eternal"? Did you prove this in some way? Did some higher power tell you that this must be true? Is it just intuitive?

I'm not trying to goad you into an argument. I'm just curious about your answer, and I wanted to illustrate something.

 

[HallsofIvy]

The laws and the facts about logic are eternal and no author and his preferred formalism
can change that

And this is true because you SAY so?

 

[evagelos]

The F----->T, F----->F implications are just part of the conditional definition of two statements p and q......p----->q.

So the ' vacuously true' expression is not an axiom is not a theorem in the axiomatic
foundation of the predicate calculus.

It can be used in a semantical type of proof and not in a syntactical type where all the mathematical proofs are.

Anyway to get rid of the 'vacuously true' myth here is a proof that the empty set is bounded

We want to prove that:...Ea(x)[ xεΦ====>absvalue(a)>=x]........

Let...xεΦ====>( xεΦ v absvalue(a)>=x) <=====> ( ~xεΦ===> absvalue(a)>=x) and
since ~xεΦ ( because no object belongs to the empty set) we have...absvalue(a)>=x.

Thus......xεΦ====>absvalue(a)>=x ..........

And......Ea(x)[ xεΦ====>absvalue(a)>=x].........

Where (x) means for all x, Ea means there exists an a, ~xεΦ means x does not belong to the empty set.

Perhaps an easier and shorter proof is the following:

We know...(x)[ ~xεΦ]====> ~xεΦ====>(~xεΦ v absvalue(a)>=x) <=====>
........xεΦ====>absvalue(a)>=x .........

And.......Ea(x)[ xεΦ====>absvalue(a)>=x]........

 

[Focus]

p -> q is a shorthand way of writing $$\neg p \vee q$$. The other operators (such as "not" and "or") are defined, you can call them axioms but its a bit pointless because they really do not tell you anything. As for the rest of the post with the proof, I can't read it, please use http://en.wikibooks.org/wiki/LaTeX/Mathematics" to write it, ascii is not made for maths, it makes my eyes bleed.

 

[HallsofIvy]

pardon me but i like to postmortem mathematics.

Who the hell is that Lucas guy anyway

So you think mathematics is dead? That's explains a lot about your posts!

 

[evagelos]

the law that i used in my proof is the tautology or theorem of the propositional calculus :

[(pvq)<----->(~p---->q)].............1

If we give values to p and q , (T,F) all the values of 1 will be true,thus 1 is a tautology,

meaning that pvq and ~p--->q are logically equivalent ,meaning that pvq logically implies
~p--->q and ~p---->q logically implies pvq,written as pvq <====> ~p---->q,notice the double line in the double implication.

If we now put p= ~xεΦ and q= absvalue(x)=<a, we have:

we have p......a fact not an assumption
but p===>pvq ( because of the law called disjunction introduction) <=====> ~p---->q (because of the above law).

hence...~p=====>q and if we substitute p and q we we have that the empty set is

bounded

 

[evagelos]

Another maybe understandable proof is to reason by contradiction:

Assume that the empty set is not bounded ,then we must have : Αn x belonging to the empty set, xεΦ,and an absolute x bigger than a, absvalue(x)> a.

But since no x belongs to the empty set ,~xεΦ, we have a contradiction :

xεΦ & ~xεΦ

Thus,,,,,,,,,,,,,, the empty set is bounded