MHB Proving 0<1 with Axioms: A+B=B+A, A.B=B.A, and More!

[solakis1]

Given the following axioms:

For all A,B,C:

1) A+B=B+A

2) A+(B+C) =(A+B)=C

3) A.B=B.A

4) A.(B.C) = (A.B).C

5) A.(B+C)= A.B+A.C

6) A+0=A

7) A.1=A

8) A+(-A)=1

9) A.(-A)=0

10) Exactly one of the following:
A<B or B<A or A=B

11) A<B => A.C<B.C

12 [math] 1\neq 0 [/math]

Then prove using only the above axioms: 0<1

 

[Evgeny.Makarov]

9) A.(-A)=0

Do you mean $A\cdot0=0$? Also, don't you have an axiom that addition respects the order?

Then prove using only the above axioms: 0<1

This page has some proof. Also, several proof assistants have this theorem in their libraries, but they may use a number of lemmas, i.e., their proofs may not be the shortest.

Why are such problems interesting to you? After looking at several examples, they seem routine. It may be interesting to develop a new, somehow better axiomatization of integers, for example, but axiomatization of rings and fields seems good enough.

 

[solakis1]

Do you mean $A\cdot0=0$? Also, don't you have an axiom that addition respects the order?

This page has some proof. Also, several proof assistants have this theorem in their libraries, but they may use a number of lemmas, i.e., their proofs may not be the shortest.

Why are such problems interesting to you? After looking at several examples, they seem routine. It may be interesting to develop a new, somehow better axiomatization of integers, for example, but axiomatization of rings and fields seems good enough.

No.

A.(-A) =0 ,but you can prove : A.0 =0

The above is a mix of axiomatics.

And the question is :

Can we prove : 0<1 w.r.t the above axiomatics,since w.r.t the axiom 10 we can have i<0 ,o<1,1=0??