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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 \(\displaystyle 1\neq 0 \)

Then prove using only the above axioms: 0<1