- #1

solakis1

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For all A,B,C...we have:

1) A=A

2) A=B <=> B=A

3) A=B & B=C => A=C

4) A=B => A+C= B+C

5) A=B=> AC =BC ( NOTE :Instead of writing A.C or B.C e.t.c we write AB.BC e.t.c)

6) A+B= B+A..........AB=BA

7) A+(B+C) = (A+B)+C............A(BC)=(AB)C

10) A+0=A...............1A=A

11) A+(-A)=0...............A=/=0 => (1/A)A=1

12).......(A+B)C= AC+BC.........

13) 1=/= 0

Then prove: (By using only the axioms above)

1) 0=0A

2) A=/=0 & B=/= 0 => (1/A)(1/B)= 1/AB