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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 that: 0A=0, (-A)B=-AB, (-A)(-B) =AB are equivalent to each other

You may use as arule for substitution the following :

If S is a formula ,from S and t=s,or from S and s=t we mayderive T,provided that T is the result from S by replacing one or more occurances of t in S by s

The above was taken from the book : formal proofs in maths