Proving Equivalence of 0A=0, (-A)B=-AB, (-A)(-B)=AB

[solakis1]

Given the following axioms:

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

 

[jvicens]

First, let's define what we mean by "equivalent." In this context, it means that if we can prove one statement, we can derive the other two from it. So, we need to show that if we can prove 0A=0, then we can also prove (-A)B=-AB and (-A)(-B)=AB, and vice versa.

To prove 0A=0, we can use axiom 12, which states that (A+B)C=AC+BC. Let's let A=0 and B=0. This gives us (0+0)C=0C+0C. By axiom 10, we know that 0+0=0, so we can substitute that in and get 0C=0C. We can then use axiom 1 to say that C=C. Therefore, 0A=0 is proven.

Next, we can use the same approach to prove (-A)B=-AB. Let's let A=-A and B=0. This gives us (-A+0)C=(-A)C+0C. By axiom 11, we know that (-A)+0=0, so we can substitute that in and get 0C=(-A)C. We can then use axiom 1 to say that C=C. Therefore, (-A)B=-AB is proven.

Finally, we can use the same approach to prove (-A)(-B)=AB. Let's let A=-A and B=-B. This gives us (-A+-B)C=(-A)C+(-B)C. By axiom 11, we know that (-A)+(-B)=0, so we can substitute that in and get 0C=(-A)C+(-B)C. We can then use axiom 1 to say that C=C. Therefore, (-A)(-B)=AB is proven.

Now, to show that these three statements are equivalent, we just need to show that if we can prove one, we can derive the other two from it. So, if we can prove 0A=0, then we can also prove (-A)B=-AB and (-A)(-B)=AB using the same substitution method as above. Similarly, if we can prove (-A)B=-AB, we can also prove 0A=0 and (-A)(-B)=AB, and if