Proving the Equality of A and B: A+B=0 or A=B

[solakis1]

Given for all A,B,C...

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

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

3) A+0=A..........1A=A

4) $$\forall A\exists B(A+B=0)$$.....$$\forall A(\neg(A=0)\Longrightarrow\exists B(AB=1))$$

5) A(B+C)= AB+AC

Then prove: $$AA=BB\Longrightarrow (A+B)=0\vee A=B$$

Note: AB means A.B ...e.t.c ,e.t.c

 

[MarkFL]

Given for all A,B,C...

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

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

3) A+0=A..........1A=A

4) $$\forall A\exists B(A+B=0)$$.....$$\forall A(\neg(A=0)\Longrightarrow\exists B(AB=1))$$

5) A(B+C)= AB+AC

Then prove: $$AA=BB\Longrightarrow (A+B)=0\vee A=B$$

Note: AB means A.B ...e.t.c ,e.t.c

Please post the solution you have ready. :)

 

[solakis1]

Given for all A,B,C...

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

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

3) A+0=A..........1A=A

4) $$\forall A\exists B(A+B=0)$$.....$$\forall A(\neg(A=0)\Longrightarrow\exists B(AB=1))$$

5) A(B+C)= AB+AC

Then prove: $$AA=BB\Longrightarrow (A+B)=0\vee A=B$$

Note: AB means A.B ...e.t.c ,e.t.c

[sp] Proof:

1) AA=BB................Given

2) AA+AB=BB+AB..........additive property of equality

3) A(A+B)=B(B+A)............by axiom 5

4) A(A+B)=B(A+B)............by axiom 1

5) $$A+B\neq 0$$................assumption

6)$$\exists C[C(A+B)=1]$$..............by axiom 4

7) [C(A+B)]=1......................fix C

8) C[A(A+B)]= C[B(A+B)].................by the multiplicative property of equality

9) C[(A+B)A]=C[(A+B)B]..................by axiom 1

10) [C(A+B)]A= [C(A+B)]B..................by axiom 2

11) 1.A= 1.B......................by substituting 7 into 10

12) A=B........................by axiom 3

13) $$A+B\neq 0\Longrightarrow A=B$$..........Closing the assumption that we started at step 5

14) A+B=0 V A=B................This is equivalent to formula (13)Hence: If AA=BB ,then A+B=0 or A=B [/sp]