Solve Boolean Algebra: A \oplus B=C, C \oplus B=A

[needhelp83]

if A \oplus B=C, then C \oplus B=A, and A \oplus C=B (use substitution)

C \oplus B=A

C \overline{B} + \overline{C} B = A

(A \oplus B) \overline{B} + (\overline {A \oplus B}) B = A

(A \overline{B} + \overline {A} B ) \overline {B} + (AB+ \overline{A} \overline{B})B=A

A \overline{B} \overline {B} + \overline {A} B \overline{B} + ABB + \overline{A} \overline{B}B=A

AB + 0 + AB + 0 = A

AB = A

I don't know how to get A = A

Any help?

 

[needhelp83]

if A \oplus B=C, then C \oplus B=A, and A \oplus C=B (use substitution)

C \oplus B=A

C \overline{B} + \overline{C} B = A

(A \oplus B) \overline{B} + (\overline {A \oplus B}) B = A

(A \overline{B} + \overline {A} B ) \overline {B} + (AB+ \overline{A} \overline{B})B=A

A \overline{B} \overline {B} + \overline {A} B \overline{B} + ABB + \overline{A} \overline{B}B=A

AB + 0 + AB + 0 = A

AB = A

Am I doing the steps correctly?

 

[needhelp83]

Anybody know how to solve

 

[CEL]

if A \oplus B=C, then C \oplus B=A, and A \oplus C=B (use substitution)

C \oplus B=A

C \overline{B} + \overline{C} B = A

(A \oplus B) \overline{B} + (\overline {A \oplus B}) B = A

(A \overline{B} + \overline {A} B ) \overline {B} + (AB+ \overline{A} \overline{B})B=A

A \overline{B} \overline {B} + \overline {A} B \overline{B} + ABB + \overline{A} \overline{B}B=A

AB + 0 + AB + 0 = A

AB = A

I don't know how to get A = A

Any help?

You made a mistake in line 6. You should have:
A\overline{B} + 0 + AB + 0 = A

A(\overline{B} + B) = A

A.1 = A

A = A