Question

Prove (A ⊕ B) ⊕ C = A ⊕ (B ⊕ C) using boolean algebra.
I made the truth tables and found the sum of products, but couldnt figure how to show their equal.

I then tried doing

(a xor b) xor c  (a' - is NOT(A)/inverse)
(a'b + ab') ⊕ C
c' (a'b + ab') + c[(a'b + ab')']  

Couldn’t go from there,

Answer 1

A^B = (AB'+A'B)

(AB)' = (A'+B')

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(A^B)^C

= (AB'+A'B)C' + (AB'+A'B)'C

= (AB'C'+A'BC')+((AB')'(A'B)')C

= (AB'C'+A'BC')+(A'+B)(A+B')C

= (AB'C'+A'BC')+(A'(A+B')+B(A+B'))C

= (AB'C'+A'BC')+(A'B' + AB)C

= (AB'C'+A'BC'+A'B'C + ABC)

= A(B'C'+BC)+A'(BC'+B'C)

= A(B'C'+BC)+A'(B^C) (1)

(B^C)'

=(BC'+B'C)'

= (BC')'(B'C)'

= (B'+C)(B+C')

= (B'C'+BC) (2)

From (2), the (1) = A(BC'+B'C)' + A'(B^C) = A(B^C)' + A'(B^C) = A^(B^C) #