Question

Simplify the following boolean algebra equation only with the xor logic gate!

Please answer clearly

a. (AB+C'D')(A'+B')(AB)+(C+B)(C'+B')+(A+1)A(B+C+D)AB(A’+B') b. (((CD'+C'D)'+((A’B)'+(AB'))*(((A’B)'+(AB')')+ (CD’+C'D)))'

Answer 1

@ " A.B = Ãtà AB + C D (Ā+B)(AB) + (C+B) (C+B) +(A+1) A (B+C+D) 0 @ AB(A+B) Demorgan's law AB=Ā+B Ā.B = A+B A+BC = (A+B) (A+C) Axor B - Ā B +BA ② (C+B) (C+B) => > (+5A)(B+BĀ) from o -> (Ata)lē +) (B+b) (877) (C+B)(C+B)= COB => 1. (A+B) (6+) 0 ta xos By B)(A+B) O» (AB+TD (A+) (AB) - => (AB+TD)(A.B) (AB) A. Ā=0 - ARTTO) (O) 0 da * (A+1)A (B+C+D) AB (Ā+B) » A+1) A(B+C+D) AB (AB) -> (A+) A(B+C+D) (0) to A+ Ā=) Simplihed ex poession is =) O + COB+O -> COB

® ((CD + CD + (ñ8+AZ)( 26+) +[C0+))) A + B = AB CD+CD = CAD ĀB+AB AB. AB Ā B + A B. AŘ + - * (@O)1) (1 + (HD) CODTD+CÒ COD= TD+CD = COD (COD + o) (1) IA =) 2) COD - COD * 6 have small mistake in given there is an extra open bracket question

In given question b there is extra bracket

If I was wrong anywhere please comment so I can improve answer