To verify (A + C)(A' + B) = AB + A'C, the answer expands:

The right side and simplifies to AB + A'C

The left side by adding all minterms

2069 Bhadra · 2+2 marks

Verify Boolean identities

Short answer

i) Expanding (A+C)(A'+B) gives AB + A'C, so AB + A'C = (A+C)(A'+B). ii) XY + X'Z + YZ = XY + X'Z by consensus; multiplying YZ by (X+X') and simplifying shows YZ is redundant.

Verify the following: i) AB + A'C = (A+C)(A'+B) ii) XY + X'Z + YZ = XY + X'Z

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

Take the right-hand side (RHS):

(A + C)(A' + B) = AA' + AB + CA' + CB
                = 0 + AB + A'C + BC
                = AB + A'C + BC(A + A')
                = AB + A'C + ABC + A'BC
                = AB(1 + C) + A'C(1 + B)
                = AB + A'C

RHS = LHS. Verified.

ii) XY + X'Z + YZ = XY + X'Z (consensus theorem)

Take the left-hand side (LHS) and multiply YZ by (X + X') = 1:

XY + X'Z + YZ = XY + X'Z + YZ(X + X')
              = XY + X'Z + XYZ + X'YZ
              = XY(1 + Z) + X'Z(1 + Y)
              = XY + X'Z

LHS = RHS. Verified.

✓ Answer checked by Claude Opus 5.5 (high effort)

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Verify Boolean identities

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

ii) XY + X'Z + YZ = XY + X'Z (consensus theorem)

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