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K-map: A'B + BC' + AC'
Short answer
F(A,B,C) = A'B + BC' + AC' simplifies to A'B + AC'. Using a Karnaugh map, the minterms Σm(2,3,4,6) are grouped into two prime implicants: A'B (covers m2,m3) and AC' (covers m4,m6).
Simplify the expression using K-Map, F(A,B,C) = A'B + BC' + AC'.
Given F(A,B,C) = A'B + BC' + AC'. Expand to minterms:
- A'B = A'BC' + A'BC = m2 + m3
- BC' = A'BC' + ABC' = m2 + m6
- AC' = ABC' + AB'C' = m6 + m4
So F = Σm(2, 3, 4, 6).
K-map
BC
00 01 11 10
A = 0 | 0 | 0 | 1 | 1 | m2, m3 -> A'B
A = 1 | 1 | 0 | 0 | 1 | m4, m6 -> AC'
Groups:
- Pair (m2, m3): B = 1, C varies, A = 0 gives A'B.
- Pair (m4, m6): A = 1, C = 0, B varies gives AC'.
- The pair (m2, m6) = BC' is redundant, because both its minterms are already covered.
Answer: F = A'B + AC'
✓ Answer checked by Claude Opus 5.5 (high effort)
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