Computer Science · real student question

Simplify the Boolean expression F = A'B'C'D' + A'B'CD' + A'BC'D + A'BCD' + AB'C'D' + AB'C'D + ABC'D' + ABCD using the laws of Boolean algebra.

Question

Simplify the Boolean expression

F=ABCD+ABCD+ABCD+ABCD+ABCD+ABCD+ABCD+ABCDF = A'B'C'D' + A'B'CD' + A'BC'D + A'BCD' + AB'C'D' + AB'C'D + ABC'D' + ABCD

using the laws of Boolean algebra.

Step-by-step solution

  1. List the minterms before touching the algebra. Reading ABCDABCD as a 4-bit number (with AA the most significant bit and a primed literal meaning 00), the eight products are

    m0, m2, m5, m6, m8, m9, m12, m15m_0,\ m_2,\ m_5,\ m_6,\ m_8,\ m_9,\ m_{12},\ m_{15}

    Having the index list makes it possible to verify the final answer instead of trusting the manipulation.

  2. Combine pairs that differ in exactly one literal. The only rule doing real work is X+X=1X + X' = 1 applied after factoring. For example

    ABCD+ABCD=ABD(C+C)=ABDA'B'C'D' + A'B'CD' = A'B'D'(C'+C) = A'B'D'

    and

    ABCD+ABCD=ABC(D+D)=ABCAB'C'D' + AB'C'D = AB'C'(D'+D) = AB'C'

  3. Reuse minterms — this is the step that is usually skipped. Idempotence, X+X=XX + X = X, lets a minterm appear in more than one grouping at no cost. Duplicating m6m_6 and m12m_{12} produces two more three-literal terms:

    ABCD+ABCD=ACD(B+B)=ACDA'BCD' + A'B'CD' = A'CD'(B+B') = A'CD'
    ABCD+ABCD=ACD(B+B)=ACDABC'D' + AB'C'D' = AC'D'(B+B') = AC'D'

    Stopping at the first pairing leaves ABCDA'BCD' and ABCDABC'D' at four literals each, which is why the "obvious" grouping is not minimal.

  4. Handle the two minterms with no neighbour. m5=ABCDm_5 = A'BC'D and m15=ABCDm_{15} = ABCD differ from every other listed minterm in more than one bit, so neither can be combined. They survive as essential prime implicants in full four-literal form.

  5. Assemble the minimal sum of products.

    F=ABD+ACD+ABCD+ABC+ACD+ABCDF = A'B'D' + A'CD' + A'BC'D + AB'C' + AC'D' + ABCD

    Six product terms, 20 literals in total. (An equally minimal alternative swaps ABDA'B'D' for BCDB'C'D'; both cover m0m_0, so the minimum is not unique.)

  6. Verify by exhaustive evaluation, not by inspection. Evaluating the simplified expression over all 1616 assignments of (A,B,C,D)(A,B,C,D) gives 11 exactly on {0,2,5,6,8,9,12,15}\{0,2,5,6,8,9,12,15\} — the same set as the original. Every one of the eight original minterms is covered and no extra row turns on, so the two expressions are equal.

Answer

F=ABD+ACD+ABCD+ABC+ACD+ABCDF = A'B'D' + A'CD' + A'BC'D + AB'C' + AC'D' + ABCD

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