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Positive/negative logic; verify 3-variable De Morgan

Short answer

Positive logic assigns HIGH=1, LOW=0; negative logic reverses. De Morgan’s law for three variables: (A+B+C)′ = A′B′C′ and (ABC)′ = A′+B′+C′. Truth table verification shows the corresponding columns are identical for all eight input combinations, confirming both theorems.

Define the positive and negative logic. State and verify 3-bits De Morgan's theorem.

Positive and negative logic

  • Positive logic: HIGH voltage = logic 1, LOW voltage = logic 0 (e.g. TTL: +5 V = 1, 0 V = 0).
  • Negative logic: HIGH voltage = logic 0, LOW voltage = logic 1 (e.g. +5 V = 0, 0 V = 1).

The same physical gate performs a different logic function in the two systems, because changing the convention complements every input and output. Example: a gate whose output is HIGH only when both inputs are HIGH is an AND gate in positive logic but an OR gate in negative logic.

De Morgan's theorem for 3 variables

Theorem 1: A+B+C‾=Aˉ⋅Bˉ⋅Cˉ\overline{A + B + C} = \bar A \cdot \bar B \cdot \bar C

Theorem 2: A⋅B⋅C‾=Aˉ+Bˉ+Cˉ\overline{A \cdot B \cdot C} = \bar A + \bar B + \bar C

Verification by truth table

ABCA+B+C‾\overline{A+B+C}AˉBˉCˉ\bar A \bar B \bar CABC‾\overline{ABC}Aˉ+Bˉ+Cˉ\bar A + \bar B + \bar C
0001111
0010011
0100011
0110011
1000011
1010011
1100011
1110000

Columns 4 and 5 are identical, and columns 6 and 7 are identical for all 8 combinations, so both theorems hold for 3 variables.

Algebraic check (using the 2-variable law with X=B+CX = B + C):

A+B+C‾=A+X‾=Aˉ⋅Xˉ=Aˉ⋅B+C‾=AˉBˉCˉ\begin{aligned} \overline{A + B + C} &= \overline{A + X} = \bar A \cdot \bar X \\ &= \bar A \cdot \overline{B + C} = \bar A \bar B \bar C \end{aligned}

Similarly, A(BC)‾=Aˉ+BC‾=Aˉ+Bˉ+Cˉ\overline{A(BC)} = \bar A + \overline{BC} = \bar A + \bar B + \bar C.

 3-input NOR = bubbled 3-input AND
 3-input NAND = bubbled 3-input OR

✓ Answer checked by Claude Opus 5.5 (high effort)

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