布尔表达式$(b + d)(a' + b' + c)$化简步骤疑问求解

Got it, let's break down how to get from your intermediate expression to the simplified result the tool provided. The key here is using the consensus theorem (a core rule for eliminating redundant terms in Boolean algebra) and some basic Boolean simplification steps.

First, let's recap where you left off:
Your simplified expression:
a'b + bc + a'd + b'd + cd

We need to eliminate the a'd and cd terms. Here's how to do it step by step:

Step 1: Expand the redundant term using complement law #

Since $b + b' = 1$ (any OR of a variable and its complement equals 1), we can multiply cd by $(b + b')$ without changing the expression's value:

= a'b + bc + a'd + b'd + cd(b + b')

Expand this out:

= a'b + bc + a'd + b'd + bcd + b'cd

Step 2: Group and simplify terms using absorption law #

The absorption law states that $X + XY = X$ (adding a term that's a subset of another doesn't change the result). Let's group terms:

• Group bc + bcd: $bc(1 + d) = bc$ (because $1 + d = 1$)
• Group b'd + b'cd: $b'd(1 + c) = b'd$ (same logic, $1 + c = 1$)

Substituting these back in, our expression becomes:

= a'b + bc + b'd + a'd

Step 3: Eliminate the final redundant term with the consensus theorem #

The consensus theorem tells us that for terms $XY + X'Z + XZ$, the term $XZ$ is redundant and can be removed. In our current expression, look at a'b + b'd + a'd:

• Let $X = a'$, $Y = b$, $Z = d$
• This fits the form $XY + Y'Z + XZ$, so the $XZ$ term (a'd) is redundant.

Removing a'd leaves us with:

a'b + bc + b'd

Which matches the result from the online tool!

Quick alternative: Using a Karnaugh Map (for verification) #

If you want to visualize why those terms are redundant, a 4-variable K-map shows that the minterms covered by a'd and cd are already fully covered by a'b, bc, and b'd. No extra minterms are added by keeping a'd or cd, so they can safely be removed.

内容的提问来源于stack exchange，提问作者Parker Queen