Final answer:
To convert the Boolean expression (A+B'+C)(B'+C+D')(A+B'+C'+D) to the standard expanded form, we use the distributive law repeatedly. We start by expanding the first two terms, simplify them, and then proceed with the third term. Throughout, we simplify by combining like terms and applying the idempotent law.
Step-by-step explanation:
Converting Boolean Expressions to Standard Expanded Form
To convert the given Boolean expression (A+B'+C)(B'+C+D')(A+B'+C'+D) to its standard expanded form, we will apply the distributive law of Boolean algebra multiple times. The distributive law states that A(B+C) = AB + AC. This process is similar to algebraic expansion in arithmetic, except that in Boolean algebra there is no concept of subtraction or carrying over; the same variable squared (or to any power) is equal to the variable itself (e.g., AA = A).
The first step is to expand the first two terms:
- (A+B'+C)(B'+C+D') becomes A(B'+C+D') + B'(B'+C+D') + C(B'+C+D')
- Then simplify each product:
- A(B'+C+D') = AB' + AC + AD'
- B'(B'+C+D') = B'B' + B'C + B'D' (Note that B'B'=B')
- C(B'+C+D') = CB' + CC + CD' (Note that CC=C)
Now we have: AB' + AC + AD' + B'B' + B'C + B'D' + CB' + CC + CD'. Simplify the terms where possible by combining like terms and eliminating terms using the idempotent law (AA = A).
After simplification, we are left with: AB' + AC + AD' + B'C + B'D' + CB' + C + CD'. Notice that we can remove the term CB' as C alone satisfies the expression.
Finally, multiply the result with the third term (A+B'+C'+D) and repeat the distributive process until all terms are expanded and simplified, resulting in the standard expanded form of the original Boolean expression.
Checking the Result
After expanding the terms, eliminate terms wherever possible to simplify the algebra. Lastly, check the answer to see if it is reasonable and aligns with the rules of Boolean algebra. This practice ensures the integrity of the expanded form.