POS expression
(A + B + C)(A + B + D)(A + B + C + D)
A standard POS expression ?
Yes, (A + B + C)(A + B + D)(A + B + C + D) is a standard POS expression.
A standard POS (Product of Sums) expression is the complement of a standard SOP (Sum of Products) expression. In other words, it represents the logic function where all the inputs are complemented and the operations are changed (i.e., AND becomes OR, and OR becomes AND).
To convert the given expression into a standard POS expression, we need to apply De Morgan's Law and distribute the OR operator over the AND operator. Here are the steps:
Step 1: Apply De Morgan's Law to each term inside parentheses:
(A + B + C)(A + B + D)(A + B + C + D)
= (A' · B' · C')(A' · B' · D')(A' · B' · C' · D')
Step 2: Use the distributive property to expand the expression:
(A' · B' · C')(A' · B' · D')(A' · B' · C' · D')
= (A' · A' · A' · B' · B' · B' · C' · C' · D')(D')
= A' · A' · B' · B' · B' · C' · C' · D' · D'
= A' · B' · C' · D'
So, the standard POS expression for (A + B + C)(A + B + D)(A + B + C + D) is A' · B' · C' · D'.
A standard POS (Product of Sums) expression is an expression in Boolean algebra that is in the form of a product of sums.
To determine if the given expression, (A + B + C)(A + B + D)(A + B + C + D), is a standard POS expression, we need to first check if each term within the parentheses is a sum of literals.
In this case, each term is indeed a sum of literals: (A + B + C), (A + B + D), and (A + B + C + D). Each term consists of variables (A, B, C, D) connected with the logical OR operator (+).
Next, we need to check if the overall expression is a product of these sums. Since the given expression is in the form of (term1)(term2)(term3), where each term is a sum of literals, the expression fits the criteria of a standard POS expression.
Therefore, the given expression, (A + B + C)(A + B + D)(A + B + C + D), can be considered as a standard POS expression.