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A8.5 Simplifying expressions

Boolean algebra and logic circuits · A level · OCR H446 1.4.3, AQA 7517 4.6.5.1, Eduqas A500QS 1.2 · about 25 min

BugBotLab

What this lesson is about

A method for simplifying by algebra, worked exam-style examples, and checking the result.

Questions 5 marks in all

  1. [1 mark]Simplify (A ∨ B) ∧ (A ∨ ¬B).

    1. AA
    2. BB
    3. CA ∨ B
    4. D1
    Answer: A. Distribution gives A ∨ (B ∧ ¬B) = A ∨ 0 = A.

  2. [1 mark]Simplify ¬(A ∨ B) ∨ (¬A ∧ B).

    1. A¬A
    2. B¬B
    3. C¬A ∧ ¬B
    4. DA ∨ B
    Answer: A. De Morgan gives (¬A ∧ ¬B) ∨ (¬A ∧ B) = ¬A ∧ (¬B ∨ B) = ¬A.

  3. [1 mark]Simplify (A ∧ B ∧ C) ∨ (A ∧ B ∧ ¬C) ∨ (A ∧ ¬B).

    1. AA
    2. BA ∧ B
    3. CA ∧ C
    4. DB ∨ C
    Answer: A. The first two terms combine to A ∧ B, and (A ∧ B) ∨ (A ∧ ¬B) = A.

  4. [1 mark]Put the steps simplifying A · B + A · B̅ + A̅ · B in order.

    Number the lines 1 to 4 to put them in the right order.

    1. A + A̅ · B
    2. A · 1 + A̅ · B
    3. A + B
    4. A · (B + B̅) + A̅ · B
    Answer:
    A · (B + B̅) + A̅ · B
A · 1 + A̅ · B
A + A̅ · B
A + B

    Take out A, use the complement, use AND with 1, then X + X̅ · Y = X + Y.

  5. [1 mark]A student writes ¬A ∨ ¬B = ¬A ∧ ¬B as a step. What is wrong?

    1. AThe operator was changed without the bar over the whole that De Morgan needs
    2. BNothing, it is De Morgan's law
    3. CCommutation was used instead of association
    4. DThe NOTs should have cancelled
    Answer: A. ¬A ∨ ¬B is ¬(A ∧ B), not ¬A ∧ ¬B; the two differ when exactly one input is 1.

The task: simplify, then prove it

The function original(a, b, c) computes Q = ¬(A ∨ ¬B) ∨ (A ∧ B) ∨ (A ∧ ¬B ∧ C). Its inputs are bits (0 or 1) and it returns 0 or 1. Simplify Q on paper, then write simplified(a, b, c), taking and returning the same, as a single return line that uses at most two of and, or and not in total. It may not call original or use the bitwise operators &, |, ^ or ~. Print eight lines, one per row in binary order (A outermost, then B, then C), in exactly this form, with every value 0 or 1: A=0 B=1 C=0 original=1 simplified=1

# the two lines every program starts with: the commands, then the robot
from bugbot import *
connect()

def original(a, b, c):
    return int((not (a or not b)) or (a and b) or (a and not b and c))

def simplified(a, b, c):
    return 0

The hint students can ask for: Work on paper first. Apply De Morgan to the bracket with NOT in front, then look for two terms that differ only in one letter being inverted, and for a letter that can absorb or cancel part of the last term. Check your answer against the truth table the program prints.

A solution

# the two lines every program starts with: the commands, then the robot
from bugbot import *
connect()

def original(a, b, c):
    return int((not (a or not b)) or (a and b) or (a and not b and c))

def simplified(a, b, c):
    return int(b or (a and c))

for a in [0, 1]:
    for b in [0, 1]:
        for c in [0, 1]:
            print(f"A={a} B={b} C={c} original={original(a, b, c)} simplified={simplified(a, b, c)}")

Any program that meets the task's checks is marked correct in the simulator; this is one way, not the only way.