Proving a Logic Rule with Boolean Algebra: Step-by-Step Guide

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Discussion Overview

The discussion revolves around proving a logic rule using Boolean algebra. Participants explore various methods of simplification and seek clarification on steps involved in the proof, including the use of distributive property and other Boolean laws. The conversation includes hints and corrections as participants work through the problem step-by-step.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation
  • Exploratory

Main Points Raised

  • One participant asks for help in proving a logic rule and expresses uncertainty about whether to draw a digital circuit.
  • Another participant suggests using the distributive property and emphasizes the importance of showing attempts at solutions.
  • There is a discussion about the expression A.B + A.C, with one participant questioning the presence of C in the expression.
  • Participants discuss simplifying expressions using the Idempotent Law and the distributive property, with one suggesting A + A.B and another proposing A + B.A.
  • Hints are provided regarding the simplification of expressions, including the suggestion to factor using the reverse of the distributive law.
  • There is a discussion about the simplification of the expression 1 + B, with one participant expressing confusion about the outcome and its implications.
  • Another participant mentions the use of a Karnaugh map as a method for handling more complex logical expressions.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the final steps of the proof, and multiple viewpoints on the simplification process are presented. Some participants express confusion about certain steps, indicating that the discussion remains unresolved.

Contextual Notes

Limitations include the lack of clarity on certain assumptions and the need for participants to show their work to facilitate assistance. The discussion also reflects varying levels of understanding of Boolean algebra concepts.

Mason98
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No Effort - Member warned that some effort must be shown
Homework Statement
Boolean algebra help
Relevant Equations
A.(a+b) = a
1589225716790.png
Hello, can anyone help how i prove this logic rule? I am not sure whether i have to draw a digital circuit or something. if someone could help me solve it showing the steps they took i'd appreciate thanks
 
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What do you get when you use the distributive property here?

PS: Forum rules require you to show your attempt at a solution. We can only help by providing hints and pointing mistakes in your work.
 
cnh1995 said:
What do you get when you use the distributive property here?

PS: Forum rules require you to show your attempt at a solution. We can only help by providing hints and pointing mistakes in your work.
I thought it could possibly be A.B + A.C but I am not sure tbh
 
Mason98 said:
I thought it could possibly be A.B + A.C but I am not sure tbh
Where is C in that expression?
 
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A.A + A.B i mean sorry, thanks for the help, i tried multiplying the brackets out
 
Mason98 said:
A.A + A.B i mean sorry
Right.

You need to use the laws of boolean algebra to simplify this expression. The distributive property was one of these laws. How will you reduce this further?
Can you find anything in your lecture notes?
 
cnh1995 said:
Right.

You need to use the laws of boolean algebra to simplify this expression. The distributive property was one of these laws. How will you reduce this further?
Can you find anything in your lecture notes?
Would i now use the Idempotent Law, which would change A.A to just A which would leave me with A +A.B and the A.B would change to B.A, so A + B.A ?
 
Mason98 said:
Would i now use the Idempotent Law, which would change A.A to just A which would leave me with A +A.B?
Right.
Then?
 
Mason98 said:
and the A.B would change to B.A, so A + B.A ?
No need to change A.B to B.A.

Hint (if you haven't got the next step yet): A=A.1
 
  • #10
cnh1995 said:
No need to change A.B to B.A.

Hint (if you haven't got the next step yet): A=A.1
Hmm thanks for the help by the way appreciate it :), I'm thinking it could be, A.1 + A.B?
 
  • #11
Mason98 said:
Hmm thanks for the help by the way appreciate it :), I'm thinking it could be, A.1 + A.B?
Now factor the expression, using the reverse of the distributive law. What do you get?
 
  • #12
Mark44 said:
Now factor the expression, using the reverse of the distributive law. What do you get?
A.(1+B)?
 
  • #13
Mason98 said:
A.(1+B)?
Right, and how can ##1 + B## be simplified? Remember that + is used for OR, so in terms of sets, this would be ##U \cup B##, where U is the universal set.
 
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  • #14
1 + B is basically 1 or B? So, it can be simplified down to just 1? I'm so confused
 
  • #15
Mason98 said:
1 + B is basically 1 or B? So, it can be simplified down to just 1? I'm so confused

Yes it can. If you think about the logic table for the expression 1 + B, what is the outcome if B = 1? What if B = 0? Does it matter what B is? That should help prove to you the identity X + 1 = 1 for a boolean variable X.

Another method when facing these problems is to use a Karnaugh map. These help when the logical expressions become more complex.
 
  • #16
Do you know how to make a logic table. If so, you should make one and if you have trouble, show your work.
 

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