In boolean algebra if b+c=c+a and b+a`=c+a` then prove that b=c
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In Boolean algebra, is there any proof of A+BC=(A+B) (A+C) other than through truth tables?
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Howard Ludwig, Ph.D. Physics, Northwestern University (1982)
Answered Oct 23, 2017 · Author has 976 answers and 813.3k answer views
The 7 answers plus 4 collapsed answers I currently see are making this far too complicated, and I think I know why and will explain after my one-step proof.
First I am going to rewrite the expression a bit by making all operations explicit and using parentheses to specify intended order of operations. Digital electronics engineers and some other people are notoriously lazy and sloppy in their notation and it gets them in trouble. As I will explain later, I would like to make even more changes to the notation.
Prove: A + (B · C) = (A + B) · (A + C).
A + (B · C) = (A + B) · (A + C) by distributivity of + over · .
QED
That is all there is to it—one step. There are many ways to define a Boolean algebra—as a special type of distributive lattice, as a Huntington algebra, as a Robbins algebra, or as an abstract algebra built from scratch based on satisfying certain properties.
To be a distributive lattice requires having a lattice with two distributive properties:
A + (B · C) = (A + B) · (A + C) and A · (B + C) = (A · B) + (A · C) for all elements A, B, and C of the defining set for the Boolean algebra (which may have more than the two elements T and F, or 1 and 0. Thus, if you know you have a Boolean algebra, then you know that you have a distributive lattice, which means that you know you have both distributive properties, one of which is exactly what you are trying to prove.
Itemizing all the basic properties from scratch for a Boolean algebra involves both binary operations being commutative and having an identity element, the existence of a complement of each element of the defining set, and the same two distributive properties mentioned in the distributive lattice approach (each binary operation being distributive over the other), again proving your assertion in the same one step.
Now, why are you and all of the respondents having problems with this? The notation is very likely misleading you. Most people are very used to a · (x + y) = a · x + a · y from the ordinary arithmetic of real numbers. Ordinary multiplication is distributive over ordinary addition. However, it does not work the other way round for ordinary arithmetic: in general, a + x · y ≠ (a + x) · (a + y). However, in Boolean algebras, both directions work, by definition. Most people are so used to the latter not working that they have a hard time remembering that it does in Boolean algebra. When there is an opportunity to apply distributivity of + over ·, they fail to see it because of the bias generated by past experience. When operands are juxtaposed instead of using an explicit ·, this opportunity is even harder to see—people are subconsciously reluctant to break apart 2 items that are juxtaposed. That is one of the reasons that I never use implicit operation notation in Boolean algebra problems, and, consequently, I said up-front that I would explicitly include the ·. Always including the parentheses to indicate explicitly which of a + operation or a · operation also helps a lot.
YOUR ANSWER IS
In Boolean algebra, is there any proof of A+BC=(A+B) (A+C) other than through truth tables?
Still have a question? Ask your own!
What is your question?
Promoted by Commonlounge
Want to learn deep learning in 2 weeks for free?
25 crisp tutorials on neural networks, convolutional networks, word embeddings and adversarial networks.
Start Now
11 ANSWERS

Howard Ludwig, Ph.D. Physics, Northwestern University (1982)
Answered Oct 23, 2017 · Author has 976 answers and 813.3k answer views
The 7 answers plus 4 collapsed answers I currently see are making this far too complicated, and I think I know why and will explain after my one-step proof.
First I am going to rewrite the expression a bit by making all operations explicit and using parentheses to specify intended order of operations. Digital electronics engineers and some other people are notoriously lazy and sloppy in their notation and it gets them in trouble. As I will explain later, I would like to make even more changes to the notation.
Prove: A + (B · C) = (A + B) · (A + C).
A + (B · C) = (A + B) · (A + C) by distributivity of + over · .
QED
That is all there is to it—one step. There are many ways to define a Boolean algebra—as a special type of distributive lattice, as a Huntington algebra, as a Robbins algebra, or as an abstract algebra built from scratch based on satisfying certain properties.
To be a distributive lattice requires having a lattice with two distributive properties:
A + (B · C) = (A + B) · (A + C) and A · (B + C) = (A · B) + (A · C) for all elements A, B, and C of the defining set for the Boolean algebra (which may have more than the two elements T and F, or 1 and 0. Thus, if you know you have a Boolean algebra, then you know that you have a distributive lattice, which means that you know you have both distributive properties, one of which is exactly what you are trying to prove.
Itemizing all the basic properties from scratch for a Boolean algebra involves both binary operations being commutative and having an identity element, the existence of a complement of each element of the defining set, and the same two distributive properties mentioned in the distributive lattice approach (each binary operation being distributive over the other), again proving your assertion in the same one step.
Now, why are you and all of the respondents having problems with this? The notation is very likely misleading you. Most people are very used to a · (x + y) = a · x + a · y from the ordinary arithmetic of real numbers. Ordinary multiplication is distributive over ordinary addition. However, it does not work the other way round for ordinary arithmetic: in general, a + x · y ≠ (a + x) · (a + y). However, in Boolean algebras, both directions work, by definition. Most people are so used to the latter not working that they have a hard time remembering that it does in Boolean algebra. When there is an opportunity to apply distributivity of + over ·, they fail to see it because of the bias generated by past experience. When operands are juxtaposed instead of using an explicit ·, this opportunity is even harder to see—people are subconsciously reluctant to break apart 2 items that are juxtaposed. That is one of the reasons that I never use implicit operation notation in Boolean algebra problems, and, consequently, I said up-front that I would explicitly include the ·. Always including the parentheses to indicate explicitly which of a + operation or a · operation also helps a lot.
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