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I have constructed truth tables to prove that: $ABC + ABC'+ AB'C +A'BC = AB+AC+BC$

How do I prove it by simplifying the expression? I know that I can simplify: $ABC + ABC' = AB(C+C')=AB$. However I can't repeat this for $AB'C$ or $A'BC$, to get the answer I desire.

I'm fairly new to boolean algebra and have tried to use the basic identities to figure it out, but can't seem to get there.

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$ \begin{align} ABC + ABC' + AB'C + A'BC = \\ ABC + ABC + ABC + ABC' + AB'C + A'BC = & & & \text{(2x Idempotency)}\\ AB(C + C') + ABC + ABC + AB'C + A'BC = & & &\text{(Distribution)}\\ AB + ABC + ABC + AB'C + A'BC = & & & \text{(Negation)}\\ AB + (AB + AB')C + (AB + A'B)C = & & & \text{(2x Distribution)} \\ AB + (A(B+B'))C + ((A+A')B)C = & & & \text {(2x Distribution)} \\ AB + AC + BC & & & \text{(2x Negation)} \end{align} $

I'm not proficient at this kind of derivations so please correct any notation/terminology

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Idempotency: ABC = ABC + ABC.

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