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    • Thread Starter
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    Hey, I'm currently constructing some simple proofs as an exercise, I was wondering if someone from TSR could give me a hand with them.

    This is the first one, I will probably put more up.

    Code:
    Theory: A/(A/B)=B/(B/A)
    
    Proof: Let x∊A/(A/B) then x∊A and x∉A/B, therefore x∊ and (x∉A or x∊B). Using the dissociative law we have two cases to consider either x∊ A and x∉A which leads to a contradiction so we need not consider it, or x∊A and  x∊B. If x∊A and  x∊B then x∉B/A. As  x∊B and x∉B/A then x∊B/(B/A) and thus A/(A/B)⊆B/(B/A).
    Conversely let y∊B/(B/A) then y∊B and y∉B/A. Therefore we can once again reach the conclusion that y∊B and y∊A. Therefore y∉A/B and further y∊A/(A/B) and thus we conclude B/(B/A)⊆A/(A/B) as required.
    • PS Helper
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    For what it's worth, shouldn't those "/"s be "\"s?

    I think your proof is alright, although probably slightly longer than it needs to be. You can also use the rules A\backslash B = A \cap \overline{B} ,\ A \cap \overline{A} = \emptyset and \overline{(A \cap B)} = \overline{A} \cup \overline{B} to neaten it up.

    [Also, avoid the [ code ] tag, it makes it harder to read!]
 
 
 
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