How Does the Indicator Function Prove 1A U (B∩C) = 1A.(1B + 1C - 1B.1C)?

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The discussion centers on clarifying the proof of the equation 1A U (B∩C) = 1A.(1B + 1C - 1B.1C) from a book by Corbae, Stinchcombe, and Zemam. A participant expresses confusion over the equation, suggesting it should instead be 1A∩(B∪C) = 1A.(1B + 1C - 1B.1C). This correction is based on the definitions of indicator functions for intersections and unions, indicating a possible misprint in the original text. The conversation emphasizes the importance of correctly applying set theory principles in the context of indicator functions. Overall, the thread seeks to clarify a potential error in the mathematical representation.
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Hi Everyone,
So here is a prove given in book by Corbae S tinchcombe Zemam so can someone clear this line of prove to me that
1AU(B∩C)=1A.(1 B +1C -1B .1C )

I don't get it please help
thanks in advance
 
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vandanak said:
Hi Everyone,
So here is a prove given in book by Corbae S tinchcombe Zemam so can someone clear this line of prove to me that
1AU(B∩C)=1A.(1 B +1C -1B .1C )

I don't get it please help
thanks in advance

Seems to me that it should be $$1_{A\cap(B\cup C)}=1_A\cdot(1_B+1_C-1_B\cdot1_C)$$ which follows from $$1_{X\cap Y}=1_X\cdot 1_Y$$ and $$1_{X\cup Y}=1_X+1_Y-1_X\cdot1_Y$$
 
that means it must be misprint or something like that because union is between A and (B intersection C) and you have taken union in B and C where there was a intersection given
 
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...
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