How do $P( A | B , C) < P(A| B^c,C)$ and $P( A | B , C^c ) < P(A| B^c,C^c) \Longrightarrow P( A|B) > P(A|B^c)$?

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In general, Simpson's Paradox occurs because situation such as following occurs for some arbitrary events $A,B,$ and $C$:

$P( A | B , C) < P(A| B^c,C) \tag{1}$

$P( A | B , C^c ) < P(A| B^c,C^c) \tag{2}$

Can someone show me a step-by-step way to arrive at $P( A|B) > P(A|B^c)$ from (1), (2)?

The Law of Total Probability

$P( A | B ) = P( A | B , C ) P( C | B) + P( A | B, C^c) P(C^c | B)$

appears somehow involved but I don't see how. Any help would be appreciated.

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There is no derivation of the third equation from the first two. If there were, then it would be the case that Simpson's paradox occurs for $all$ $A,B,C$. This is clearly not true. The correct statement is that $there$ $exists$ $A,B,C$ such that those three conditions hold. The fact that this occurs highly depends on the events $A,B,C$