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I am trying to use RCC8 algebra relations and axioms in order to tackle the following problem :

We consider the below configuration and square $x$ such that "$x$ NTPP $a$" and "$x$ TPP $c$"

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and we must prove that :

  • a) $x$ DC $b$
  • b) For $y$ such that "$y$ EC $a$", prove that it is not possible to have "$y$ EC $x$"

using the tree

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My difficulty is that I do not know how to "formally" prove the above statements using the tree.

For instance, regarding question (a), since "$x$ TPP $c$" and "$c$ DC $b$", by the composition table we have that "$x$ DC $b$". But, this argument does not use "Interior" and "Boundary" statements as the tree suggests. Could you please assist with this ?

nick
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1 Answers1

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According to the tree:

  1. "x TPP c" implies "x IntInt c" and "x BndBnd c" and "x BndInt c" and "not x IntBnd c".
  2. "c DC b" implies "not c IntInt b" and "not c BndBnd b".

From "x IntInt c" and "not x IntBnd c" we can say "not x IntExt c". This relation together to "not c IntInt b" proves that "not x IntInt b".

In similar way, we can infer "not x BndBnd b".

Following tree for "not x IntInt b" and "not x BndBnd b" we conclude "x DC c".

In other words, you need to use the composition of the relations "IntInt", "IntBnd", ... . Look for "9-intersection model", "4-intersection model" and/or "3-intersection model".

pasaba por aqui
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