Prove A is not a subset of C, or else B is not a subset of A
- Thread starter Enh0702
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What have you tried? How far have you gotten?Enh0702 said:
Since A-union-B is not equal to A-intersect-C, there must be an element "x" in A-union-B that is not in A-intersect-C. Then, by definition of "union", x is either in A or in B.
Suppose x is in A. Since x is not in A-intersect-C, then x cannot be in C. What does this tell you about A being a subset of C?
Suppose x is not in A. Where can you go from there? And so forth.
Eliz.
pka
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This is more complicated than I thought at first.
Given \(\displaystyle A \cup B \not= A \cap C\). Assume that \(\displaystyle \left[ {A \not\subset C \vee B \not\subset A} \right]\) is false.
Then we have \(\displaystyle \left[ {A \subseteq C \wedge B \subseteq A} \right]\) is true.
If \(\displaystyle \left[ {A \subseteq C \right]\) then \(\displaystyle \left[ {A \cap C} \right] = A\).
So from the given \(\displaystyle A \cup B \not= A\), that implies that \(\displaystyle {B \not\subset A}\) which is in contradiction to the assumption.
Do you see how to finish?
Given \(\displaystyle A \cup B \not= A \cap C\). Assume that \(\displaystyle \left[ {A \not\subset C \vee B \not\subset A} \right]\) is false.
Then we have \(\displaystyle \left[ {A \subseteq C \wedge B \subseteq A} \right]\) is true.
If \(\displaystyle \left[ {A \subseteq C \right]\) then \(\displaystyle \left[ {A \cap C} \right] = A\).
So from the given \(\displaystyle A \cup B \not= A\), that implies that \(\displaystyle {B \not\subset A}\) which is in contradiction to the assumption.
Do you see how to finish?