Answer and Solution:
As per the question:
Given:
If
![A\subseteq B\cup C](https://img.qammunity.org/2020/formulas/mathematics/college/s4c2xuono648xd7x4gwmqs2pt5cm8l96wk.png)
![A\cap B = \phi](https://img.qammunity.org/2020/formulas/mathematics/college/r27up73r1eet91a2jca9hidwccre47ne9o.png)
To prove:
![A\subseteq \C](https://img.qammunity.org/2020/formulas/mathematics/college/75uaknqwggoiorncx00i0z30kl1cdeyzaq.png)
Proof:
Suppose
![t\in A](https://img.qammunity.org/2020/formulas/mathematics/college/rnrjamziynwnwtzmey37p0cri9jejuenfl.png)
As we know that:
![A\subseteq B\cup C](https://img.qammunity.org/2020/formulas/mathematics/college/s4c2xuono648xd7x4gwmqs2pt5cm8l96wk.png)
Therefore,
or
![t\in C](https://img.qammunity.org/2020/formulas/mathematics/college/u6fi7bzfzkxxhhf73nnbl4fchestvv1a6j.png)
Now, if we assume that
![t\in B](https://img.qammunity.org/2020/formulas/mathematics/college/ch5aawhofsqhsobpy87cplp8opk183w6ap.png)
Then
![t\in A\cap B](https://img.qammunity.org/2020/formulas/mathematics/college/vmt6hc7i6hvmu1vu0l4qaqo50edhitarca.png)
Since,
and
![t\in B](https://img.qammunity.org/2020/formulas/mathematics/college/ch5aawhofsqhsobpy87cplp8opk183w6ap.png)
But
A and B are disjoint set and
![A\cap B = \phi](https://img.qammunity.org/2020/formulas/mathematics/college/r27up73r1eet91a2jca9hidwccre47ne9o.png)
Therefore, this is contradictory.
Thus
![t\\otin B](https://img.qammunity.org/2020/formulas/mathematics/college/4r93zwwwz2pkzpthnlxhox8a64gi5jbib1.png)
So,
![t\in C](https://img.qammunity.org/2020/formulas/mathematics/college/u6fi7bzfzkxxhhf73nnbl4fchestvv1a6j.png)
Every element in the set A is also present in the set C
Therefore,
![A\subseteq \C](https://img.qammunity.org/2020/formulas/mathematics/college/75uaknqwggoiorncx00i0z30kl1cdeyzaq.png)
Hence, proved.