We are given sets `P_1, P_2, S_2, S_2` and know
that:
1. `P_2 nnS_1 subset
P_1`
2. `P_1 nn S_2 subset
P_2`
3. `S_1 nn S_2 subset P_1 uu
P_2`
We must show that `S_1 nn S_2 subset P1 nn
P_2`
(In English: show every everything in both S1 and S2
is also in both P1 and
P2)
Proof:
Let `x` be an
element of `S_1 nn S_2` . This implies that `x in S_1` and `x in S_2` . By (3), we also
know that either `x in P_1` or `x in P_2` .
Case one:
suppose `x in P_1` : By (2), we see `x in S_2 ` and `x in P_1` and by (2) `P_1 nn S_2
subset P_2` , therefore `x in P_2` .
Case two: suppose` x
in P_2` . Since` x in S_1` and `P_2 nn S_1 subset P_1` , it follows that `x in P_1`
.
Since each of these cases shows x must exist in both P1
and P2, it follows that `S_1 nn S_2 subset P_1 nn P_2`
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