lower bound for sum of the n factors of the inclusion exclusion principle

Question

Suppose the following relation is established:

$P\Bigl(A\cup B \cup C\cup D\Bigr) < P\Bigl(E\cup F\cup G\Bigr)$

based on boole's inequality, for each of the above probabilities we can have the following lower bounds:

$ P(A)+P(B)+P(C)+P(D)$ $ - P(A\cap B)- P(A\cap C)- P(A\cap D)- P(B\cap C)- P(B\cap D)- P(C\cap D) \le P\Bigl(A\cup B \cup C \cup D\Bigr) $

and

$ P(E)+P(F)+P(G)$ $ - P(E\cap F)- P(E\cap G)- P(F\cap G)\le P\Bigl(E\cup F\cup G\Bigr) $

Given these lower bounds, can we claim that the following relation holds?

$ P(A)+P(B)+P(C)+P(D)- P(A\cap B)- P(A\cap C)- P(A\cap D)- P(B\cap C)- P(B\cap D)- P(C\cap D)\le P(E)+P(F)+P(G)- P(E\cap F)- P(E\cap G)- P(F\cap G) $

Edit

Or can the opposite case be claimed? In other words if:

$ P(A)+P(B)+P(C)+P(D)- P(A\cap B)- P(A\cap C)- P(A\cap D)- P(B\cap C)- P(B\cap D)- P(C\cap D)\le P(E)+P(F)+P(G)- P(E\cap F)- P(E\cap G)- P(F\cap G) $

can we claim that:

$P\Bigl(A\cup B \cup C\cup D\Bigr) < P\Bigl(E\cup F\cup G\Bigr)$

Answer 1

Neither of the two questions you ask have an affirmative answer. Examples are easy to find.

For completeness sake, let only 2-wise intersections be nonzero, for both set families, and let the first family be denoted $$ \{A_i\}_{i=0,\ldots,3} $$ while the second family is $$ \{B_i\}_{i=0,\ldots,2} $$

Further let each set have a nonzero intersection only with its cyclic neighbours, i.e., $$ A_i \cap A_j = \emptyset,\quad \mathrm{if~~} i\notin \{j-1,j+1\}, $$ where the indices are taken modulo $4.$ Similarly, $$ B_i \cap B_j = \emptyset,\quad \mathrm{if~~} i\notin \{j-1,j+1\}, $$ where the indices are taken modulo $3.$ Let $\#(A_i \cap A_j)=a>0,$ whenever it is nonzero and similarly let $\#(B_i \cap B_j)=b>0,$ whenever it is nonzero. Also let $\#(A_i \setminus (A_{i+1} \cup A_{i-1}))=a'$ and let $\#(B_i \setminus (B_{i+1} \cup B_{i-1}))=b'.$

Then you can find appropriate integers $a,a',b,b'$ to force either of the implications you want to be false.