Logic proposition for partition conditions

Proposition

If

1. \( \forall s \in S, s\in A_1 \cup A_2 \, \land \, A_1 \cup A_2 = \varnothing \)
2. \( s\in A_1 \implies s \in B_1 \, \land \, s\in A_2 \implies s \in B_2 \)
3. \( \forall s \in S, s\in B_1 \cup B_2 \, \land \, B_1 \cup B_2 = \varnothing \)

Then \( \forall s \in S, s\in B_1 \implies s \in A_1 \, \land \, s\in B_2 \implies s \in A_2 \)

Proof.

\( (s\in A_2 \implies s \in B_2) \iff \{ \lnot (s \in B_2) \implies \lnot (s\in A_2) \} \)

Since \( B_1, B_2 \) partition \( S \), \( s \not\in B_2 \iff s\in B_1 \). Similarly, \( s \not\in A_2 \iff s\in A_1 \)

Thus, \( (s\in A_2 \implies s \in B_2) \iff (s\in B_1 \implies s \in A_1) \)

Similarly, \( (s\in A_1 \implies s \in B_1) \iff (s\in B_2 \implies s \in A_2) \)

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Corollary

If

1. \( \forall s \in S, s\in A_1 \cup A_2 \cup \cdots \cup A_n \, \land \, A_1,A_2,\ldots,A_n \) are pairwise disjoint
2. \( s\in A_i \implies s \in B_i \)
3. \( \forall s \in S, s\in B_1 \cup B_2 \cup \cdots \cup B_n \, \land \, B_1,B_2,\ldots,B_n \) are pairwise disjoint

Then \( \forall s \in S, s\in B_i \implies s \in A_i \)