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Problem 23

# Let $$B$$ be any language over the alphabet $$\Sigma$$. Prove that $$B=B^{+}$$iff $B B \subseteq B$.

Expert verified
In summary, to prove that a language B is equal to its concatenation closure, B^+, if and only if the concatenation of B with itself (BB) is a subset of B, we demonstrated two directions of the equivalence: 1. Assuming B = B^+, we showed that BB ⊆ B, as B^+ contains all concatenations of any length of words from B, including BB. 2. Assuming BB ⊆ B, we showed that B = B^+, because B ⊆ B^+ and B^+ ⊆ B (since B contains all elements from B^n where n goes from 1 to infinity). Hence, we have proved the equivalence: B = B^+ ⟺ BB ⊆ B.
See the step by step solution

## Step 1: Assuming B = B^+

To prove that BB ⊆ B if B = B^+, let's first analyze what it means for a language B to be equal to its concatenation closure B^+. By definition, the concatenation closure is the smallest superset of B that includes concatenations of any length of words from B. Formally, $$B^+ = \bigcup_{n=1}^{\infty} B^n$$, where each B^n is the n-fold concatenation of the language B with itself, and B^1 = B. Since B = B^+, it contains all concatenations of any length of words from B. Therefore, the concatenation of B with itself, denoted as BB or B^2, must also be a part of B. In other words, BB ⊆ B.

## Step 2: Assuming BB ⊆ B

To prove that B = B^+ if BB ⊆ B, let's analyze what it means for BB to be a subset of B. By definition, every element (word) within BB must also be an element of B. Since BB ⊆ B, we know that every element in B^2 (the concatenation of B with itself) is also in B. Now, let's consider the concatenation closure B^+ again. We know that $$B^+ = \bigcup_{n=1}^{\infty} B^n$$ and that B^1 = B. It's clear that B ⊆ B^+ because B^+ includes all concatenations of any length of words from B. Since we've shown that BB = B^2 ⊆ B, and B ⊆ B^+, we can now confidently state that B = B^+. Specifically, B ⊆ B^+ and B^+ ⊆ B (since B contains all elements from B^n where n goes from 1 to infinity). Thus, we have shown that if BB ⊆ B, then B = B^+. By demonstrating both directions (B = B^+ implies BB ⊆ B, and BB ⊆ B implies B = B^+), we have proved the given equivalence: B = B^+ ⟺ BB ⊆ B.

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