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windsylph
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Discrete Math: Formal Languages: Suppose that A is a subset of V*, where V is an alphabet. Prove that if A = A^2, then the empty string is in A.
First, my question is that how can A = A^2?
 2 years ago
 2 years ago
windsylph Group Title
Discrete Math: Formal Languages: Suppose that A is a subset of V*, where V is an alphabet. Prove that if A = A^2, then the empty string is in A. First, my question is that how can A = A^2?
 2 years ago
 2 years ago

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satellite73 Group TitleBest ResponseYou've already chosen the best response.0
what is an alphabet?
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
oh i see and multiplication is given by concatenation. so i guess your alphabets have to be either empty or each string is infinitely long
 2 years ago

windsylph Group TitleBest ResponseYou've already chosen the best response.0
Yes, well each string does not have to be infinitely long. It can be any length of string. Here's my response, please review it if you'd like: False if $A = \emptyset$, but true otherwise. First, if $A = \emptyset$, then the premise is true, but the conclusion is false, since $\lambda \notin \emptyset$ and so $\lambda \notin A$. Therefore the claim is false for $A = \emptyset$. Second, if $A \neq \emptyset$, and if $A = \left\{\lambda\right\}$, then $A = A^2$ indeed and obviously, $\lambda \in A$. Thus the claim is true for this case. Lastly, for all other cases, the premise is always false, since $A$ can never be equal to $A^2$ unless $A = \emptyset$ or $A = \lambda$. Thus, the claim is vacuously true if and only if $A \neq \emptyset$.
 2 years ago

windsylph Group TitleBest ResponseYou've already chosen the best response.0
Correction on last statement Thus, the claim is vacuously true for all other cases, and so the claim is true except when A = \[\emptyset\]
 2 years ago
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