For the language $\{w \mid w \in \Sigma^*, w \neq \lambda\}$, is the following regular expression correct?

For the language $\{w \mid w \in \Sigma^*, w \neq \lambda\}$, is the following regular expression correct?

$(a+b)^+$ . Which is $(a+b)$ to the power of $+$ which I think accepts all combinations of $a$ and $b$ except for the empty string.

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Sorry, that was a typo. I've fixed the question except I don't know what the format is for the not equal to symbol. So, you can read that as w is not equal to the empty string. –  zeqof Oct 18 '12 at 7:24
You are correct. –  martini Oct 18 '12 at 7:28

Some people don't allow $r^+$ in their definition of regular expressions, in which case you'd have to settle for using $(\mathtt{a}+\mathtt{b})(\mathtt{a}+\mathtt{b})^*$, but otherwise, I'd agree with May: you got it.