- How important to require users to put details on About Us?
- Is “one” always a quantity if not a pronoun?
- “While” versus “whilst”?
- Is fight always of a negative connotation?
- Sending mails day by day
- Nouns that can be both count and noncount nouns
- From when did the phrase “Africa” become the name of the whole continent?
- Implied or historical volatility to calculate theoretical options price with black scholes?
- Black & Scholes article : option pricing
- Sweden: Base salary with taxable benefits
- Backdating employment start date
- Who can help me understand my credit report?
- Difference between Black-Scholes, Binomial models and Market price in European index options?
- How are stock options priced?
- Why IV and stock price are inversely related
- Effect of company issued options on share price
- How does a delta signify the probability of expiring in the money
- How to calculate the standard deviation of stock returns?
- In option pricing formulas, is the volatility and short rate a decimal or a percentage?
- Can the Delta be used to calculate the option premium given a certain target?

# Does showing a problem and its complement are not Turing-decidable means that the language & its complement are not Turing-recognizable?

I was reading the Sipser's book on the Theory of Computation, 3rd edition and came up with a question. "Does showing a problem and its complement are not Turing-decidable means that the language & its complement are not Turing-recognizable?" I believe that the answer is NO, however, the Theorem 5.30 states something different.

There are two problems concerned in this question. One is $A_{TM} = \{

On Page 238, the Theorem 5.30 is stated as follows:

Theorem 5.30 $EQ_{TM}$ is neither Turing-recognizable nor co-Turing-recognizable.

The proof is by mapping reduction of $A_{TM}$ to $\overline{EQ_{TM}}$, and at the same time, reduction from $A_{TM}$ to $EQ_{TM}$. This way, it has shown:

$\overline{EQ_{TM}}$ is Turing-undecidable.

$EQ_{TM}$ is Turing-undecidable.

Note that this reduction does not show that either of $\overlin