Semantics of E and A operators in CTL*

Question

I'm referring to a book where E and A were defined in text as follows

The formula $A \phi$ states that all the executions out of the current state satisfy property $\phi$ whereas $E\phi$ states that from the current state, there exists and execution satisfying $\phi$

And formally, they were defined as

$\sigma, i \models E\phi $ iff there exists as $\sigma'$ such that $\sigma(0)\ldots\sigma(i) = \sigma'(0)\ldots\sigma'(i)$ and $\sigma', i \models phi$ $\sigma, i \models A\phi $ iff for all $\sigma'$ such that $\sigma(0)\ldots\sigma(i) = \sigma'(0)\ldots\sigma'(i)$, we have $\sigma', i \models phi$

$\sigma(i)$ was previously defined to be the ith state of an execution $\sigma$.

Here's where my confusion lies. Doesn't $\sigma(0)\ldots\sigma(i) = \sigma'(0)\ldots\sigma'(i)$ imply that the executions $\sigma$ and $\sigma'$ have both the same start and end state, therefore $\sigma', i \models \phi$ is same as $\sigma, i \models \phi$? That seems to be wrong.

Answer 1

I believe it means that $\sigma,\sigma$ should share the same prefix up to position $i$. Then the quantification ranges over all possible extensions of this prefix from state $\sigma(i) = \sigma(i)'$.

Answer 2

I think you misinterpret the expression $$\sigma(0) \cdots \sigma(i) = \sigma'(0) \cdots \sigma'(i')$$ and I agree that it is written in an unintuitive way.

The author means that $\sigma(i) = \sigma'(0)$ and that the final path is $$\sigma(0) \cdots \sigma(i) \; \sigma'(0) \cdots \sigma'(i').$$

The equality ensures that the path $\sigma'$ starts where $\sigma$ ends.

Are you sure there isn't a prime on the $i$?