How can i know the full duplex does work from the Shannon's theorem?
You think the Shannon-Hartley Theorem is in conflict with Full Duplex (in the sense you describe), and luckily it's not!

The problem really is that your $\eqref{SNR}$ formula isn't relevant for the Shannon-Hartley Theorem (SHT). With respect to the SHT, your $t_A$ isn't noise!

That's easy to explain. SHT makes a statement about the channel capacity $C$, i.e. the maximum mutual information between $B$ and $A$, given the best possible source symbol probability distribution:

$$\label{C} C= \max_{P(B)} I(B;A)$$

Mutual information between the information source $B$ and the information sink $A$ is, in the end, nothing but the statement


  How much uncertainty about $B$ can I remove by observing $A$?


Now, even from reading that, you'll notice that in a perfect transceiver $A$, you simply don't add any uncertainty by adding your own transmission – you know what you're sending, and thus could simply subtract it from what you observe!

Formally, it's easy to show that

\begin{align}
I(B;A) &amp;= H(B) - H(B|A)\\
&amp;=H(t_B) - H(t_B|r_A)\\
&amp;=H(t_B) - H(t_B|t_B+Z+t_A)\text.\label{crucial}\tag{a}
\end{align}

Now, that $\eqref{crucial}$ line is crucial – what's written there literally means


  The amount of information you get about the transmit signal $t_B$ by observing the received signal $r_a=t_B+Z+t_A$ is exactly the expected amount of information (==entropy) of $t_B$, minus the remaining uncertainty about $t_B$ given that you've observed $r_B$, which is the sum of A's own signal $t_A$, independent noise $Z$ and our signal of interest, $B$'s $t_B$.


And that conditional entropy is what the influence of your noise it. Let's look at it a bit closer, $r_A$ be the receive signal at $A$

$$r_A=t_B + Z  + t_A,$$

with $Z$ being your "true" noise.

And hence, since these three are independent:

$$H(r_A) = H(t_B) + H(Z) + H(t_A)\text.$$

Little problem: We're at the end $A$. So, there's nothing uncertain about $t_A$ (we're the one sending that!) and the formula reduces to 

$$H(r_A)_A = H(t_B) + H(Z)\text.\tag{ö}\label{known}$$

(Read $\eqref{known}$ as "The entropy of the received signal $r_A$, being $A$, i.e. knowing $t_A$ already").

Automatically that means that $\eqref{crucial}$ also simplifies when we, as $A$,  know our own transmit signal $t_A$:

\begin{align}
I(B;A)_A 
&amp;= H(t_B) - H(t_B|t_B+Z+t_A,t_A)\\
&amp;= H(t_B) - H(t_B|t_B + Z)\label{read}\tag{r}\\
&amp;= H(t_B) - H(Z)\\
&amp;= H(B) - H(Z)&amp;\rule{0.5em}{0.5em}
\end{align}

The trick is that in line $\eqref{read}$, the second term is "the amount of uncertainty about one element of a sum, given the value of the sum", and that simply reduces to the amount of uncertainty in the other element.

When you look at that, you'll notice that your Full-Duplex mutual information formula is the same as your (normal) simplex mutual information formula. Hence, the capacity is the same.

In other words: For Shannon-Hartley, any operation that you can mathematically reverse as the receiver, like adding a signal you know, makes absolutely no difference for the channel capacity. Yay!

The fact that there's nevertheless a lot of things to calculate for the Full Duplex use case is because adding a strong signal to a received signal is typically not fully reversible, because the two amplitudes physically don't add up exactly, due to imperfections of the real world (receive amplifier saturation, changed quantization noise, feedback path uncertainties…). But: Your simple SNR mode $\eqref{SNR}$ can't describe that, so it's neither suitable for the ideal case I described above, nor suitable for the non-ideal case. So, if you want to understand in detail why Full Duplex works, and what the limitations are: You'll need to learn quite a bit about information theory!