P = a_{i_1,\dots,i_d}$. Hence, we deduce by iterating $$a_{i_1,\dots,i_n} = (\Delta_{i_d} \dots \Delta_{i_1} P)(X) = 0,$$ which finally leads to a contradiction. Therefore, we must have $P=0$, which concludes the proof.

Combining the above Proposition \[key-prop-1\] with Proposition \[key-prop-2\] leads us now directly to the following theorem.

\[main1\] As before, let $(M,\tau)$ be a tracial $W^\ast$-probability space. Let $X_1,\dots,X_n\in M$ be self-adjoint and assume that there are elements $\xi_1,\dots,\xi_n\in L^2(M,\tau)$, such that $(\xi_1,\dots,\xi_n)$ satisfies the conjugate relations for $(X_1,\dots,X_n)$, i.e. holds for $j=1,\dots,n$. Then we have the following statements:

-   $X_1,\dots,X_n$ do not satisfy any non-trivial algebraic relation, i.e. there exists no non-zero polynomial $P\in{\mathbb{C}}\langle Z_1,\dots,Z_n\rangle$ such that $P(X_1,\dots,X_n)=0$.

-   For $j=1,\dots,n$, there is a unique derivation $$\hat{\partial}_j:\ {\mathbb{C}}\langle X_1,\dots,X_n\rangle \rightarrow {\mathbb{C}}\langle X_1,\dots,X_n\rangle \otimes {\mathbb{C}}\langle X_1,\dots,X_n\rangle$$ which satisfies $\hat{\partial}_j(X_i) = \delta_{j,i} 1 \otimes 1$ for $i=1,\dots,n$.

Note that part (b) is an immediate consequence of part (a): Since (a) tells us that the evaluation homomorphism ${\operatorname{ev}}_X$ is in fact an isomorphism, we can immediately define a non-commutative derivation $$\hat{\partial}_j:\ {\mathbb{C}}\langle X_1,\dots,X_n\rangle \rightarrow {\mathbb{C}}\langle X_1,\dots,X_n\rangle \otimes {\mathbb{C}}\langle X_1,\dots,X_n\rangle,$$ where the terminology derivation has to be understood with respect to the ${\mathbb{C}}\langle X_1,\dots,X_n\rangle$-bimodule structure of ${\mathbb{C}}\langle X_1,\dots,X_n\rangle^{\otimes 2}$. The uniqueness can be 