CAR- & Weyl-Algebra: Uniqueness? Given a Hilbert space: $\mathcal{h}$

Consider representations of the CAR-algebra: $\mathcal{A}_\text{CAR}^{(\prime)}(\mathcal{h})$

In Bratelli &amp; Robinson it is stated the uniqueness:
$$\Phi:\mathcal{A}_\text{CAR}(\mathcal{h})\to\mathcal{A}_\text{CAR}'(\mathcal{h})$$

My guess would be: $\Phi[a(f)]:=a'(f)$

Extended to the algebra via:
$$\Phi[A+B]:=\Phi[A]+\Phi[B],\,\Phi[\lambda A]:=\lambda\Phi[A]$$
$$\Phi[AB]:=\Phi[A]\Phi[B],\,\Phi[A^*]:=\Phi[A]^*,\Phi[1]:=1$$
Especially, it is an isometry then: $\|\Phi[A]\|=\|A\|$

Similarly for the Weyl algebra: $\mathcal{W}^{(\prime)}(\mathcal{h})$

Why do these construction work?
Ok, I think I got it now...

Both work perfectly fine as they are always nondegenerate:
$$\mathcal{A}_\text{CAR}:\quad a(f\neq0)\neq0$$
$$\mathcal{W}:\quad W(f)\neq0$$

A counterexample is provided by the angular momentum algebra:
$$\mathcal{J}:\quad [J_i,J_j]=\imath\varepsilon_{ijk}J_k$$
There one has one trivial representation: $J_x=J_y=J_z=0$
Ok, I think I got it now...

Both work perfectly fine as they are always nondegenerate:
$$\mathcal{A}_\text{CAR}:\quad a(f\neq0)\neq0$$
$$\mathcal{W}:\quad W(f)\neq0$$

A counterexample is provided by the angular momentum algebra:
$$\mathcal{J}:\quad [J_i,J_j]=\imath\varepsilon_{ijk}J_k$$
There one has one trivial representation: $J_x=J_y=J_z=0$