Is Quantum Measurement Projective in GHZ States?

[neu]

Homework Statement

Have two Greenberger-Horne-Zeilinger (GHZ) states of qubits A, B, C and
D, E, F as follows:

\mid GHZ \rangle_{ABC} = \frac{1}{\sqrt{2}} \left( \mid 0 \rangle_{A}\mid 0 \rangle_{B}\mid 0 \rangle_{C} + \mid 1 \rangle_{A}\mid 1 \rangle_{B}\mid 1 \rangle_{C}\right)

and

\mid GHZ \rangle_{DEF} = \frac{1}{\sqrt{2}} \left( \mid 0 \rangle_{D}\mid 0 \rangle_{E}\mid 0 \rangle_{F} + \mid 1 \rangle_{D}\mid 1 \rangle_{E}\mid 1 \rangle_{F}\right)

If you perform a measurement in the Bell basis on the qubits A and D, and obtain
the outcome: \mid \Psi^{+} \rangle_{AD} = \frac{1}{\sqrt{2}} \left(\mid 0 \rangle_{A}\mid 1 \rangle_{D} + \mid 1 \rangle_{A}\mid 0 \rangle_{D}\right)

Write down the state to which qubits B, C, D and F are projected?

Attempted Solution

Total state is : \mid GHZ \rangle_{ABCDEF}= \mid GHZ \rangle_{ABC} \mid GHZ \rangle_{DEF}

Projector operator for measurement of A and D is \mid Bell \rangle \langle Bell \mid_{AD}

so \mid Bell \rangle \langle Bell \mid_{AD} \mid GHZ \rangle_{ABCDEF} = \mid Bell \rangle_{AD} \mid \Psi^{+} \rangle_{AD} \mid \Psi^{+} \rangle_{BCEF}; is this right?

Here I get confused. I think the method is simply:

\langle \Psi^{+} \mid_{AD}\mid GHZ \rangle_{ABCDEF} = \mid \Psi^{+} \rangle_{BCEF}

But if so, why?

 

[neu]

one bump then I'll give up.

I'm still stuck on this one, no takers?