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Write $c=⟨e_0|e_1⟩=v{e}^{i\alpha }$. The Cauchy-Schwarz inequality gives $0\le v\le 1$ because the two environment states are normalized. Immediately after the interaction, the joint quantum state isApplying the last Hadamard gate givesSince the environment is unobserved, the Born rule uses the squared norm of the environment-valued coefficient of $|0⟩$:Thus the Hadamard interference with environment overlap formula isThe plus sign follows from the explicitly chosen phase gate and the order $⟨e_0|e_1⟩$ of the inner product. Equivalently, before the last Hadamard gate the partial trace over the environment givesThe complex environment overlap in qubit dephasing both reduces the quantum coherence by $v$ and adds a relative quantum phase $\alpha$. Orthogonal environment states have $v=0$ and give $P_0=1/2$ for every $\phi$. If $v=1$, the quantum interference fringe has full contrast and is only shifted; the qubit has not become mixed. For $v=0$, $\alpha$ is immaterial.

Moving the interaction to between the first Hadamard gate and the phase gate givesthe identical joint quantum state. Therefore the expression is unchanged. This is the commutation of computational-basis dephasing and phase gates; it uses the specified population-preserving interaction with the same environment states, rather than an arbitrary model of quantum decoherence.