Yes. Bell inequality relies on energy to do the work of non-local correlation. The cost to resolve entangled spin states is one unit of gravitational energy, which allows A to resolve B's spin, and one Hawking photon as a receipt for a cosmic horizon Planck area erasure. Everything conserves.

(S+, E-)A -> (S-, E+)B = 0

Where the first particle to be measured and whose cosmic horizon handles the entropy is labeled A.

E- negative energy bit on the cosmic horizon that resolves B's spin and spin locks it until the next measurement. Local emission means spins were resolved on the cosmic horizon.

S+ is the first to be measured emergent spin state, set by the observer particle that measured it, and this spin state sets S- to opposite of whatever A shows.

E+ the Hawking photon, at the same energy level as spin and gravity, emitted into A's observable universe to conserve the now smaller cosmic horizon, one unit smaller in volume, one extra bit of energy emitted into space to conserve total energy.

So, Bell's inequality is a function of E- and E+ and two measurements. The particles resolve spin on the holoraphic horizon because all particle spin information is holgographically encoded on cosmic horizons, so no matter how far apart A and B are in 3-D space, on the cosmic horizon all particles are adjaccent to each other. Entanglement energy causes A and B to load up virtual gravity and hawking photon bits, and the first particle measured gets to use that entanglement energy to pull up a bit of gravity from the horizon and resolve B's spin, and B emits a Hawking photon to balance out the equation. See how everything conserves? The spins conserve from S+ and S-. We have conservation of information, energy and entropy. Every spooky action at a distance event costs one unit of gravity, one Planck area of the particles cosmic horizon, but the Hawking photon conserves.