And it seems we do actually need AssumeS1(C)&AssumeS2(C)≃AssumeS1∪S2(C) in the proof to justify:
Thus it suffices to show that (D1&1R2∪R3)⊗(D2&1R1∪R3)≃D1&D2&1R3.
Without it, we have to show (D1&1R2∪R3)⊗(D2&1R1∪R3)≃AssumeS1∪S2(C)&1R3 instead.
UPDATE: I was able to prove AssumeS1(C)&AssumeS2(C)≃AssumeS1∪S2(C) in general whenever S1 and S2 are disjoint and both in Obs(C), with help from Rohin Shah, following the “restrict attention to world S1∪S2” approach I hinted at earlier.
And it seems we do actually need AssumeS1(C)&AssumeS2(C)≃AssumeS1∪S2(C) in the proof to justify:
Without it, we have to show (D1&1R2∪R3)⊗(D2&1R1∪R3)≃AssumeS1∪S2(C)&1R3 instead.
UPDATE: I was able to prove AssumeS1(C)&AssumeS2(C)≃AssumeS1∪S2(C) in general whenever S1 and S2 are disjoint and both in Obs(C), with help from Rohin Shah, following the “restrict attention to world S1∪S2” approach I hinted at earlier.