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Proof step 16 from proof embed-split (
previous step
)
⊦ (P0 * P1)[rs, f] ↔ (∃ rs0, rs1. rs0 ∪ rs1 = rs ∧ rs0 ∩ rs1 = ∅ ∧ P0[rs0, f] ∧ P1[rs1, f])
A spec-lemma