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Proof step 61 from proof path-split-wand (
previous step
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⊦ dirs ⊥ dirs0 ∧ dirs0 ⊥ dirs1 ∧ dirs1 ⊥ dirs → (dirs ∪ dirs0) ∪ dirs1 = dirs ∪ dirs0 ∪ dirs1
A spec-lemma