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Proof step 240 from proof hasSharing-put-index-refval (
previous step
)
⊦ ¬ (∃ sks0, sks1. sks = sks0 + (sk + sks1) ∧ st[r -- sks0] .refval = r0 ∧ (∀ sks2. sks2 + sk ⊑ sks0 → st[r -- sks2] .refval ≠ r0)) → (st[r0 -- sk, val][r -- sks]) = st[r -- sks]
A spec-lemma