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Proof step 79 from proof btree-clean-fwd-strong (
previous step
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⊦ n < m ∧ m ≤ # zbrar0 ∧ # zbrar0 = # zbrar1 ∧ (∀ n0. n0 ≤ n ∧ n0 ≤ m → zbrar0[n0].key = zbrar1[n0].key) → key-range(zbrar0, n, m, ri) = key-range(zbrar1, n, m, ri)
A spec-lemma