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Proof step 47 from proof prepost-read-superblock (previous step)

n + m ≤ # buf, n0 = # buf ⊦ subarray(buf, n, m) = buf0 ↔ (∃ buf1, buf2. buf = buf1 + buf0 + buf2 ∧ # buf1 = n ∧ # buf2 = n0 - n + m)

A spec-lemma