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Proof step 1 from proof abs-fillcount-empty-echdr (conclusion)

⊦ n < # pebs ∧ apebs[n] = mkpeb(empty, avhdr, buf, m, false) ∧ pebs[n] = mkpeb(buf0, m0, false) ∧ abs(pebs, apebs) → m0 = 0

System applied rule normalize

And obtained the following new goals

Rule argument

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