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Proof step 1 from proof abs-bad (conclusion)

⊦ n < # apebs ∧ abs(pebs, apebs) ∧ pebs[n] = mkpeb(buf0, m0, boolvar0) ∧ apebs[n] = mkpeb(aehdr, avhdr, buf, m1, boolvar1) → (boolvar0 ↔ boolvar1)

System applied rule normalize

And obtained the following new goals

Rule argument

Additional information