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Proof step 5 from proof ebm-cons-above-fillcount (previous step)

avols[v][m] = mapped(buf, m0), m < # avols[v], m0 + m1 < LEB_SIZE, v ∈ avols, ∀ v. v ∈ avols → ebm-cons(avols[v]) ⊦ buf[m1 + m0] = empty

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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