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Proof step 5 from proof crash-neutral-unmap (previous step)

# lpt = # avol, adr = adr0, m = m1, m0 = m2, ns = ns0, plog = plog0, m2 < # avol, ebm-vol-inv(avol), avol ⊆ avol0 ⊦ ⟨avol[m0] := unmapped⟩ (avol ⊆ avol0)

Heuristic symbolic execution applied rule assign right

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ avol ⊆ avol0 ∧ ebm-vol-inv(avol) → avol[m, unmapped] ⊆ avol0; (s)