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Proof step 17 from proof refine-unmap (previous step)

# apebs = # wlar, vols[v][vtblpnum2] = ⌜n⌝ , vtblpnum = vtblpnum2, avols[v][vtblpnum2].mapped?, vtblpnum2 < # avols[v], vtblpnum2 < # vols[v], n < # wlar, abs(avols, apebs, vols, q), bflipset-cons(bflipset, wlar), ebm-cons(avols), ubi-io-inv(apebs), unique-sqns(apebs), valid-ebatbl(apebs, wlar, vols, vtblpnum1), valid-eraseq(apebs, q), valid-eraseq(wlar, q), valid-eraseq(q), valid-global-sqn(apebs, sqnum), valid-vtbl(apebs, vols, wlar, vtblpnum1), valid-wl(apebs, wlar), v ∈ avols, v ∈ vols ⊦ ⟨avols[v][vtblpnum] := unmapped⟩ abs(avols, apebs, vols[v, vols[v][vtblpnum2, unmapped]], q + eq-entry(vols[v][vtblpnum2].pnum, v × vtblpnum2) '), VTBL_VOLID ∈ vols

Heuristic symbolic execution applied rule assign right

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ v ∈ vols ∧ m < # vols[v] ∧ abs(avols, apebs, vols, q) → m < # avols[v]; (ss)% ⊦ v ∈ vols ∧ abs(avols, apebs, vols, q) → v ∈ avols; (ss)% ⊦ ⌜m⌝ .pnum = m; (s)%⊦ v ∈ vols ∧ m < # vols[v] ∧ vols[v][m] = ⌜n⌝ ∧ abs(avols, apebs, vols, q) → abs (avols[v, avols[v][m, unmapped]], apebs, vols[v, vols[v][m, unmapped]], q + eq-entry(n, v × m) '); (ss)%a: ⊦ (q + q0) + q1 = q + q0 + q1; from specification list[erasequeue-base]