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Proof step 47 from proof recovery (previous step)

# avols0[v] = # vols0[v], # apebs0 = # wlar0, # wlar = # wlar0, # vols[v] = # vols0[v], vols0[v][m] = unmapped, avols0[v][m] = erased, apebs0[unmapped.pnum] = mkpeb(aehdr, avhdr, buf, m0, boolvar), vols[v][m] = unmapped, m < # vols0[v], bflipset-cons(bflipset, wlar), bflipset-cons(bflipset0, wlar0), ebm-cons(avols0), is-max(apebs0, vols0, wlar0, q0, bflipset0, vtblpnum3, sqnum), ubi-io-inv(apebs0), unique-sqns(apebs0), valid-ebatbl(apebs0, wlar, vols, vtblpnum), valid-ebatbl(apebs0, wlar0, vols0, vtblpnum3), valid-eraseq(q), valid-eraseq(wlar, q), valid-eraseq(apebs0, q), valid-eraseq(q0), valid-eraseq(wlar0, q0), valid-eraseq(apebs0, q0), valid-global-sqn(apebs0, vtblpnum0), valid-global-sqn(apebs0, sqnum), valid-vtbl(apebs0, vols, wlar, vtblpnum), valid-vtbl(apebs0, vols0, wlar0, vtblpnum3), valid-wl(apebs0, wlar), valid-wl(apebs0, wlar0), v ∈ vols0, v ∈ vols, v ∈ avols0, v × m ∈ q0, vols0⌈vols, q⌉, ∀ lref. lref ∈ q0 ∧ lref ≠ none → lref ∈ q ⊦ v × m ∈ q, VTBL_VOLID ∈ vols0, VTBL_VOLID ∈ vols

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 35 terms: v × m

Additional information

Used simplifier rules: ⊦ none ≠ v × m; (s) Used terms: v × m