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Proof step 190 from proof prepost-write-once (previous step)

# apebs0 = # wlar1, # apebs2 = # wlar0, q0.head = eq-entry(m4, lref0), LEB_SIZE = (n1 + m0) + m1, vols0[v][m] = unmapped, wlar0[vtblpnum] = wl-entry(m5, free), apebs2[vtblpnum] = mkpeb(aehdr, empty, mkempbuf(LEB_SIZE), 0, false), lref = none, q = q1, wlar = wlar0, aehdr.valid?, vtblpnum < # wlar0, m < # vols0[v], bflipset-cons(bflipset, wlar1), bflipset-cons(bflipset, wlar0), id(apebs2, apebs0, mapped-pebs(vols0)), is-aligned(m1, EB_PAGE_SIZE), is-aligned(m0, EB_PAGE_SIZE), ubi-io-inv(apebs0), ubi-io-inv(apebs2), ubi-io-inv(apebs2[vtblpnum, mkpeb(aehdr, avhdr, mkempbuf(LEB_SIZE), 0, false)]), unique-sqns(apebs0), unique-sqns(apebs2), valid-ebatbl(apebs0, wlar1, vols0, vtblpnum2), valid-ebatbl(apebs2, wlar0, vols0, vtblpnum2), valid-eraseq(apebs0, q0), valid-eraseq(wlar1, q0), valid-eraseq(q0), valid-eraseq(q1), valid-eraseq(wlar0, q1), valid-eraseq(apebs2, q1), valid-global-sqn(apebs0, sqnum), valid-global-sqn(apebs2, sqnum), valid-vtbl(apebs0, vols0, wlar1, vtblpnum2), valid-vtbl(apebs2, vols0, wlar0, vtblpnum2), valid-wl(apebs0, wlar1), valid-wl(apebs2, wlar0), v ∈ vols0, m1 ≤ LEB_SIZE, m1 + m2 ≤ # buf, m0 + m1 ≤ LEB_SIZE, q1.lebrefs ⊆ q0.lebrefs, ⌊err0⌋ , apebs0 = apebs2 ∧ wlar1 = wlar0 ∧ q0 = q1 ∨ id○(apebs0, apebs2, vtblpnum) ∧ id○(wlar1, wlar0, vtblpnum) ∧ q0 = eq-entry(vtblpnum, lref0) ' + q1 ⊦ ⟪wlar[vtblpnum].status := erasing; q := q + eq-entry(vtblpnum, lref)⟫ ( # apebs2 = # wlar ∧ unique-sqns(apebs2[vtblpnum, apebs2[vtblpnum].vidhdr:= avhdr]) ∧ valid-global-sqn(apebs2[vtblpnum, apebs2[vtblpnum].vidhdr:= avhdr], sqnum + 1) ∧ valid-wl(apebs2[vtblpnum, apebs2[vtblpnum].vidhdr:= avhdr], wlar) ∧ valid-eraseq(q) ∧ valid-eraseq(wlar, q) ∧ valid-eraseq(apebs2[vtblpnum, apebs2[vtblpnum].vidhdr:= avhdr], q) ∧ bflipset-cons(bflipset, wlar) ∧ valid-ebatbl(apebs2[vtblpnum, apebs2[vtblpnum].vidhdr:= avhdr], wlar, vols0, vtblpnum2) ∧ valid-vtbl(apebs2, vols0, wlar, vtblpnum2) ∧ id(apebs2[vtblpnum, apebs2[vtblpnum].vidhdr:= avhdr], apebs0, mapped-pebs(vols0)) ∧ q.lebrefs ⊆ q0.lebrefs ++ (v × m) ∧ (m1 = 0 → q.lebrefs ⊆ q0.lebrefs)), vtblpnum2 = vtblpnum, avhdr.valid?, vtblpnum2 ∈ q0, VTBL_VOLID ∈ vols0, vtblpnum2 ∈ q1

Heuristic symbolic execution applied rule assign right

And obtained the following new goals

Rule argument

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