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Proof step 61 from proof inv-create-volume (previous step)

# apebs1 = # wlar0, # apebs = # wlar, vols0 = vols, bflipset-cons(bflipset, wlar0), bflipset-cons(bflipset, wlar), id(apebs1, apebs, mapped-pebs(vols)), ubi-io-inv(apebs1), ubi-io-inv(apebs), unique-sqns(apebs1), unique-sqns(apebs), valid-ebatbl(apebs1, wlar0, vols, vtblpnum3), valid-ebatbl(apebs, wlar, vols, vtblpnum3), valid-eraseq(q1), valid-eraseq(wlar0, q1), valid-eraseq(apebs1, q1), valid-eraseq(apebs, q), valid-eraseq(wlar, q), valid-eraseq(q), valid-global-sqn(apebs1, sqnum), valid-global-sqn(apebs, sqnum), valid-vtbl(apebs1, vols, wlar0, vtblpnum3), valid-vtbl(apebs, vols, wlar, vtblpnum3), valid-wl(apebs1, wlar0), valid-wl(apebs, wlar), 2 * ENCODED_NAT_SIZE + (ENCODED_VOLID_SIZE + ENCODED_NAT_SIZE) * # vols + ENCODED_VOLID_SIZE ≤ LEB_SIZE, q1 ⊆ q, err = ESUCCESS → q1 = [], err = ESUCCESS ∨ ⌊err⌋ ⊦ ⟪skip⟫ ( # apebs1 = # wlar0 ∧ unique-sqns(apebs1) ∧ valid-global-sqn(apebs1, sqnum) ∧ valid-wl(apebs1, wlar0) ∧ valid-eraseq(q1) ∧ valid-eraseq(wlar0, q1) ∧ valid-eraseq(apebs1, q1) ∧ ¬ VTBL_VOLID ∈ vols ∧ bflipset-cons(bflipset, wlar0) ∧ valid-ebatbl(apebs1, wlar0, vols, vtblpnum3) ∧ valid-vtbl(apebs1, vols, wlar0, vtblpnum3) ∧ ubi-io-inv(apebs1) ∧ (err = ESUCCESS ∨ ⌊err⌋ ) ∧ id(apebs1, apebs, mapped-pebs(vols)) ∧ (err = ESUCCESS → ¬ v ∈ vols0 ∧ vols = vols0[v, fill(mkebatbl(n), unmapped)] ∧ q1.lebrefs ⊆ ∅ ++ (VTBL_VOLID × VTBL_LNUM)) ∧ (err ≠ ESUCCESS → q1.lebrefs ⊆ q.lebrefs ∧ vols = vols0)), err = ESUCCESS, v = VTBL_VOLID, vtblpnum3 ∈ q1, vtblpnum3 ∈ q, VTBL_VOLID ∈ vols, v ∈ vols

Heuristic symbolic execution applied rule skip right

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ ⌊err⌋ → err ≠ ESUCCESS; (s)% ⊦ q0 ⊆ q1 → q0.lebrefs ⊆ q1.lebrefs; (ss)