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Proof step 23 from proof inv-recover (previous step)

# apebs0 = # wlar1, err = EINVAL, vols = ∅, vtblpnum2 = LEB_SIZE, bflipset-cons(bflipset1, wlar1), ubi-io-inv(apebs0), unique-sqns(apebs0), valid-ebatbl(apebs0, wlar1, vols1, vtblpnum1), valid-eraseq(apebs0, q1), valid-eraseq(wlar1, q1), valid-eraseq(q1), valid-global-sqn(apebs0, vtblpnum5), valid-vtbl(apebs0, vols1, wlar1, vtblpnum1), valid-wl(apebs0, wlar1), err0 = ESUCCESS → # wlar1 = # wlar ∧ ¬ dups(nx) ∧ nx < # wlar1 ∧ indices-erasing(wlar, nx) ∧ valid-global-sqn(apebs0, sqnum) ∧ valid-eraseq(wlar, q) ∧ bflipset-cons(bflipset, wlar) ∧ valid-eraseq-stronger(apebs0, q) ∧ valid-wl(apebs0, wlar) ∧ valid-recs-sqns(apebs0, recs0) ∧ valid-recs(apebs0, wlar, recs0), err0 = ESUCCESS ∨ ⌊err0⌋ ⊦ ⟪skip⟫ ( ( err = ESUCCESS → # wlar1 = # wlar ∧ valid-global-sqn(apebs0, sqnum) ∧ valid-wl(apebs0, wlar) ∧ valid-eraseq(q) ∧ valid-eraseq(wlar, q) ∧ valid-eraseq(apebs0, q) ∧ ¬ VTBL_VOLID ∈ vols ∧ bflipset-cons(bflipset, wlar) ∧ valid-ebatbl(apebs0, wlar, vols, vtblpnum3) ∧ valid-vtbl(apebs0, vols, wlar, vtblpnum3) ∧ vtblpnum2 = LEB_SIZE) ∧ (err = ESUCCESS ∨ ⌊err⌋ ) ∧ (err = ESUCCESS → is-max(apebs0, vols, wlar, q, bflipset, vtblpnum3, sqnum) ∧ q.lebrefs ⊆ q1.lebrefs ∧ vols1 ⊆ vols ∧ vols⌈vols1, q1⌉)), VTBL_VOLID × VTBL_LNUM ∈ recs0, VTBL_VOLID ∈ vols1, vtblpnum1 ∈ q1

Heuristic symbolic execution applied rule skip right

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ ESUCCESS ≠ EINVAL; (s)% ⊦ ⌊EINVAL⌋ ; (s)