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Proof step 6 from proof inv-erase-worker (previous step)

# apebs = # wlar, q ≠ [], bflipset-cons(bflipset, wlar), ubi-io-inv(apebs), unique-sqns(apebs), valid-ebatbl(apebs, wlar, vols, vtblpnum), valid-eraseq(q), valid-eraseq(wlar, q), valid-eraseq(apebs, q), valid-global-sqn(apebs, sqnum), valid-vtbl(apebs, vols, wlar, vtblpnum), valid-wl(apebs, wlar), ¬ vtblpnum ∈ q, ¬ VTBL_VOLID ∈ vols ⊦ ⟪ubi_erase_worker_helper#(q.head.pnum; wlar, apebs; err); if err = ESUCCESS then q := q.tail⟫ ( # apebs = # wlar ∧ unique-sqns(apebs) ∧ valid-global-sqn(apebs, sqnum) ∧ valid-wl(apebs, wlar) ∧ valid-eraseq(q) ∧ valid-eraseq(wlar, q) ∧ valid-eraseq(apebs, q) ∧ bflipset-cons(bflipset, wlar) ∧ valid-ebatbl(apebs, wlar, vols, vtblpnum) ∧ valid-vtbl(apebs, vols, wlar, vtblpnum) ∧ ubi-io-inv(apebs))

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Rule argument

spec: list-data, inst: erasequeue-base name: elim-list-c seq: q ≠ [] ⊦ eqent = q.head ∧ q0 = q.tail ↔ q = eqent + q0 subst: [q, eqent, q0] → [q, eqent, q0]

Additional information

Used simplifier rules: q ≠ [] ⊦ eqent = q.head ∧ q0 = q.tail ↔ q = eqent + q0; (elim) Used terms: q, eqent, q0