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Proof step 42 from proof prepost-init-vols-mappings (previous step)

# wlar = # apebs, avhdr.sqn = valid-sqns(VTBL_VOLID, VTBL_LNUM, apebs).max, avhdr0.sqn = avhdr.sqn, wlar[n] = wl-entry(m0, used), wlar[n0] = wl-entry(m4, used), apebs[n0] = mkpeb(aechdr(m4), avhdr0, buf0, m1, false), apebs[n] = mkpeb(aechdr(m0), avhdr, buf, m, false), recs0[VTBL_VOLID × VTBL_LNUM] = recovery-entry(n0, m3), avhdr0.valid?, n < # apebs, n0 < # apebs, bflipset-cons(bflipset, wlar), is-encoded(to-vtbl(vols), buf), unique-sqns(apebs), unmapped-vols(vols), valid-eraseq(wlar, q), valid-eraseq-stronger(apebs, q), valid-peb(VTBL_VOLID, VTBL_LNUM, mkpeb(aechdr(m4), avhdr0, buf0, m1, false)), valid-peb(VTBL_VOLID, VTBL_LNUM, mkpeb(aechdr(m0), avhdr, buf, m, false)), VTBL_VOLID × VTBL_LNUM ∈ recs0, n < # apebs ∧ n0 < # apebs ∧ n ≠ n0 ∧ valid-peb(VTBL_VOLID, VTBL_LNUM, apebs[n]) ∧ valid-peb(VTBL_VOLID, VTBL_LNUM, apebs[n0]) → (∃ bag. valid-sqns(VTBL_VOLID, VTBL_LNUM, apebs) = bag ++ apebs[n].vidhdr.sqn ++ apebs[n0].vidhdr.sqn) ⊦ n0 = n, n ∈ q, VTBL_VOLID ∈ vols

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 43:
# wlar = # apebs, avhdr0.sqn = avhdr.sqn, avhdr.sqn = (avhdr.sqn < (bag = ∅ ⊃ avhdr.sqn;(avhdr.sqn < bag.max ⊃ bag.max;avhdr.sqn)) ⊃ (bag = ∅ ⊃ avhdr.sqn;bag.max);avhdr.sqn), recs0[VTBL_VOLID × VTBL_LNUM] = recovery-entry(n0, m3), apebs[n] = mkpeb(aechdr(m0), avhdr, buf, m, false), apebs[n0] = mkpeb(aechdr(m4), avhdr0, buf0, m1, false), wlar[n0] = wl-entry(m4, used), wlar[n] = wl-entry(m0, used), valid-sqns(VTBL_VOLID, VTBL_LNUM, apebs) = bag ++ avhdr.sqn ++ avhdr.sqn, avhdr0.valid?, n0 < # apebs, n < # apebs, bflipset-cons(bflipset, wlar), is-encoded(to-vtbl(vols), buf), unique-sqns(apebs), unmapped-vols(vols), valid-eraseq(wlar, q), valid-eraseq-stronger(apebs, q), valid-peb(VTBL_VOLID, VTBL_LNUM, mkpeb(aechdr(m0), avhdr, buf, m, false)), valid-peb(VTBL_VOLID, VTBL_LNUM, mkpeb(aechdr(m4), avhdr0, buf0, m1, false)), VTBL_VOLID × VTBL_LNUM ∈ recs0 ⊦ n0 = n, VTBL_VOLID ∈ vols, n ∈ q

Rule argument

Additional information

Used simplifier rules: ls-max: ⊦ (bag ++ n).max = (bag = ∅ ⊃ n;(bag.max > n ⊃ bag.max;n));% ⊦ ¬ n < n; (s)% ⊦ m > n ↔ n < m; (s)% ⊦ ∅ ≠ bag ++ n; (s)% ⊦ mkpeb(aehdr, avhdr, buf, m, boolvar).vidhdr = avhdr; (s)%⊦ valid-peb(v0, m0, apeb) → (valid-peb(v1, m1, apeb) ↔ v0 = v1 ∧ m0 = m1); (s)