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Proof step 31 from proof ubi-cons-same-vtbl (previous step)

# wlar0 = # wlar1, # apebs = # wlar1, avhdr0.sqn = avhdr.sqn, wlar0[n0] = wl-entry(m4, used), apebs[n0] = mkpeb(aechdr(m4), avhdr0, buf0, m3, false), apebs[m0] = mkpeb(aechdr(m2), avhdr, buf, m, false), wlar1[m0] = wl-entry(m2, used), valid-sqns(VTBL_VOLID, VTBL_LNUM, apebs) = bag ++ avhdr.sqn ++ avhdr.sqn, n0 < # wlar1, m0 < # wlar1, bflipset-cons(bflipset1, wlar1), bflipset-cons(bflipset0, wlar0), is-encoded(to-vtbl(vols0), buf0), is-encoded(to-vtbl(vols1), buf), valid-ebatbl(apebs, wlar1, vols1, m0), valid-ebatbl(apebs, wlar0, vols0, n0), valid-eraseq(q1), valid-eraseq(wlar1, q1), valid-eraseq(apebs, q1), valid-eraseq(q0), valid-eraseq(wlar0, q0), valid-eraseq(apebs, q0), valid-global-sqn(apebs, m1), valid-global-sqn(apebs, n1), valid-peb(VTBL_VOLID, VTBL_LNUM, mkpeb(aechdr(m4), avhdr0, buf0, m3, false)), valid-peb(VTBL_VOLID, VTBL_LNUM, mkpeb(aechdr(m2), avhdr, buf, m, false)), (bag ++ avhdr.sqn).max ≤ avhdr.sqn, ∀ v, m. unique-elems(valid-sqns(v, m, apebs)) ⊦ n0 = m0, VTBL_VOLID ∈ vols1, VTBL_VOLID ∈ vols0

User applied rule all left

And closed the goal!

Rule argument

pos: left 28 terms: VTBL_VOLID, VTBL_LNUM abort: F discard: T computed: F

Additional information

Used simplifier rules: ⊦ unique-elems(bag0) → bag0 ≠ bag1 ++ n ++ n; (s) Used terms: VTBL_VOLID, VTBL_LNUM