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Proof step 18 from proof lf-valid-eraseq-stronger-notin-recs (previous step)

# apebs = # wlar, avhdr.sqn = valid-sqns(v0, m1, apebs).max, wlar[n] = wl-entry(m2, used), apebs[n] = mkpeb(aechdr(m2), avhdr, buf, m0, false), n < # wlar, valid-eraseq(wlar, q), valid-eraseq-stronger(apebs, q), valid-peb(v0, m1, mkpeb(aechdr(m2), avhdr, buf, m0, false)), ∀ ladr, m. ladr ∈ recs ∧ recs[ladr].pnum = m ↔ wlar[m].status = used ∧ m < # wlar ∧ apebs[m].vidhdr.vol = ladr.vol ∧ apebs[m].vidhdr.leb = ladr.leb ⊦ avhdr = empty, n ∈ q, v0 × m1 ∈ recs

User applied rule all left

And closed the goal!

Rule argument

pos: left 9 terms: v0 × m1, n abort: F discard: T computed: T

Additional information

Used simplifier rules: ⊦ (v × m).vol = v; (s)% ⊦ (v × m).leb = m; (s)% ⊦ wl-entry(m, stat).status = stat; (s)% ⊦ mkpeb(aehdr, avhdr, buf, m, boolvar).vidhdr = avhdr; (s)%⊦ valid-peb(v, m, mkpeb(aehdr, avhdr, buf, m0, boolvar)) → avhdr.vol = v; (s)%⊦ valid-peb(v, m, mkpeb(aehdr, avhdr, buf, m0, boolvar)) → avhdr.leb = m; (s) Used terms: v0 × m1, n