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Proof step 8 from proof valid-wl-use (previous step)

# apebs = # wlar, apebs[n, mkpeb(aehdr, avhdr, buf, m0, boolvar)][n0] = mkpeb(aehdr0, avhdr0, buf0, m1, boolvar0), apebs[n0] = mkpeb(aehdr1, avhdr1, buf1, m2, boolvar1), n0 < # wlar, n < # wlar, valid-peb(mkpeb(aehdr, avhdr, buf, m0, boolvar)), valid-sqns(avhdr.vol, avhdr.leb, apebs) ≠ ∅ → valid-sqns(avhdr.vol, avhdr.leb, apebs).max < avhdr.sqn, ∀ n. n < # wlar → (wlar[n].status = erroneous ↔ apebs[n].bad) ∧ (wlar[n].status = free → apebs[n].vidhdr = empty) ∧ (wlar[n].status ≠ erroneous ∧ wlar[n].status ≠ erasing → apebs[n].echdr = aechdr(wlar[n].ec)) ∧ (wlar[n].status = used → (∃ v, m. valid-peb(v, m, apebs[n]) ∧ apebs[n].vidhdr.sqn = valid-sqns(v, m, apebs).max)), ∀ m. m < # wlar ∧ m ≠ n ∧ wlar[m].status = used → apebs[m].vidhdr.vol ≠ avhdr.vol ∨ apebs[m].vidhdr.leb ≠ avhdr.leb ⊦ (wlar[n, wl-entry(aehdr.ec, used)][n0].status = erroneous ↔ boolvar0) ∧ (wlar[n, wl-entry(aehdr.ec, used)][n0].status = free → avhdr1 = empty) ∧ ( wlar[n, wl-entry(aehdr.ec, used)][n0].status ≠ erroneous ∧ wlar[n, wl-entry(aehdr.ec, used)][n0].status ≠ erasing → aehdr0 = aechdr(wlar[n, wl-entry(aehdr.ec, used)][n0].ec)) ∧ ( wlar[n, wl-entry(aehdr.ec, used)][n0].status = used → (∃ v, m. valid-peb(v, m, mkpeb(aehdr0, avhdr0, buf0, m1, wlar[n, wl-entry(aehdr.ec, used)][n0].status = erroneous)) ∧ avhdr0.sqn = valid-sqns(v, m, apebs[n, mkpeb(aehdr, avhdr, buf, m0, boolvar)]).max))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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