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Proof step 6 from proof valid-wl-write-valid (previous step)

# apebs = # wlar, apebs[m] = mkpeb(aehdr, avhdr, buf0, m0, boolvar0), wlar[m] = wl-entry(m2, used), apebs[n] = mkpeb(aehdr0, avhdr0, buf, m3, boolvar), n < # wlar, m < # wlar, valid-peb(mkpeb(aehdr, avhdr, buf0, m0, boolvar0)), valid-peb(mkpeb(aehdr, avhdr, buf1, m1, boolvar1)), ∀ n. n < # wlar → (wlar[n].status = erroneous ↔ apebs[m, mkpeb(aehdr, avhdr, buf1, m1, boolvar1)][n].bad) ∧ (wlar[n].status = free → apebs[m, mkpeb(aehdr, avhdr, buf1, m1, boolvar1)][n].vidhdr = empty) ∧ (wlar[n].status ≠ erroneous ∧ wlar[n].status ≠ erasing → apebs[m, mkpeb(aehdr, avhdr, buf1, m1, boolvar1)][n].echdr = aechdr(wlar[n].ec)) ∧ ( wlar[n].status = used → (∃ v, m0. valid-peb(v, m0, apebs[m, mkpeb(aehdr, avhdr, buf1, m1, boolvar1)][n]) ∧ apebs[m, mkpeb(aehdr, avhdr, buf1, m1, boolvar1)][n].vidhdr.sqn = valid-sqns(v, m0, apebs[m, mkpeb(aehdr, avhdr, buf1, m1, boolvar1)]).max)) ⊦ (wlar[n].status = erroneous ↔ boolvar) ∧ (wlar[n].status = free → avhdr0 = empty) ∧ (wlar[n].status ≠ erroneous ∧ wlar[n].status ≠ erasing → aehdr0 = aechdr(wlar[n].ec)) ∧ (wlar[n].status = used → (∃ v, m. valid-peb(v, m, apebs[n]) ∧ avhdr0.sqn = valid-sqns(v, m, apebs).max))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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