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Proof step 6 from proof valid-wl-next-valid-erase (previous step)

# apebs = # wlar, apebs[n0] = mkpeb(aehdr, avhdr, buf, m2, boolvar), apebs[m] = mkpeb(aehdr0, avhdr0, buf0, m3, boolvar0), m < # wlar, n0 < # wlar, valid-peb(v, n, apebs[m]), n0 ≤ m, valid-sqns(v, n, apebs, m) ≠ ∅ → avhdr0.sqn < valid-sqns(v, n, apebs, m).max, ∀ n0. n0 < m → (wlar[n0].status = erroneous ↔ apebs[n0].bad) ∧ (wlar[n0].status = free → apebs[n0].vidhdr = empty) ∧ (wlar[n0].status ≠ erroneous ∧ wlar[n0].status ≠ erasing → apebs[n0].echdr = aechdr(wlar[n0].ec)) ∧ (wlar[n0].status = used → (∃ v0, m1. valid-peb(v0, m1, apebs[n0]) ∧ apebs[n0].vidhdr.sqn = valid-sqns(v0, m1, apebs, m).max)) ⊦ (wlar[m, wl-entry(m0, erasing)][n0].status = erroneous ↔ boolvar) ∧ (wlar[m, wl-entry(m0, erasing)][n0].status = free → avhdr = empty) ∧ (wlar[n0].status ≠ erroneous ∧ wlar[m, wl-entry(m0, erasing)][n0].status ≠ erasing → aehdr = aechdr(wlar[n0].ec)) ∧ ( wlar[m, wl-entry(m0, erasing)][n0].status = used → (∃ v0, m1. valid-peb(v0, m1, mkpeb(aehdr, avhdr, buf, m2, wlar[n0].status = erroneous)) ∧ avhdr.sqn = (valid-sqns(v0, m1, apebs, m) ∪ (v = v0 ∧ n = m1 ⊃ {avhdr0.sqn} ;∅)).max))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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