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Proof step 5 from proof valid-wl-next-valid-use (previous step)

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

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: right 1

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