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Proof step 11 from proof valid-wl-next-valid-erase (previous step)

# apebs = # wlar, apebs[m] = mkpeb(aehdr0, avhdr0, buf0, m3, boolvar0), apebs[n0] = mkpeb(aehdr, avhdr, buf, m2, boolvar), wlar[n0] = wl-entry(m4, stat), n0 + 1 < # wlar, n0 < # wlar, m < # wlar, n0 < m, valid-peb(v, n, mkpeb(aehdr0, avhdr0, buf0, m3, boolvar0)), 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)) ⊦ n0 = m, (stat = erroneous ↔ boolvar) ∧ (stat = free → avhdr = empty) ∧ (stat ≠ erroneous ∧ stat ≠ erasing → aehdr = aechdr(m4)) ∧ ( stat = used → (∃ v0, m0. valid-peb(v0, m0, mkpeb(aehdr, avhdr, buf, m2, false)) ∧ avhdr.sqn = (valid-sqns(v0, m0, apebs, m) ∪ (v = v0 ∧ n = m0 ⊃ {avhdr0.sqn} ;∅)).max))

User applied rule all left

And obtained the following new goals

Proof step 12:
# apebs = # wlar, wlar[n0] = wl-entry(m4, stat), apebs[n0] = mkpeb(aehdr, avhdr, buf, m2, boolvar), apebs[m] = mkpeb(aehdr0, avhdr0, buf0, m3, boolvar0), n0 < m, m < # wlar, n0 < # wlar, n0 + 1 < # wlar, valid-peb(v, n, mkpeb(aehdr0, avhdr0, buf0, m3, boolvar0)), valid-sqns(v, n, apebs, m) ≠ ∅ → avhdr0.sqn < valid-sqns(v, n, apebs, m).max, 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, (stat = erroneous ↔ boolvar) ∧ (stat = free → avhdr = empty) ∧ (stat ≠ erroneous ∧ stat ≠ erasing → aehdr = aechdr(m4)) ∧ ( stat = used → (∃ v0, m0. valid-peb(v0, m0, mkpeb(aehdr, avhdr, buf, m2, false)) ∧ avhdr.sqn = (valid-sqns(v0, m0, apebs, m) ∪ (v = v0 ∧ n = m0 ⊃ {avhdr0.sqn} ;∅)).max))

Rule argument

pos: left 11 terms: n0 abort: F discard: T computed: T

Additional information

Used terms: n0