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Proof step 20 from proof valid-wl-use (previous step)

# apebs = # wlar, apebs[n] = mkpeb(aehdr1, avhdr1, buf1, m2, boolvar1), avhdr0.sqn < (valid-sqns(v, m, apebs) \ (valid-peb(v, m, mkpeb(aehdr1, avhdr1, buf1, m2, boolvar1)) ⊃ {avhdr1.sqn} ;∅)).max, n < # wlar, valid-peb(v, m, mkpeb(aehdr0, avhdr0, buf0, m1, false)), valid-sqns(v, m, apebs) ≠ ∅ → valid-sqns(v, m, apebs).max < avhdr0.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)), ∀ m0. m0 < # wlar ∧ m0 ≠ n ∧ wlar[m0].status = used → apebs[m0].vidhdr.vol ≠ v ∨ apebs[m0].vidhdr.leb ≠ m ⊦ valid-sqns(v, m, apebs) \ (valid-peb(v, m, mkpeb(aehdr1, avhdr1, buf1, m2, boolvar1)) ⊃ {avhdr1.sqn} ;∅) = ∅

Heuristic if-then-else split applied rule case distinction

And closed the goal!

Rule argument

pos: right 1

Additional information

Used simplifier rules: ⊦ bag \ ∅ = bag; (s)% ⊦ m < n → ¬ n < m; (s)% ⊦ bag \ {n} = bag -- n; (s)%⊦ valid-peb(v0, m0, apeb) → (valid-peb(v1, m1, apeb) ↔ v0 = v1 ∧ m0 = m1); (s)% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ m < n → (n ≤ m ↔ false); (s)%n < # apebs ⊦ apebs[n] = apeb ∧ valid-peb(v, m, apeb) → valid-sqns(v, m, apebs) ≠ ∅; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic%transitivity: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification oelem[nat-set]% ⊦ m < n → n ≠ 0; (s)% ⊦ m +1 = m + 1; (s)% ⊦ bag -- n ≠ ∅ → ¬ bag.max < (bag -- n).max; (s)