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Proof step 47 from proof valid-wl-use (previous step)

# apebs = # wlar, avhdr1.sqn = valid-sqns(v, m, apebs).max, wlar[n0] = wl-entry(m3, used), apebs[n0] = mkpeb(aechdr(m3), avhdr1, buf1, m2, false), n0 < # wlar, n < # wlar, valid-peb(v, m, mkpeb(aechdr(m3), avhdr1, buf1, m2, false)), valid-peb(mkpeb(aehdr, avhdr, buf, m0, boolvar)), valid-sqns(avhdr.vol, avhdr.leb, apebs) ≠ ∅ → valid-sqns(avhdr.vol, avhdr.leb, apebs).max < avhdr.sqn, v ≠ avhdr.vol ∨ m ≠ avhdr.leb ⊦ avhdr1.sqn = (valid-sqns(v, m, apebs) ∪ (valid-peb(v, m, mkpeb(aehdr, avhdr, buf, m0, boolvar)) ⊃ {avhdr.sqn} ;∅)).max, n0 = n, valid-peb(v, m, apebs[n])

Heuristic if-then-else split applied rule case distinction

And closed the goal!

Rule argument

pos: right 1

Additional information

Used simplifier rules: ⊦ ∅ ∪ bag0 = bag0; (s)%union-ass: ⊦ bag ∪ bag3 ∪ bag4 = (bag ∪ bag3) ∪ bag4; from specification bag-union[nat-bag]%union-com: ⊦ bag ∪ bag2 = bag2 ∪ bag; from specification bag-union[nat-bag]% ⊦ bag ∪ {n} = bag ++ n; (s)%⊦ valid-peb(v, m, mkpeb(aehdr, avhdr, buf, m0, boolvar)) → avhdr.vol = v; (s)%⊦ valid-peb(v, m, mkpeb(aehdr, avhdr, buf, m0, boolvar)) → avhdr.leb = m; (s)