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Proof step 23 from proof unique-sqns-update (previous step)

apebs[m] = mkpeb(aehdr0, avhdr0, buf0, m2, boolvar0), m < # apebs, unique-elems(valid-sqns(v, m0, apebs)), valid-peb(v, m0, mkpeb(aehdr0, avhdr0, buf0, m2, boolvar0)), ¬ valid-peb(mkpeb(aehdr, avhdr, buf, m1, boolvar)) ∨ ¬ avhdr.sqn ∈ valid-sqns(avhdr.vol, avhdr.leb, apebs) \ (v = avhdr.vol ∧ m0 = avhdr.leb ⊃ {avhdr0.sqn} ;∅) ⊦ unique-elems((valid-sqns(v, m0, apebs) -- avhdr0.sqn) ∪ (valid-peb(v, m0, mkpeb(aehdr, avhdr, buf, m1, boolvar)) ⊃ {avhdr.sqn} ;∅))

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]% ⊦ unique-elems(bag) → unique-elems(bag -- n); (ss)% ⊦ bag ∪ {n} = bag ++ n; (s)% ⊦ unique-elems(bag ++ m) ↔ unique-elems(bag) ∧ ¬ m ∈ bag; (s)% ⊦ unique-elems(bag) → (m ∈ bag -- n ↔ m ≠ n ∧ m ∈ bag); (s)% ⊦ valid-peb(v, m, apeb) → valid-peb(apeb); (ss)%⊦ 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)% ⊦ bag \ {n} = bag -- n; (s)