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Proof step 5 from proof valid-global-sqn-invalidate (previous step)

m < # apebs, n ≤ ((valid-sqns(v, m0, apebs) \ (valid-peb(v, m0, apebs[m]) ⊃ {apebs[m].vidhdr.sqn} ;∅)) ∪ (valid-peb(v, m0, apeb) ⊃ {apeb.vidhdr.sqn} ;∅)).max, ∀ v, m0. valid-sqns(v, m0, apebs) ≠ ∅ → valid-sqns(v, m0, apebs).max < n ⊦ (valid-sqns(v, m0, apebs) \ (valid-peb(v, m0, apebs[m]) ⊃ {apebs[m].vidhdr.sqn} ;∅)) ∪ (valid-peb(v, m0, apeb) ⊃ {apeb.vidhdr.sqn} ;∅) = ∅, valid-peb(apeb)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
m < # apebs, n ≤ (valid-sqns(v, m0, apebs) \ (valid-peb(v, m0, apebs[m]) ⊃ {apebs[m].vidhdr.sqn} ;∅)).max, ∀ v, m0. valid-sqns(v, m0, apebs) ≠ ∅ → valid-sqns(v, m0, apebs).max < n ⊦ valid-sqns(v, m0, apebs) \ (valid-peb(v, m0, apebs[m]) ⊃ {apebs[m].vidhdr.sqn} ;∅) = ∅, valid-peb(apeb)

Rule argument

Additional information

Used simplifier rules: ⊦ ∅ ∪ bag0 = bag0; (s)% ⊦ valid-peb(v, m, apeb) → valid-peb(apeb); (ss)%union-com: ⊦ bag ∪ bag2 = bag2 ∪ bag; from specification bag-union[nat-bag]%union-ass: ⊦ bag ∪ bag3 ∪ bag4 = (bag ∪ bag3) ∪ bag4; from specification bag-union[nat-bag]% ⊦ bag ∪ {n} = bag ++ n; (s)%⊦ valid-peb(v0, m0, apeb) → (valid-peb(v1, m1, apeb) ↔ v0 = v1 ∧ m0 = m1); (s)