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Proof step 11 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} ;∅)).max, ∀ v, m0. valid-sqns(v, m0, apebs) ≠ ∅ → valid-sqns(v, m0, apebs).max < n ⊦ valid-sqns(v, m0, apebs) \ ∅ = ∅, valid-peb(apeb), valid-peb(v, m0, apebs[m])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
m < # apebs, n ≤ valid-sqns(v, m0, apebs).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]), valid-peb(apeb)

Rule argument

Additional information

Used simplifier rules: ⊦ bag \ ∅ = bag; (s)