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Proof step 7 from proof valid-global-sqn-invalidate (previous step)

m < # apebs, valid-peb(v, m0, apebs[m]), 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) \ {apebs[m].vidhdr.sqn} = ∅, valid-peb(apeb)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
m < # apebs, valid-peb(v, m0, apebs[m]), n ≤ (valid-sqns(v, m0, apebs) -- apebs[m].vidhdr.sqn).max, ∀ v, m0. valid-sqns(v, m0, apebs) ≠ ∅ → valid-sqns(v, m0, apebs).max < n ⊦ valid-sqns(v, m0, apebs) -- apebs[m].vidhdr.sqn = ∅, valid-peb(apeb)

Rule argument

Additional information

Used simplifier rules: ⊦ valid-peb(v0, m0, apeb) → (valid-peb(v1, m1, apeb) ↔ v0 = v1 ∧ m0 = m1); (s)% ⊦ bag \ {n} = bag -- n; (s)