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Proof step 9 from proof valid-global-sqn-next (previous step)

apebs[m] = mkpeb(aehdr, avhdr, buf, m1, boolvar), m < # apebs, n ≤ avhdr.sqn, ∀ v, m0. valid-sqns(v, m0, apebs, m) ≠ ∅ → valid-sqns(v, m0, apebs, m).max < n, ∀ v, m0. valid-sqns(v, m0, apebs, m) ∪ (valid-peb(v, m0, mkpeb(aehdr, avhdr, buf, m1, boolvar)) ⊃ {avhdr.sqn} ;∅) ≠ ∅ → (valid-sqns(v, m0, apebs, m) ∪ (valid-peb(v, m0, mkpeb(aehdr, avhdr, buf, m1, boolvar)) ⊃ {avhdr.sqn} ;∅)).max < n, ∃ v, m. valid-peb(v, m, mkpeb(aehdr, avhdr, buf, m1, boolvar)) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
apebs[m] = mkpeb(aehdr, avhdr, buf, m1, boolvar), m < # apebs, valid-peb(v, m0, mkpeb(aehdr, avhdr, buf, m1, boolvar)), n ≤ avhdr.sqn, ∀ v, m0. valid-sqns(v, m0, apebs, m) ≠ ∅ → valid-sqns(v, m0, apebs, m).max < n, ∀ v0, m2. valid-sqns(v0, m2, apebs, m) ∪ (v = v0 ∧ m0 = m2 ⊃ {avhdr.sqn} ;∅) ≠ ∅ → (valid-sqns(v0, m2, apebs, m) ∪ (v = v0 ∧ m0 = m2 ⊃ {avhdr.sqn} ;∅)).max < n ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ valid-peb(v0, m0, apeb) → (valid-peb(v1, m1, apeb) ↔ v0 = v1 ∧ m0 = m1); (s)