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Proof step 4 from proof valid-sqns-extract-sqn-h (previous step)

n < # apebs, n < m1 +1, valid-peb(v, m, apebs[n]), m1 +1 ≤ # apebs, ∀ n, apebs, v, m. n < m1 ∧ n < # apebs ∧ valid-peb(v, m, apebs[n]) ∧ m1 ≤ # apebs → (∃ bag. valid-sqns(v, m, apebs, m1) = bag ++ apebs[n].vidhdr.sqn) ⊦ ∃ bag. valid-sqns(v, m, apebs, m1 +1) = bag ++ apebs[n].vidhdr.sqn

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
n < # apebs, m1 < # apebs, valid-peb(v, m, apebs[n]), n ≤ m1, ∀ n, apebs, v, m. n < m1 ∧ n < # apebs ∧ valid-peb(v, m, apebs[n]) ∧ m1 ≤ # apebs → (∃ bag. valid-sqns(v, m, apebs, m1) = bag ++ apebs[n].vidhdr.sqn) ⊦ ∃ bag. valid-sqns(v, m, apebs, m1) ∪ (valid-peb(v, m, apebs[m1]) ⊃ {apebs[m1].vidhdr.sqn} ;∅) = bag ++ apebs[n].vidhdr.sqn

Rule argument

Additional information

Used simplifier rules: valid-sqns-rec: ⊦ n < # apebs → valid-sqns(v, m, apebs, n + 1) = valid-sqns(v, m, apebs, n) ∪ (valid-peb(v, m, apebs[n]) ⊃ {apebs[n].vidhdr.sqn} ;∅);% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ m < n + 1 ↔ ¬ n < m; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + 1 ≤ n ↔ m < n; (s)% ⊦ m +1 = m + 1; (s)