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Proof step 7 from proof valid-sqns-subarray-h (previous step)

n1 < n, n1 < # apebs, n1 < # subarray(apebs, 0, n), n ≤ # apebs, ∀ n, apebs, v, m. n ≤ # apebs ∧ n1 ≤ n ∧ n1 ≤ # apebs → valid-sqns(v, m, subarray(apebs, 0, n), n1) = valid-sqns(v, m, apebs, n1) ⊦ valid-sqns(v, m, subarray(apebs, 0, n), n1) ∪ (valid-peb(v, m, subarray(apebs, 0, n)[n1]) ⊃ {subarray(apebs, 0, n)[n1].vidhdr.sqn} ;∅) = valid-sqns(v, m, apebs, n1) ∪ (valid-peb(v, m, apebs[n1]) ⊃ {apebs[n1].vidhdr.sqn} ;∅)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
n1 < # apebs, n1 < n, n ≤ # apebs, ∀ n, apebs, v, m. n ≤ # apebs ∧ n1 ≤ n ∧ n1 ≤ # apebs → valid-sqns(v, m, subarray(apebs, 0, n), n1) = valid-sqns(v, m, apebs, n1) ⊦ valid-sqns(v, m, subarray(apebs, 0, n), n1) = valid-sqns(v, m, apebs, n1)

Rule argument

Additional information

Used simplifier rules: ⊦ # subarray(apebs, m, n) = n; (s)%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]% ⊦ bag0 ∪ bag3 = bag0 ∪ bag4 ↔ bag3 = bag4; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ n + 0 = n; (s)% ⊦ n0 < n → subarray(apebs, m, n)[n0] = apebs[m + n0]; (s)