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

n1 +1 ≤ # apebs, n1 +1 ≤ 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 +1) = valid-sqns(v, m, apebs, n1 +1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
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 + 1) = valid-sqns(v, m, apebs, n1) ∪ (valid-peb(v, m, apebs[n1]) ⊃ {apebs[n1].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 ≠ 0; (s)%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%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)