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Proof step 6 from proof valid-sqns-subseteq-h (previous step)

n < # apebs, n0 < n, n0 ≤ # apebs, n0 ≤ n + 1, ∀ apebs, n0, v, m. n ≤ # apebs ∧ n0 ≤ n ∧ n0 ≤ # apebs → valid-sqns(v, m, apebs, n0) ⊆ valid-sqns(v, m, apebs, n) ⊦ valid-sqns(v, m, apebs, n0) ⊆ valid-sqns(v, m, apebs, n) ∪ (valid-peb(v, m, apebs[n]) ⊃ {apebs[n].vidhdr.sqn} ;∅)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
n0 < # apebs, n < # apebs, n0 < n, n0 + 1 < # apebs, n0 ≤ n + 1, ∀ apebs, n0, v, m. n ≤ # apebs ∧ n0 ≤ n ∧ n0 ≤ # apebs → valid-sqns(v, m, apebs, n0) ⊆ valid-sqns(v, m, apebs, n) ⊦ valid-sqns(v, m, apebs, n0) ⊆ valid-sqns(v, m, apebs, n) ∪ (valid-peb(v, m, apebs[n]) ⊃ {apebs[n].vidhdr.sqn} ;∅)

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → m ≤ n; (s)% ⊦ m +1 = m + 1; (s)% ⊦ m < n → n ≠ 0; (s)%trans: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic