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Proof step 2 from proof valid-global-sqn-notin (previous step)

valid-global-sqn(apebs, n), n ∈ valid-sqns(v, m, apebs) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
n ∈ valid-sqns(v, m, apebs), ∀ v0, m0. valid-sqns(v0, m0, apebs) ≠ ∅ → valid-sqns(v0, m0, apebs).max < n ⊦

Rule argument

Additional information

Used simplifier rules: valid-global-sqn: ⊦ valid-global-sqn(apebs, n) ↔ (∀ v, m. valid-sqns(v, m, apebs) ≠ ∅ → n > valid-sqns(v, m, apebs).max);% ⊦ m > n ↔ n < m; (s)