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Proof step 2 from proof valid-global-sqn-not-ls (previous step)

n < valid-sqns(v, m, apebs).max, valid-global-sqn(apebs, 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).max, ∀ v0, m0. valid-sqns(v0, m0, apebs) ≠ ∅ → valid-sqns(v0, m0, apebs).max < n ⊦ valid-sqns(v, m, apebs) = ∅

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 ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m > n ↔ n < m; (s)