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Proof step 3 from proof valid-global-sqn-not-leq (
previous step
)
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) = ∅
Heuristic quantifier closing applied rule all left
And closed the goal!
Rule argument
pos: left 2 terms: v, m
Additional information
Used simplifier rules: ⊦ m ≤ n → (n < m ↔ false); (s) Used terms: v, m