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Proof step 3 from proof valid-global-sqn-notin (previous step)

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

User applied rule all left

And closed the goal!

Rule argument

pos: left 2 terms: v, m abort: F discard: T computed: F

Additional information

Used simplifier rules: ⊦ n ∈ bag → (bag = ∅ ↔ false); (s)% ⊦ n ∈ bag → ¬ bag.max < n; (s) Used terms: v, m