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Proof step 3 from proof firstN-lastN-in-01 (previous step)

n < # x, a ∈ x ⊦ x = [], a ∈ firstn(# x - n, x), a ∈ lastn(n, x)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ n ≤ # x ∧ a ∈ x ∧ ¬ a ∈ lastn(n, x) → (a ∈ firstn(# x - n, x) ↔ true)

Rule argument

spec: <toplevel> name: firstN-lastN-in seq: ⊦ n ≤ # x ∧ a ∈ x ∧ ¬ a ∈ lastn(n, x) → (a ∈ firstn(# x - n, x) ↔ true) subst: [n, x, a] → [n, x, a] flags: <all paths>

Additional information

Used simplifier rules: ⊦ m ≤ n ↔ ¬ n < m; (ss)% ⊦ m < n → ¬ n < m; (s) Used terms: n, x, a