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Proof step 5 from proof firstN-in (previous step)

a ∈ firstn(n, a0 ' + x0), n ≤ 1 + # x0, ∀ a, n. a ∈ firstn(n, x0) ∧ n ≤ # x0 → a ∈ x0 ⊦ a = a0, a ∈ x0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ n ≠ 0 → firstn(n, a ' + x) = a ' + firstn(n -1, x)
Proof step 7:
a ∈ a0 ' + firstn(n -1, x0), n ≤ 1 + # x0, ∀ a, n. a ∈ firstn(n, x0) ∧ n ≤ # x0 → a ∈ x0 ⊦ a = a0, n = 0, a ∈ x0

Rule argument

spec: <toplevel> name: firstN-rec-01 seq: ⊦ n ≠ 0 → firstn(n, a ' + x) = a ' + firstn(n -1, x) subst: [x, a, n] → [x0, a0, n] flags: <all paths>

Additional information

Used simplifier rules: firstN-zero: ⊦ firstn(0, x) = [];% ⊦ 0 ≤ m; (s)% ⊦ ¬ a ∈ []; (s) Used terms: x0, a0, n