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Proof step 7 from proof firstN-in (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
a ∈ 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

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a ≠ b → (a ∈ b ' + x ↔ a ∈ x); (s)% ⊦ n -1 = n - 1; (s)