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Proof step 4 from proof firstN-in (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
a ∈ firstn(n, a0 ' + x0), n ≤ 1 + # x0, ∀ a, n. a ∈ firstn(n, x0) ∧ n ≤ # x0 → a ∈ x0 ⊦ a = a0, a ∈ x0

Rule argument

Additional information

Used simplifier rules: ⊦ a ∈ b ' + x ↔ ¬(a ≠ b ∧ ¬ a ∈ x); (ss)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ # a ' = 0 +1; (s)% ⊦ #(x + y) = # x + # y; (s)% ⊦ a + x = a ' + x; (s)