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Proof step 9 from proof firstN-in (previous step)

n = m + 1, a ∈ firstn(m, 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 10:
a ∈ firstn(m, x0), m ≤ # x0, ∀ a, n. a ∈ firstn(n, x0) ∧ n ≤ # x0 → a ∈ x0 ⊦ a = a0, a ∈ x0

Rule argument

Additional information

Used simplifier rules: ⊦ m + n ≤ m + n0 ↔ n ≤ n0; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + 1 ≠ 0; (s)