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Proof step 10 from proof firstN-in (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Rule argument

spec: <toplevel> name: elim-firstn-restn seq: ⊦ n ≤ # x → (y = firstn(n, x) ∧ z = restn(n, x) ↔ x = y + z ∧ # y = n) subst: [x, n, y, z] → [x0, m, y, z] flags: <rotated> <all paths>

Additional information

Used simplifier rules: ⊦ #(x + y) = # x + # y; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ m ≤ m + n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ a ∈ x → a ∈ x + y; (s) Used terms: x0, m, y, z