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Proof step 8 from proof pos-get (previous step)

n ≤ # x0, ∀ n0. n0 < # x0 → pos(x0[n0], x0) = n0 ∨ x0 = [] ⊦ n = 0, pos(x0[n - 1], a ' + x0) = n, dups(x0), a ∈ x0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 9:
n = m + 1, n ≤ # x0, ∀ n0. n0 < # x0 → pos(x0[n0], x0) = n0 ∨ x0 = [] ⊦ n = 0, pos(x0[m], a ' + x0) = n, dups(x0), a ∈ x0

Rule argument

spec: nat name: elim-pred-c seq: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1 subst: [n, m] → [n, m]

Additional information

Used simplifier rules: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: n, m