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Proof step 13 from proof diffgetn-nodupssize (previous step)

n ≤ # x0, ∀ m, n0. # x0 = m + 1 ∧ n0 ≤ m → #(x0 \ x0[n0] ') = m ⊦ x0 = [], #(x0 \ x0[n - 1] ') + 1 = # x0, a = x0[n - 1], n = 0, dups(x0), a ∈ x0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 14:
n = m0 + 1, n ≤ # x0, ∀ m, n0. # x0 = m + 1 ∧ n0 ≤ m → #(x0 \ x0[n0] ') = m ⊦ x0 = [], #(x0 \ x0[m0] ') + 1 = # x0, a = x0[m0], n = 0, dups(x0), a ∈ x0

Rule argument

spec: nat inst: name: elim-pred-c seq: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1 subst: [n, m] → [n, m0] flags:

Additional information

Used simplifier rules: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: n, m0