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Proof step 15 from proof dups-def (previous step)

(a ' + x0)[m] = (a ' + x0)[n], n ≤ # x0, m ≤ # x0 ⊦ m = n, dups(x0), a ∈ x0, ∃ m, n. m < # x0 ∧ n < # x0 ∧ m ≠ n ∧ x0[m] = x0[n], m - 1 < # x0 ∧ n - 1 < # x0 ∧ m - 1 ≠ n - 1 ∧ x0[m - 1] = x0[n - 1]

User applied rule insert elim lemma

And obtained the following new goals

Proof step 16:
(a ' + x0)[m] = (a ' + x0)[n], m = 0, m ≤ # x0, n ≤ # x0 ⊦ m = n, dups(x0), a ∈ x0, ∃ m, n. m < # x0 ∧ n < # x0 ∧ m ≠ n ∧ x0[m] = x0[n], m - 1 < # x0 ∧ n - 1 < # x0 ∧ m - 1 ≠ n - 1 ∧ x0[m - 1] = x0[n - 1]
Proof step 18:
(a ' + x0)[m] = (a ' + x0)[n], m = m0 + 1, m ≤ # x0, n ≤ # x0 ⊦ m = n, m = 0, dups(x0), a ∈ x0, ∃ m, n. m < # x0 ∧ n < # x0 ∧ m ≠ n ∧ x0[m] = x0[n], m0 < # x0 ∧ n - 1 < # x0 ∧ m0 ≠ n - 1 ∧ x0[m0] = x0[n - 1]

Rule argument

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

Additional information

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