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Proof step 4 from proof get-nodups (previous step)

(a ' + x0)[m] = (a ' + x0)[n], n ≤ # x0, m ≤ # x0, ∀ m0, n0. x0[m0] = x0[n0] ∧ n0 < # x0 ∧ m0 < # x0 → m0 = n0 ∨ x0 = [] ⊦ m = n, a ∈ x0, dups(x0)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
n ≠ 0 ⊦ (a ' + x)[(* n)] = x[n -1]
Proof step 6:
¬ m ≠ 0, (a ' + x0)[m] = (a ' + x0)[n], n ≤ # x0, m ≤ # x0, ∀ m0, n0. x0[m0] = x0[n0] ∧ n0 < # x0 ∧ m0 < # x0 → m0 = n0 ∨ x0 = [] ⊦ m = n, a ∈ x0, dups(x0)
Proof step 11:
m ≠ 0, x0[m -1] = (a ' + x0)[n], n ≤ # x0, m ≤ # x0, ∀ m0, n0. x0[m0] = x0[n0] ∧ n0 < # x0 ∧ m0 < # x0 → m0 = n0 ∨ x0 = [] ⊦ m = n, a ∈ x0, dups(x0)

Rule argument

spec: list-del inst: name: get-numrec seq: n ≠ 0 ⊦ (a ' + x)[(* n)] = x[n -1] subst: [n, x, a] → [m, x0, a] flags: rewrite, current

Additional information

Used terms: m, x0, a