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Proof step 14 from proof get-nodups (previous step)

x0[m1] = (a ' + x0)[n], m1 < # x0, n ≤ # x0, ∀ m, n0. x0[m] = x0[n0] ∧ n0 < # x0 ∧ m < # x0 → m = n0 ⊦ m1 + 1 = n, x0 = [], dups(x0), a ∈ x0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 15:
n ≠ 0 ⊦ (a ' + x)[(* n)] = x[n -1]
Proof step 16:
n ≠ 0, x0[m1] = x0[n -1], m1 < # x0, n ≤ # x0, ∀ m, n0. x0[m] = x0[n0] ∧ n0 < # x0 ∧ m < # x0 → m = n0 ⊦ m1 + 1 = n, x0 = [], dups(x0), a ∈ x0

Rule argument

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

Additional information

Used simplifier rules: get-zero-01: ⊦ (a ' + x)[0] = a;%get-in: ⊦ m < # x → (¬ x[m] ∈ x ↔ false);%a: ⊦ (x + y) + z = x + y + z; from specification list%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%lez: ⊦ 0 ≤ m; from specification nat Used terms: n, x0, a