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Proof step 3 from proof get-nodups (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
(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)

Rule argument

Additional information

Used simplifier rules: lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%rec: ⊦ dups(a ' + x) ↔ ¬ ¬(a ∈ x ∨ dups(x)); from specification list-dup%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%succ: ⊦ m +1 = m + 1; from specification nat-basic2%len-05: ⊦ #(a ' + x) = (# x)+1; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%e-01: ⊦ a ' + x ≠ []; from specification list%c: ⊦ a + x = a ' + x; from specification list