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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
(a ' + x0)[m] = (a ' + x0)[n], m ≤ # x0, n ≤ # x0, ∀ m0, n0. x0[m0] = x0[n0] ∧ m0 < # x0 ∧ n0 < # x0 → m0 = n0 ∨ x0 = [] ⊦ m = n, dups(x0), a ∈ x0

Rule argument

Additional information

Used simplifier rules: ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ dups(a ' + x) ↔ dups(x) ∨ a ∈ x; (ws)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ # a ' = 0 +1; (s)% ⊦ #(x + y) = # x + # y; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a ' + x ≠ []; (s)% ⊦ a + x = a ' + x; (s)