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

x[m] = x[n], m < # x, n < # x ⊦ m = n, x = [], dups(x)

User applied rule structural induction

And obtained the following new goals

Proof step 4:
(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)

Rule argument

indvar: x

Additional information

Used simplifier rules: list freely generated by [], +;;