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Proof step 3 from proof violations-exists (previous step)

violations(x) = 0, x[n + 1] < x[n], n + 1 < # x ⊦ x = []

User applied rule structural induction

And obtained the following new goals

Proof step 4:
violations(a + x0) = 0, n + 1 < #(a + x0), (a + x0)[n + 1] < (a + x0)[n], ∀ n. violations(x0) = 0 ∧ x0[n + 1] < x0[n] ∧ n + 1 < # x0 → x0 = [] ⊦ a + x0 = []

Rule argument

indvar: x

Additional information

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