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Proof step 1 from proof dups-def (conclusion)

⊦ dups(x) ↔ (∃ m, n. m < # x ∧ n < # x ∧ m ≠ n ∧ x[m] = x[n])

User applied rule structural induction

And obtained the following new goals

Proof step 2:
dups(x0) ↔ (∃ m, n. m < # x0 ∧ n < # x0 ∧ m ≠ n ∧ x0[m] = x0[n]) ⊦ dups(a + x0) ↔ (∃ m, n. m < #(a + x0) ∧ n < #(a + x0) ∧ m ≠ n ∧ (a + x0)[m] = (a + x0)[n])

Rule argument

indvar: x

Additional information

Used simplifier rules: list freely generated by [], +;;%e-01: ⊦ ¬ dups([]); from specification list-dup%size-01: ⊦ # [] = 0; from specification list-data%less-1: ⊦ ¬ n < 0; from specification nat-basic1