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Proof step 2 from proof dups-def (previous step)

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])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
dups(x0) ↔ (∃ m, n. m < # x0 ∧ n < # x0 ∧ m ≠ n ∧ x0[m] = x0[n]) ⊦ a ∈ x0 ∨ dups(x0) ↔ (∃ m, n. ¬ # x0 < m ∧ ¬ # x0 < n ∧ m ≠ n ∧ (a ' + x0)[m] = (a ' + x0)[n])

Rule argument

Additional information

Used simplifier rules: 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% ⊦ dups(a ' + x) ↔ a ∈ x ∨ dups(x); (s)% ⊦ a + x = a ' + x; (s)