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Proof step 1 from proof k-dups (conclusion)

⊦ dups(x) ↔ # x > 1 ∧ (x.first ∈ x.rest ∨ dups(x.rest))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
dups(x) ⊦ x = [], (x.first ∈ x.rest ∨ dups(x.rest)) ∧ # x ≠ 1

Rule argument

Additional information

Used simplifier rules: dups-in: ⊦ ¬ dups(x) ∧ x ≠ [] → ¬ x.first ∈ x.rest;%dups-rest: ⊦ ¬ dups(x) ∧ x ≠ [] → ¬ dups(x.rest);%e-01: ⊦ ¬ dups([]);%l: ⊦ # x = 0 ↔ x = []; from specification list-data%one: ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1); from specification nat-basic1%gr-def: ⊦ m > n ↔ n < m; from specification nat