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Proof step 3 from proof k-dups (previous step)

x = a + y, dups(x) ⊦ x = [], (a ∈ y ∨ dups(y)) ∧ # x ≠ 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
a ∈ [] ∨ dups([]) ⊦

Rule argument

Additional information

Used simplifier rules: rec: ⊦ dups(a ' + x) ↔ a ∈ x ∨ dups(x);%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ # x = 0 ↔ x = []; (s)% ⊦ 1 = m + 1 ↔ m = 0; (s)% ⊦ # a ' = 0 +1; (s)% ⊦ #(x + y) = # x + # y; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a ' + x ≠ []; (s)% ⊦ a + x = a ' + x; (s)