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Proof step 3 from proof dups-firstoc-rew (previous step)

dups(x0) → (∃ y, a, z. x0 = y + a ' + z ∧ a ∈ y ∧ ¬ dups(y)), dups(a + x0) ⊦ ∃ y, a0, z. a + x0 = y + a0 ' + z ∧ a0 ∈ y ∧ ¬ dups(y)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
a ∈ x0 ∨ dups(x0), dups(x0) → x0 = y + a0 ' + z ∧ a0 ∈ y ∧ ¬ dups(y) ⊦ ∃ y0, a1, z0. a ' + x0 = y0 + a1 ' + z0 ∧ a1 ∈ y0 ∧ ¬ dups(y0)

Rule argument

Additional information

Used simplifier rules: rec: ⊦ dups(a ' + x) ↔ ¬ ¬(a ∈ x ∨ dups(x));%a: ⊦ (x + y) + z = x + y + z; from specification list%c: ⊦ a + x = a ' + x; from specification list