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Proof step 15 from proof dups-firstoc-rew (previous step)

∃ y, z. x0 = y + a ' + z ∧ ¬ a ∈ y ⊦ dups(x0), ∃ y0, a1, z0. a ' + x0 = y0 + a1 ' + z0 ∧ a1 ∈ y0 ∧ ¬ dups(y0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
⊦ a ∈ y, a ∈ z, dups(y + z), ∃ y0, a1, z0. a ' + y + a ' + z = y0 + a1 ' + z0 ∧ a1 ∈ y0 ∧ ¬ dups(y0)

Rule argument

Additional information

Used simplifier rules: f-01: ⊦ dups(x + a ' + y) ↔ a ∈ x ∨ a ∈ y ∨ dups(x + y);%a: ⊦ (x + y) + z = x + y + z; from specification list