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Proof step 5 from proof dups-firstoc-rew (previous step)

dups(x0), 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)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
a0 ∈ y ⊦ dups(y), ∃ y0, a1, z0. a ' + y + a0 ' + z = y0 + a1 ' + z0 ∧ a1 ∈ y0 ∧ ¬ dups(y0)

Rule argument

Additional information

Used simplifier rules: in-04-01: ⊦ a ∈ x → dups(x + a ' + y);%a: ⊦ (x + y) + z = x + y + z; from specification list