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Proof step 7 from proof dups-firstoc-rew (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
a0 ∈ y, a ∈ y ⊦ dups(y), ∃ y0, a1, z0. a ' + y + a0 ' + z = y0 + a1 ' + z0 ∧ a1 ∈ y0 ∧ ¬ dups(y0)

Rule argument

Additional information

Used simplifier rules: rec: ⊦ dups(a ' + x) ↔ ¬ ¬(a ∈ x ∨ dups(x));%in-12: ⊦ a ∈ y → a ∈ x + y; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list