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Proof step 16 from proof dups-firstoc-rew (previous step)

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

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 4 terms: a ' + y, a, z

Additional information

Used simplifier rules: rec: ⊦ dups(a ' + x) ↔ ¬ ¬(a ∈ x ∨ dups(x));%app-05: ⊦ dups(x) → dups(x + y);%a: ⊦ (x + y) + z = x + y + z; from specification list%in-10: ⊦ ¬ a ∈ x → (a ∈ b ' + x ↔ a = b); from specification list Used terms: a ' + y, a, z