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Proof step 10 from proof dups-firstoc-rew (previous step)

a0 ∈ y, ∃ y0, z. y = y0 + a ' + z ∧ ¬ a ∈ y0 ⊦ dups(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 11:
a0 ∈ y0 ∨ a0 = a ∨ a0 ∈ z0 ⊦ a ∈ y0, a ∈ z0, dups(y0 + z0), ∃ y1, a1, z1. a ' + (y0 + a ' + z0) + a0 ' + z = y1 + a1 ' + z1 ∧ a1 ∈ y1 ∧ ¬ dups(y1)

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%in-rec: ⊦ a ∈ b ' + x ↔ ¬ ¬(a = b ∨ a ∈ x); from specification list%in-append: ⊦ a ∈ x + y ↔ ¬ ¬(a ∈ x ∨ a ∈ y); from specification list