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Proof step 1 from proof in-09 (conclusion)

⊦ x + a ' = y + b ' + z ∧ b ∈ z → dups(x + a ')

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
x + a ' = y + b ' + z, b ∈ z, a ∈ y ∨ a ∈ z ⊦ dups(x), a ∈ x

Rule argument

Additional information

Used simplifier rules: f: ⊦ dups(x + a ') ↔ a ∈ x ∨ dups(x);%in-rec: ⊦ a ∈ b ' + x ↔ ¬ ¬(a = b ∨ a ∈ x); from specification list%in-append: ⊦ a ∈ x + y ↔ ¬ ¬(a ∈ x ∨ a ∈ y); from specification list%in-15: ⊦ x + a ' = y → a ∈ y; (forward) from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list