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Proof step 1 from proof in-07 (conclusion)

⊦ ¬ dups(x + a ' + b ') → (dups(x + a ' + b ' + c ') ↔ ¬ ¬(a = c ∨ b = c ∨ c ∈ x))

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: n: ⊦ a ≠ c ∧ a ≠ b ∧ ¬ a ∈ x → (dups(x + b ' + c ' + a ') ↔ dups(x + b ' + c '));%in-06-01: ⊦ ¬ dups(x + a ') → (dups(x + a ' + b ') ↔ ¬ ¬(a = b ∨ b ∈ x));%f: ⊦ dups(x + a ') ↔ a ∈ x ∨ dups(x);%in-17: ⊦ a ∈ b ' ↔ a = b; from specification list%in-append: ⊦ a ∈ x + y ↔ ¬ ¬(a ∈ x ∨ a ∈ y); from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list