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Proof step 1 from proof in-06-01 (conclusion)

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

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: l-01: ⊦ a ≠ b ∧ ¬ b ∈ x → (dups(x + a ' + b ') ↔ dups(x + a '));%f: ⊦ dups(x + a ') ↔ a ∈ x ∨ dups(x);%l-03: ⊦ ¬ b ∈ x → (dups(x + b ') ↔ 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