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Proof step 2 from proof f (previous step)

∀ a. dups(x0 + a ') ↔ a ∈ x0 ∨ dups(x0) ⊦ dups((a0 + x0) + a ') ↔ a ∈ a0 + x0 ∨ dups(a0 + x0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
∀ a. dups(x0 + a ') ↔ a ∈ x0 ∨ dups(x0) ⊦ a0 = a ∨ dups(x0 + a ') ↔ a = a0 ∨ a ∈ x0 ∨ dups(x0), a0 ∈ x0

Rule argument

Additional information

Used simplifier rules: rec: ⊦ dups(a ' + x) ↔ ¬ ¬(a ∈ x ∨ dups(x));%in-rec: ⊦ a ∈ b ' + x ↔ ¬ ¬(a = b ∨ a ∈ x); from specification list%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%c: ⊦ a + x = a ' + x; from specification list