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Proof step 3 from proof d (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
∀ a, y. (dups(x0 + a ' + y) ↔ dups(x0 + y)) ∨ a ∈ y ∨ a ∈ x0 ⊦ a = a0, dups(x0 + a ' + y) ↔ dups(x0 + y), a0 ∈ y, a0 ∈ x0, a ∈ y, a ∈ x0

Rule argument

Additional information

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