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Proof step 3 from proof in-06 (previous step)

∀ a, y. a ∈ x0 ∧ a ∈ y → dups(x0 + y), a ∈ a0 + x0, a ∈ y ⊦ dups((a0 + x0) + y)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
a ∈ y, a ∈ x0, ∀ a, y. a ∈ x0 ∧ a ∈ y → dups(x0 + y) ⊦ a0 ∈ x0, a0 ∈ y, dups(x0 + y)

Rule argument

Additional information

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