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Proof step 6 from proof f-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
∀ a, y. a ∈ x0 ∨ a ∈ y ∨ dups(x0 + y) → dups(x0 + a ' + y) ∨ ¬ a ∈ x0 ∧ ¬ a ∈ y, a ∈ x0 ∨ a ∈ y ∨ dups(x0 + y) ⊦ a0 = a, dups(x0 + a ' + y), a0 ∈ x0, a0 ∈ y, ¬ a ∈ x0 ∧ ¬ a ∈ y

Rule argument

Additional information

Used simplifier rules: dups-cons: ⊦ dups(a ' + x) ↔ dups(x) ∨ a ∈ x;%rec: ⊦ dups(a ' + x) ↔ a ∈ x ∨ dups(x);% ⊦ a ≠ b → (a ∈ b ' + x ↔ a ∈ x); (s)% ⊦ a ∈ b ' ↔ a = b; (s)% ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a + x = a ' + x; (s)