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Proof step 10 from proof f-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
a ∈ y, ∀ a, y. a ∈ x0 ∨ a ∈ y ∨ dups(x0 + y) → dups(x0 + a ' + y) ∨ ¬ a ∈ x0 ∧ ¬ a ∈ y ⊦ a ∈ x0, a0 ∈ x0, a0 ∈ y, dups(x0 + a ' + y)

Proof step 10 from proof f-01 (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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