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Proof step 5 from proof dups-lesize (previous step)

# y ≤ # x0, ∀ y0. # y0 < # x0 ∧ (∀ a. a ∈ x0 → a ∈ y0) → x0 = [] ∨ dups(y0), a = a ∨ a ∈ x0 → a ∈ y, ∀ a0. a0 ∈ x0 → a0 ∈ y ⊦ a ∈ x0, dups(x0), dups(y)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
# y ≤ # x0, a ∈ y, ∀ a0. a0 ∈ x0 → a0 ∈ y, ∀ y0. # y0 < # x0 ∧ (∀ a. a ∈ x0 → a ∈ y0) → x0 = [] ∨ dups(y0) ⊦ a ∈ x0, dups(x0), dups(y)

Proof step 5 from proof dups-lesize (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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