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Proof step 5 from proof dups-eqsize-02 (previous step)

# x0 + 1 = # y, ∀ y0, a1. # x0 = # y0 ∧ (∀ a. a ∈ x0 → a ∈ y0) → ¬ dups(y0) ∧ (a1 ∈ y0 → a1 ∈ x0), a0 = a0 ∨ a0 ∈ x0 → a0 ∈ y, ∀ a1. a1 ∈ x0 → a1 ∈ y ⊦ a0 ∈ x0, dups(x0), ¬ dups(y) ∧ (a ∈ y → a = a0 ∨ a ∈ x0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
# x0 + 1 = # y, a0 ∈ y, ∀ a1. a1 ∈ x0 → a1 ∈ y, ∀ y0, a1. # x0 = # y0 ∧ (∀ a. a ∈ x0 → a ∈ y0) → ¬ dups(y0) ∧ (a1 ∈ y0 → a1 ∈ x0) ⊦ a0 ∈ x0, dups(x0), ¬ dups(y) ∧ (a ∈ y → a = a0 ∨ a ∈ x0)

Proof step 5 from proof dups-eqsize-02 (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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