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Proof step 23 from proof dups-eqsize-02 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 24:
# x0 = # y0 + # z, dups(y0 + a0 ' + z), a1 ∈ x0 → a1 ∈ y0 ∨ a1 ∈ z, a ∈ y0 ∨ a ∈ z → a ∈ x0, ∀ y0, a1. # x0 = # y0 ∧ (∀ a. a ∈ x0 → a ∈ y0) → ¬ dups(y0) ∧ (a1 ∈ y0 → a1 ∈ x0), ∀ a1. a1 ∈ x0 → a1 ∈ y0 ∨ a1 ∈ z, ∀ a2. a2 ∈ y0 ∨ a2 ∈ z → a2 ∈ x0 ⊦ dups(x0), dups(y0 + z), a0 ∈ x0, ¬ a0 ∈ y0 ∧ ¬ a0 ∈ z

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s)