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Proof step 10 from proof dups-eqsize-02 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
# x0 = # y0 + # z, ∀ y0, a1. # x0 = # y0 ∧ (∀ a. a ∈ x0 → a ∈ y0) → ¬ dups(y0) ∧ (a1 ∈ y0 → a1 ∈ x0), ∀ a1. a1 ∈ x0 → a1 ∈ y0 ∨ a1 ∈ z, (a1 ∈ x0 → a1 ∈ y0 ∨ a1 ∈ z) → ¬ dups(y0 + z) ∧ (a ∈ y0 ∨ a ∈ z → a ∈ x0) ⊦ a0 ∈ x0, dups(x0), ¬ dups(y0 + a0 ' + z) ∧ (a ∈ y0 ∨ a ∈ z → a = a0 ∨ a ∈ x0)

Rule argument

Additional information

Used simplifier rules: in-append: ⊦ a ∈ x + y ↔ ¬ ¬(a ∈ x ∨ a ∈ y); from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%len: ⊦ #(x + y) = # x + # y; from specification list