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Proof step 11 from proof dups-lesize (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
# y0 + # z < # x0, a0 ∈ x0, ∀ a0. a0 ∈ x0 → a0 ∈ y0 ∨ a0 ∈ z ⊦ x0 = [], dups(y0 + z), dups(x0), a ∈ z, a ∈ y0, a ∈ x0, a0 ∈ y0, a0 ∈ z

Rule argument

Additional information

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