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Proof step 10 from proof dups-lesize (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
# y0 + # z < # x0, a0 ∈ x0, ∀ a0. a0 ∈ x0 → a0 ∈ y0 ∨ a0 ∈ z ⊦ x0 = [], dups(y0 + z), a ∈ z, a ∈ y0, dups(x0), a ∈ x0, a0 ∈ y0, a0 ∈ z

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