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Proof step 3 from proof dups-eqsize-01 (previous step)

# x = # y, a ∈ y, ∀ a. a ∈ x → a ∈ y ⊦ dups(y), dups(x), a ∈ x

User applied rule structural induction

And obtained the following new goals

Proof step 4:
#(a0 + x0) = # y, a ∈ y, ∀ a. a ∈ a0 + x0 → a ∈ y, ∀ y, a. # x0 = # y ∧ a ∈ y ∧ (∀ a. a ∈ x0 → a ∈ y) → dups(y) ∨ dups(x0) ∨ a ∈ x0 ⊦ dups(a0 + x0), dups(y), a ∈ a0 + x0

Rule argument

indvar: x

Additional information

Used simplifier rules: list freely generated by [], +;;% ⊦ # [] = 0; (s)% ⊦ # x = 0 ↔ x = []; (s)