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Proof step 3 from proof dups-eqsize-02 (previous step)

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

User applied rule structural induction

And obtained the following new goals

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

Rule argument

indvar: x

Additional information

Used simplifier rules: e-01: ⊦ ¬ dups([]);%list freely generated by [], +;;% ⊦ # [] = 0; (s)% ⊦ # x = 0 ↔ x = []; (s)% ⊦ ¬ a ∈ []; (s)