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Proof step 3 from proof size-two-06 (previous step)

∀ b, a. # s0 < 2 ∧ b ∈ s0 ∧ a ∈ s0 → a = b, #(s0 ++ a0) < 2, b ∈ s0 ++ a0, a ∈ s0 ++ a0 ⊦ a = b

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
#(s0 ++ a0) < 2, ∀ b, a. # s0 < 2 ∧ b ∈ s0 ∧ a ∈ s0 → a = b, b = a0 ∨ b ∈ s0, a = a0 ∨ a ∈ s0 ⊦ a = b

Rule argument

Additional information

Used simplifier rules: rax: ⊦ a ∈ s ++ b ↔ ¬ ¬(a = b ∨ a ∈ s);