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Proof step 4 from proof size-two-03 (previous step)

a ∈ s0, # s0 ≠ 0 ∧ # s0 ≠ 1 → a0 ≠ b ∧ a0 ∈ s0 ∧ b ∈ s0 ⊦ #(s0 ++ a) = 0, #(s0 ++ a) = 1, ∃ a1, b0. a1 ≠ b0 ∧ (a1 = a ∨ a1 ∈ s0) ∧ (b0 = a ∨ b0 ∈ s0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
b ∈ s0, a0 ∈ s0, a ∈ s0 ⊦ a0 = b, # s0 = 1, # s0 = 0, ∃ a1, b0. a1 ≠ b0 ∧ a1 ∈ s0 ∧ b0 ∈ s0

Rule argument

Additional information

Used simplifier rules: eq: ⊦ a ∈ s → s ++ a = s;%ne-01: ⊦ a ∈ s → s ≠ ∅;%ne-02: ⊦ a ∈ s → (s = ∅ ↔ false);