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Proof step 11 from proof size-three-02 (previous step)

# s0 ≠ 0 → (∃ a. a ∈ s0), 2 = # s0 ⊦ a ∈ s0, ∃ a0, b, c. a0 ≠ b ∧ a0 ≠ c ∧ b ≠ c ∧ s0 ++ a = ∅ ++ a0 ++ b ++ c

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
2 = # s0, a0 ∈ s0 ⊦ a ∈ s0, ∃ a0, b, c. a0 ≠ b ∧ a0 ≠ c ∧ b ≠ c ∧ s0 ++ a = ∅ ++ a0 ++ b ++ c

Rule argument

Additional information

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