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Proof step 16 from proof size-three-02 (previous step)

1 = # s ⊦ a ∈ s ++ a0, a = a0, a0 ∈ s, a ∈ s, ∃ a1, b, c. a1 ≠ b ∧ a1 ≠ c ∧ b ≠ c ∧ s ++ a0 ++ a = ∅ ++ a1 ++ b ++ c

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
1 = # s ⊦ a = a0, a0 ∈ s, a ∈ s, ∃ a1, b, c. a1 ≠ b ∧ a1 ≠ c ∧ b ≠ c ∧ s ++ a0 ++ a = ∅ ++ a1 ++ b ++ c

Rule argument

Additional information

Used simplifier rules: in-02: ⊦ a ≠ b → (a ∈ s ++ b ↔ a ∈ s);