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Proof step 19 from proof size-three-02 (previous step)

# s ≠ 0 → (∃ a. a ∈ s), 1 = # s ⊦ 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 20:
⊦ a = a0, a0 = a1, a = a1, ∃ a2, b, c. a2 ≠ b ∧ a2 ≠ c ∧ b ≠ c ∧ ∅ ++ a1 ++ a0 ++ a = ∅ ++ a2 ++ b ++ c

Rule argument

Additional information

Used simplifier rules: size-one-02: ⊦ # s = 1 ∧ a ∈ s ∧ b ∈ s → a = b;%one-in: ⊦ # s = 1 ∧ a ∈ s → s = ∅ ++ a;%ne-01: ⊦ a ∈ s → s ≠ ∅;%size-one-01: ⊦ #(∅ ++ a) = 1;%ne: ⊦ ∅ ≠ s ++ a;%in-04: ⊦ a ∈ ∅ ++ b ↔ a = b;