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Proof step 21 from proof size-three-02 (previous step)

# s ≠ 0 → a ∈ s, ∃ a, b, c. a ≠ b ∧ a ≠ c ∧ b ≠ c ∧ s = ∅ ++ a ++ b ++ c ⊦ # s = 3

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 22:
a = c ∨ a = b ∨ a = a0 ⊦ #(∅ ++ a0 ++ b ++ c) = 0, #(∅ ++ a0 ++ b ++ c) = 3, a0 = c, a0 = b, b = c

Rule argument

Additional information

Used simplifier rules: In-insert: ⊦ a ∈ s ++ b ↔ a = b ∨ a ∈ s;%in-04: ⊦ a ∈ ∅ ++ b ↔ a = b;%size-zero-02: ⊦ # s = 0 → s = ∅;%ne: ⊦ ∅ ≠ s ++ a;%size-zero: ⊦ # s ≠ 0 → s ≠ ∅;