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Proof step 10 from proof size-three (previous step)

b ∈ s, 2 < # s, a ∈ s ⊦ #(∅ ++ a ++ b) < # s, # s = 0, b = a, ∃ a, b, c. a ∈ s ∧ b ∈ s ∧ c ∈ s ∧ a ≠ b ∧ a ≠ c ∧ b ≠ c

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
2 < #(∅ ++ a ++ b), # s ≤ #(∅ ++ a ++ b), a ∈ s, 2 < # s, b ∈ s ⊦ b = a, # s = 0, #(∅ ++ a ++ b) = 0, ∃ a, b, c. a ∈ s ∧ b ∈ s ∧ c ∈ s ∧ a ≠ b ∧ a ≠ c ∧ b ≠ c

Rule argument

Additional information

Used simplifier rules: size-zero: ⊦ # s ≠ 0 → s ≠ ∅;%ne: ⊦ ∅ ≠ s ++ a;%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat