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

2 < # s, a ∈ s, b ∈ s ⊦ b = a, #(∅ ++ a ++ b) < # s, ∃ 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 12:
2 < # s, 2 < #(∅ ++ a ++ b), b ∈ s, a ∈ s, # s ≤ #(∅ ++ a ++ b) ⊦ b = a, ∃ a, b, c. a ∈ s ∧ b ∈ s ∧ c ∈ s ∧ a ≠ b ∧ a ≠ c ∧ b ≠ c

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat% ⊦ m < n ↔ ¬ n ≤ m; (ss)