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Proof step 1 from proof ls-add (conclusion)

⊦ s < b → (¬ s ++ a < b ↔ ¬ a < b)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
∀ b0. b0 ∈ s → b0 < b ⊦ ¬(∀ b0. b0 = a ∨ b0 ∈ s → b0 < b) ↔ ¬ a < b

Rule argument

Additional information

Used simplifier rules: Greater: ⊦ s < a ↔ (∀ b. b ∈ s → b < a);%rax: ⊦ a ∈ s ++ b ↔ ¬ ¬(a = b ∨ a ∈ s); from specification set-basic