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Proof step 2 from proof ls-add (previous step)

s < b ⊦ s ++ a < b ↔ a < b

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
∀ 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);% ⊦ a ∈ s ++ b ↔ a = b ∨ a ∈ s; (s)