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Proof step 5 from proof ls-add (previous step)

¬ a < b, ∀ b0. b0 ∈ s → b0 < b ⊦ ¬(∀ b0. b0 = a ∨ b0 ∈ s → b0 < b)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
∀ b0. b0 = a ∨ b0 ∈ s → b0 < b, ∀ b0. b0 ∈ s → b0 < b ⊦ a < b

Rule argument

Additional information