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

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

User applied rule case distinction

And obtained the following new goals

Proof step 3:
¬(∀ b0. b0 = a ∨ b0 ∈ s → b0 < b), ∀ b0. b0 ∈ s → b0 < b ⊦ ¬ a < b
Proof step 5:
¬ a < b, ∀ b0. b0 ∈ s → b0 < b ⊦ ¬(∀ b0. b0 = a ∨ b0 ∈ s → b0 < b)

Rule argument

pos: right 1

Additional information