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Proof step 3 from proof sls (previous step)

⊦ a < b ↔ ¬ b < a, a = b

User applied rule insert lemma

And obtained the following new goals

Proof step 4:
⊦ a < b ∨ a = b ∨ b < a
Proof step 5:
a < b ∨ b < a ⊦ a < b ↔ ¬ b < a, a = b

Rule argument

spec: <toplevel> name: totality seq: ⊦ a < b ∨ a = b ∨ b < a subst: [a, b] → [a, b]

Additional information

Used terms: a, b