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Proof step 3 from proof lextrans (previous step)

a0 ≪1 a1 ∨ a0 = a1 ∧ b0 ≪2 b1, a ≪1 a0 ∨ a = a0 ∧ b ≪2 b0 ⊦ a ≪1 a1, a = a1 ∧ b ≪2 b1

User applied rule case distinction

And closed the goal!

Rule argument

pos: left 1

Additional information

Used simplifier rules: trans: ⊦ b ≪2 b0 ∧ b0 ≪2 b1 → b ≪2 b1; (forward) from specification wellfounded[lexical1]%trans: ⊦ a2 ≪1 a3 ∧ a3 ≪1 a4 → a2 ≪1 a4; (forward) from specification wellfounded[lexical]