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Proof step 2 from proof lextrans (previous step)

lex<(a, b, a0, b0), lex<(a0, b0, a1, b1) ⊦ lex<(a, b, a1, b1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
a0 ≪1 a1 ∨ a0 = a1 ∧ b0 ≪2 b1, a ≪1 a0 ∨ a = a0 ∧ b ≪2 b0 ⊦ a ≪1 a1, a = a1 ∧ b ≪2 b1

Rule argument

Additional information

Used simplifier rules: lex-def: ⊦ lex<(a0, b0, a, b) ↔ a0 ≪1 a ∨ a0 = a ∧ b0 ≪2 b;