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Proof step 3 from proof wflex (previous step)

⊦ ∀ a1. (∀ a2. a2 ≪1 a1 → (∀ IHPQ, a, b. a2 = a ∧ (∀ a, b. (∀ a0, b0. lex<(a0, b0, a, b) → IHPQ(a0, b0)) → IHPQ(a, b)) → IHPQ(a, b))) → (∀ IHPQ, a, b. a1 = a ∧ (∀ a, b. (∀ a0, b0. lex<(a0, b0, a, b) → IHPQ(a0, b0)) → IHPQ(a, b)) → IHPQ(a, b))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
∀ a, b1. (∀ a1, b0. a1 ≪1 a ∨ a1 = a ∧ b0 ≪2 b1 → IHPQ(a1, b0)) → IHPQ(a, b1), ∀ a1. a1 ≪1 a0 → (∀ IHPQ0, b1. (∀ a2, b2. (∀ a3, b0. a3 ≪1 a2 ∨ a3 = a2 ∧ b0 ≪2 b2 → IHPQ0(a3, b0)) → IHPQ0(a2, b2)) → IHPQ0(a1, b1)) ⊦ IHPQ(a0, b)

Rule argument

Additional information

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