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Proof step 5 from proof wflex (previous step)

(∀ a1, b0. a1 ≪1 a0 ∨ a1 = a0 ∧ b0 ≪2 b → IHPQ(a1, b0)) → IHPQ(a0, b), ∀ 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)), ∀ a, b1. (∀ a1, b0. a1 ≪1 a ∨ a1 = a ∧ b0 ≪2 b1 → IHPQ(a1, b0)) → IHPQ(a, b1) ⊦ IHPQ(a0, b)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
∀ 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)), a ≪1 a0 ∨ a = a0 ∧ b0 ≪2 b ⊦ IHPQ(a0, b), IHPQ(a, b0)

Proof step 5 from proof wflex (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information