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Proof step 1 from proof wflex (conclusion)

∀ a, b. (∀ a0, b0. lex<(a0, b0, a, b) → IHPQ(a0, b0)) → IHPQ(a, b) ⊦ IHPQ(a, b)

User applied rule induction

And obtained the following new goals

Proof step 2:
∀ a1. (∀ a2. a2 ≪1 a1 → IHP0(a2)) → IHP0(a1) ⊦ IHP0(a1)
Proof step 3:
⊦ ∀ 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))

Rule argument

spec: wellfounded, inst: lexical-param name: wellfounded seq: ∀ a1. (∀ a2. a2 ≪1 a1 → IHP0(a2)) → IHP0(a1) ⊦ IHP0(a1) subst: [IHP0, a1] → [λ a0. ∀ IHPQ, a, b. a0 = a ∧ (∀ a, b. (∀ a0, b0. lex<(a0, b0, a, b) → IHPQ(a0, b0)) → IHPQ(a, b)) → IHPQ(a, b), a]

Additional information

Used terms: λ a0. ∀ IHPQ, a, b. a0 = a ∧ (∀ a, b. (∀ a0, b0. lex<(a0, b0, a, b) → IHPQ(a0, b0)) → IHPQ(a, b)) → IHPQ(a, b), a