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Proof step 31 from proof wflex (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 32:
b ≪2 b0, b ≪2 b0 → (∀ IHPQ, a0. (∀ 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)), ∀ a1. a1 ≪1 a → (∀ 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(a, b), IHPQ(a, b0)

Rule argument

pos: left 4 terms: b abort: F discard: T computed: T

Additional information

Used terms: b