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Proof step 29 from proof wflex (previous step)

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

User applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: a2, b2 abort: F discard: T computed: T

Additional information

Used terms: a2, b2