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Proof step 10 from proof wflex (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 11:
a ≪1 a0, (∀ a2, b2. (∀ a3, b0. a3 ≪1 a2 ∨ a3 = a2 ∧ b0 ≪2 b2 → IHPQ(a3, b0)) → IHPQ(a2, b2)) → IHPQ(a, b0), ∀ a, b1. (∀ a1, b0. a1 ≪1 a ∨ a1 = a ∧ b0 ≪2 b1 → IHPQ(a1, b0)) → IHPQ(a, b1) ⊦ IHPQ(a, b0)

Rule argument

pos: left 3 terms: IHPQ, b0 abort: F discard: T computed: T

Additional information

Used terms: IHPQ, b0