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Proof step 12 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), ∀ a3, b0. a3 ≪1 a1 ∨ a3 = a1 ∧ b0 ≪2 b → IHPQ(a3, b0) ⊦ IHPQ(a, b0), IHPQ(a1, b)

User applied rule all left

And obtained the following new goals

Proof step 13:
a ≪1 a0, (∀ a0, b0. a0 ≪1 a1 ∨ a0 = a1 ∧ b0 ≪2 b → IHPQ(a0, b0)) → IHPQ(a1, b), ∀ a3, b0. a3 ≪1 a1 ∨ a3 = a1 ∧ b0 ≪2 b → IHPQ(a3, b0) ⊦ IHPQ(a1, b), IHPQ(a, b0)

Rule argument

pos: left 2 terms: a1, b abort: F discard: T computed: T

Additional information

Used terms: a1, b