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Proof step 13 from proof wflex (previous step)

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)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
a ≪1 a0, ∀ a3, b0. a3 ≪1 a1 ∨ a3 = a1 ∧ b0 ≪2 b → IHPQ(a3, b0), a2 ≪1 a1 ∨ a2 = a1 ∧ b1 ≪2 b ⊦ IHPQ(a, b0), IHPQ(a1, b), IHPQ(a2, b1)

Proof step 13 from proof wflex (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information