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Proof step 14 from proof wflex (
previous step
)
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)
User applied rule all left
And closed the goal!
Rule argument
pos: left 2 terms: a2, b1 abort: F discard: T computed: T
Additional information
Used terms: a2, b1