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Proof step 8 from proof restn-firstn-better (previous step)

m0 ≤ # y, n0 + m0 ≤ # y, ∀ n0, x. n0 + m0 ≤ # x ∧ m0 ≤ # x → restn(m0, firstn(n0 + m0, x)) = firstn(n0, restn(m0, x)) ⊦ restn(m0, firstn(n0 + m0, y)) = firstn(n0, restn(m0, y))

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: n0, y

Additional information

Used terms: n0, y