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Proof step 13 from proof firstn-firstn (previous step)

m ≤ # y, m0 ≤ m, m0 ≤ # y, ∀ n, x. m0 ≤ n ∧ n ≤ # x ∧ m0 ≤ # x → firstn(m0, firstn(n, x)) = firstn(m0, x) ⊦ firstn(m0, firstn(m, y)) = firstn(m0, y)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 4 terms: m, y

Additional information

Used terms: m, y