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Proof step 6 from proof first-append (
previous step
)
m0 ≤ # y0, ∀ x, y. m0 ≤ # x → firstn(m0, x + y) = firstn(m0, x) ⊦ firstn(m0, y0 + y) = firstn(m0, y0)
Heuristic quantifier closing applied rule all left
And closed the goal!
Rule argument
pos: left 2 terms: y0, y
Additional information
Used terms: y0, y