⍺ ack ⍵ ←→ f⍢(3∘+) ⍵   ⇒  (⍺+1)ack ⍵ ←→ f⍣(1+⍵)⍢(3∘+) 1

Using the lemma (or otherwise), it can be shown that:

0∘ack = 1∘+⍢(3∘+)
1∘ack = 2∘+⍢(3∘+)
2∘ack = 2∘×⍢(3∘+)
3∘ack = 2∘*⍢(3∘+)
4∘ack = */∘(⍴∘2)⍢(3∘+)
5∘ack = {*/∘(⍴∘2)⍣(1+⍵)⍢(3∘+) 1}

From http://xkcd.com/207