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I don't know the correct answer how this reduction should've be done. Should I simply put λfx.fx in a place of m and λzy.zzy in a place of n?

(λmnfx.mf(nfx) λfx.fx λzy.zzy)

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  • $\begingroup$ You're on the right track. Pay special attention to the names of the variables as there is another reduction which must be performed before you can apply terms. $\endgroup$ – c.w.chambers Jun 22 '14 at 13:57
  • $\begingroup$ (λfx.f(f(fx))) I've figured answer like that, is that correct? Please praise my effort with correct answer or solution because I'am dying out of this lambda :( $\endgroup$ – Sheil Jun 22 '14 at 15:50
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You have placed parenthesis around the two final variables in the body. I don't see that right-associativity supported in the stated problem where the third abstraction is \zy.zzy not \zy.z(zy).

I believe the answer is \fx.f(ffx) which is equivalent to \fx.f((ff)x).

Also, this is a duplicate question but I'm not sure how to link to the other one.

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