I have following expression to reduce:
$(λmnfx.mf(nfx) λfx.fx λzy.zzy)$
After some substitutions i get the result:
$(λfx. f(f(f x)))$
Is it correct answer? If not, please tell me what is correct one so i could try to get into it. Thanks in advance.
I have following expression to reduce:
$(λmnfx.mf(nfx) λfx.fx λzy.zzy)$
After some substitutions i get the result:
$(λfx. f(f(f x)))$
Is it correct answer? If not, please tell me what is correct one so i could try to get into it. Thanks in advance.
In order to keep the variables straight, I'm going to do an alpha conversion to the first abstraction, replacing f with a and x with b which leaves me with...
Note the third abstraction contains no parenthesis and is therefore equivalent to \zy.(zz)y but is not equivalent to \zy.z(zy)
The second abstraction will replace the m variable and the third abstraction will replace the n variable. The a and b variables are going to be part of the solution abstraction.
Now I resolve the \fx abstraction, replacing f with a and x with ((\zy.zzy)ab).
Finally, I resolve \zy, replacing z with a and y with b.
This is not the same as the answer you have provided because the parenthetical portion is equivalent to "((aa)b)", not "(a(ab))" which would make it a Church Numeral. The variables in Church Numeral abstractions are right associative. Are you certain the problem is stated correctly?