text: # Is there a systematic way to know when to alpha-transform free variables?
  
  So, using Church numerals, we define
  
  $$3 = {\lambda} f. {\lambda}x.f(f(f(x)))$$,
  
  and
  
  $$4 = {\lambda} f. {\lambda}x.f(f(f(f(x))))$$.
  
  We can then add with an expression like
  
  $$3\ g\ (4\ g\ z)$$
  
  And this reduces to:
  
  $$(g (g (g (g (g (g (g\ z)))))))$$
  
  ... but why?
  
  $$g$$ is a free variable in each expression, and my understanding is that you must $${\alpha}$$-convert free standing variables in unrelated expressions. Shouldn't we instead end up with something like
  
  $$(g (g (g (g_2 (g_2 (g_2 (g_2\ z)))))))$$?
  
  • You never $\alpha$-convert free variables. Only bound variables can be $\alpha$-converted. – chi Jan 28 at 9:41
  • @chi Thank you for that clarification! That was a real element of my misunderstanding. You should write that up into a partial answer so that it becomes part of the record for future visitors. – Ben I. Jan 28 at 13:03
  
  The answer here is the same as in the other question: one thing is missing here!
  
  $$3 + 4 = \lambda g . \lambda z . 3 g (4 g z) = \lambda g . \lambda z . 7 g z$$
  
  Note that $$g$$ is now a lambda parameter, not a free variable! So now if you want to apply this to something, it'll get substituted in the same everywhere:
  
  $$7 q r = (\lambda g . \lambda z . 7 g z) q r = q q q q q q q r$$
  
  • This could be misunderstood, though. Free variables are never renamed. Variable $g$ above occurs under a $\lambda$ so we could rename it if we wished: $\lambda g.\lambda z.\ 7 g z$ could be converted to $\lambda h.\lambda x.\ 7 h x$. – chi Jan 28 at 9:43
  • @chi Fair; edited – Draconis Jan 28 at 16:04
  
  Free variables never get $$\alpha$$-converted, only bound variables can.
  
  In the term $$(\lambda x.\ xy)$$ we can rename the bound variable $$x$$ to any other variable (except $$y$$, since that would cause a name clash). For instance, we can obtain $$(\lambda z.\ zy)$$. Instead, we can never rename $$y$$, since that is free, not being under any $$\lambda y$$.
  
  By contrast, in $$(\lambda y.\ \lambda x.\ xy)$$ we can $$\alpha$$-convert both variables to any other pair of (distinct) variables. This is because now they are both bound.
  
  In the OP's example, $$g$$ is free, so no renaming is possible. At most, we can $$\alpha$$-convert $$f$$ and $$x$$, but not $$g$$.
  
  • So, could you think about bound variables as local variables, and free variables as globals? Meaning that a free $g$ in one location would represent the same idea as another free $g$ elsewhere in the problem, while a bound $x$ would only have a designated meaning within its own function. – Ben I. Feb 1 at 13:33
  • @BenI. Yes, that's a rough way to think about it. In an expression/command/piece of syntax $e$, bound variables are exactly the variables which are declared/defined inside $e$ -- the "local" variables, if you want. The others must instead be defined "outside" $e$. E.g. in imperative code {int x=5; print(x+y);} we have x bound and y free. If we "insert" that snippet in a large context, e.g. {int y=4; if(true) { int x=5; print(x+y);}} now the free y is "captured" by the outer declaration, so y is now bound in this larger command. – chi Feb 1 at 13:59
  • @BenI. Note that in $\lambda$-calculus we do not have global definitions, so it's best to think of a free variable as "something which has to be defined from outside the $\lambda$-term". – chi Feb 1 at 14:02
  • Sure, that makes sense. I was using "global" a little carelessly, I meant it more as you said. :) – Ben I. Feb 1 at 17:44
explanation: The text extract discusses the concept of alpha-conversion in the context of lambda calculus, specifically focusing on the treatment of free variables. It presents a question about why free variables are not alpha-converted and includes a discussion and an answer clarifying that only bound variables can be alpha-converted, not free variables.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: # Is there a systematic way to know when to alpha-transform free variables?
  
  So, using Church numerals, we define
  
  $$3 = {\lambda} f. {\lambda}x.f(f(f(x)))$$,
  
  and
  
  $$4 = {\lambda} f. {\lambda}x.f(f(f(f(x))))$$.
  
  We can then add with an expression like
  
  $$3\ g\ (4\ g\ z)$$
  
  And this reduces to:
  
  $$(g (g (g (g (g (g (g\ z)))))))$$
  
  ... but why?
  
  $$g$$ is a free variable in each expression, and my understanding is that you must $${\alpha}$$-convert free standing variables in unrelated expressions. Shouldn't we instead end up with something like
  
  $$(g (g (g (g_2 (g_2 (g_2 (g_2\ z)))))))$$?
  
  • You never $\alpha$-convert free variables. Only bound variables can be $\alpha$-converted. – chi Jan 28 at 9:41
  • @chi Thank you for that clarification! That was a real element of my misunderstanding. You should write that up into a partial answer so that it becomes part of the record for future visitors. – Ben I. Jan 28 at 13:03
  
  The answer here is the same as in the other question: one thing is missing here!
  
  $$3 + 4 = \lambda g . \lambda z . 3 g (4 g z) = \lambda g . \lambda z . 7 g z$$
  
  Note that $$g$$ is now a lambda parameter, not a free variable! So now if you want to apply this to something, it'll get substituted in the same everywhere:
  
  $$7 q r = (\lambda g . \lambda z . 7 g z) q r = q q q q q q q r$$
  
  • This could be misunderstood, though. Free variables are never renamed. Variable $g$ above occurs under a $\lambda$ so we could rename it if we wished: $\lambda g.\lambda z.\ 7 g z$ could be converted to $\lambda h.\lambda x.\ 7 h x$. – chi Jan 28 at 9:43
  • @chi Fair; edited – Draconis Jan 28 at 16:04
  
  Free variables never get $$\alpha$$-converted, only bound variables can.
  
  In the term $$(\lambda x.\ xy)$$ we can rename the bound variable $$x$$ to any other variable (except $$y$$, since that would cause a name clash). For instance, we can obtain $$(\lambda z.\ zy)$$. Instead, we can never rename $$y$$, since that is free, not being under any $$\lambda y$$.
  
  By contrast, in $$(\lambda y.\ \lambda x.\ xy)$$ we can $$\alpha$$-convert both variables to any other pair of (distinct) variables. This is because now they are both bound.
  
  In the OP's example, $$g$$ is free, so no renaming is possible. At most, we can $$\alpha$$-convert $$f$$ and $$x$$, but not $$g$$.
  
  • So, could you think about bound variables as local variables, and free variables as globals? Meaning that a free $g$ in one location would represent the same idea as another free $g$ elsewhere in the problem, while a bound $x$ would only have a designated meaning within its own function. – Ben I. Feb 1 at 13:33
  • @BenI. Yes, that's a rough way to think about it. In an expression/command/piece of syntax $e$, bound variables are exactly the variables which are declared/defined inside $e$ -- the "local" variables, if you want. The others must instead be defined "outside" $e$. E.g. in imperative code {int x=5; print(x+y);} we have x bound and y free. If we "insert" that snippet in a large context, e.g. {int y=4; if(true) { int x=5; print(x+y);}} now the free y is "captured" by the outer declaration, so y is now bound in this larger command. – chi Feb 1 at 13:59
  • @BenI. Note that in $\lambda$-calculus we do not have global definitions, so it's best to think of a free variable as "something which has to be defined from outside the $\lambda$-term". – chi Feb 1 at 14:02
  • Sure, that makes sense. I was using "global" a little carelessly, I meant it more as you said. :) – Ben I. Feb 1 at 17:44
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The text extract discusses the concept of alpha-conversion in the context of lambda calculus, specifically focusing on the treatment of free variables. It presents a question about why free variables are not alpha-converted and includes a discussion and an answer clarifying that only bound variables can be alpha-converted, not free variables.
  
  Label: question-and-answer