Help understand beta reduction example












4














I am currently reading a text book on distributed computing systems that includes a short introduction to $lambda$-calculus. There is an example of evaluating the sequence $(((if space space true) space space 4) space space 5)$ which is written below.




  1. $spacespace(((lambda b.lambda t.lambda e.((b space t) space e) space lambda x.lambda y.x) space 4) space 5) $


  2. $spacespace((lambda t.lambda e ((lambda x.lambda y.x space space t) space e) space 4) space 5) $


  3. $spacespace(lambda e ((lambda x.lambda y.x space space 4) space space e) space space 5) $


  4. $spacespace((lambda x.lambda y.x space space 4) space space 5) $


  5. $spacespace(lambda y.4 space space 5) $


  6. $spacespace 4$



The author has used $beta$-reduction on each line and I can follow up until the second to last reduction. Could someone explain how we get from line 4 to line 6?






computer-science lambda-calculus







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  • 2




    This was down-voted within 30 seconds of posting - if I can improve the question please let me know!
    – EdwardHaigh
    Nov 24 at 21:31








  • 1




    In the first line there is a typo because $(((λb.λt.λe((b t) e) λx.λt.λx) 4) 5)$ is not a $lambda$-term. In particular, what does $lambda x ) 4$ mean?
    – Taroccoesbrocco
    Nov 24 at 22:25






  • 1




    @Taroccoesbrocco - You're quite right, I made two typos. It should now be correct.
    – EdwardHaigh
    Nov 24 at 22:31
















4














I am currently reading a text book on distributed computing systems that includes a short introduction to $lambda$-calculus. There is an example of evaluating the sequence $(((if space space true) space space 4) space space 5)$ which is written below.




  1. $spacespace(((lambda b.lambda t.lambda e.((b space t) space e) space lambda x.lambda y.x) space 4) space 5) $


  2. $spacespace((lambda t.lambda e ((lambda x.lambda y.x space space t) space e) space 4) space 5) $


  3. $spacespace(lambda e ((lambda x.lambda y.x space space 4) space space e) space space 5) $


  4. $spacespace((lambda x.lambda y.x space space 4) space space 5) $


  5. $spacespace(lambda y.4 space space 5) $


  6. $spacespace 4$



The author has used $beta$-reduction on each line and I can follow up until the second to last reduction. Could someone explain how we get from line 4 to line 6?






computer-science lambda-calculus







share|cite|improve this question




















  • 2




    This was down-voted within 30 seconds of posting - if I can improve the question please let me know!
    – EdwardHaigh
    Nov 24 at 21:31








  • 1




    In the first line there is a typo because $(((λb.λt.λe((b t) e) λx.λt.λx) 4) 5)$ is not a $lambda$-term. In particular, what does $lambda x ) 4$ mean?
    – Taroccoesbrocco
    Nov 24 at 22:25






  • 1




    @Taroccoesbrocco - You're quite right, I made two typos. It should now be correct.
    – EdwardHaigh
    Nov 24 at 22:31














4












4








4







I am currently reading a text book on distributed computing systems that includes a short introduction to $lambda$-calculus. There is an example of evaluating the sequence $(((if space space true) space space 4) space space 5)$ which is written below.




  1. $spacespace(((lambda b.lambda t.lambda e.((b space t) space e) space lambda x.lambda y.x) space 4) space 5) $


  2. $spacespace((lambda t.lambda e ((lambda x.lambda y.x space space t) space e) space 4) space 5) $


  3. $spacespace(lambda e ((lambda x.lambda y.x space space 4) space space e) space space 5) $


  4. $spacespace((lambda x.lambda y.x space space 4) space space 5) $


  5. $spacespace(lambda y.4 space space 5) $


  6. $spacespace 4$



The author has used $beta$-reduction on each line and I can follow up until the second to last reduction. Could someone explain how we get from line 4 to line 6?






computer-science lambda-calculus







share|cite|improve this question















I am currently reading a text book on distributed computing systems that includes a short introduction to $lambda$-calculus. There is an example of evaluating the sequence $(((if space space true) space space 4) space space 5)$ which is written below.




  1. $spacespace(((lambda b.lambda t.lambda e.((b space t) space e) space lambda x.lambda y.x) space 4) space 5) $


  2. $spacespace((lambda t.lambda e ((lambda x.lambda y.x space space t) space e) space 4) space 5) $


  3. $spacespace(lambda e ((lambda x.lambda y.x space space 4) space space e) space space 5) $


  4. $spacespace((lambda x.lambda y.x space space 4) space space 5) $


  5. $spacespace(lambda y.4 space space 5) $


  6. $spacespace 4$



The author has used $beta$-reduction on each line and I can follow up until the second to last reduction. Could someone explain how we get from line 4 to line 6?






computer-science lambda-calculus




computer-science lambda-calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




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edited Nov 24 at 23:09









Taroccoesbrocco

4,92761838




4,92761838










asked Nov 24 at 21:29









EdwardHaigh

285




285








  • 2




    This was down-voted within 30 seconds of posting - if I can improve the question please let me know!
    – EdwardHaigh
    Nov 24 at 21:31








  • 1




    In the first line there is a typo because $(((λb.λt.λe((b t) e) λx.λt.λx) 4) 5)$ is not a $lambda$-term. In particular, what does $lambda x ) 4$ mean?
    – Taroccoesbrocco
    Nov 24 at 22:25






  • 1




    @Taroccoesbrocco - You're quite right, I made two typos. It should now be correct.
    – EdwardHaigh
    Nov 24 at 22:31














  • 2




    This was down-voted within 30 seconds of posting - if I can improve the question please let me know!
    – EdwardHaigh
    Nov 24 at 21:31








  • 1




    In the first line there is a typo because $(((λb.λt.λe((b t) e) λx.λt.λx) 4) 5)$ is not a $lambda$-term. In particular, what does $lambda x ) 4$ mean?
    – Taroccoesbrocco
    Nov 24 at 22:25






  • 1




    @Taroccoesbrocco - You're quite right, I made two typos. It should now be correct.
    – EdwardHaigh
    Nov 24 at 22:31








2




2




This was down-voted within 30 seconds of posting - if I can improve the question please let me know!
– EdwardHaigh
Nov 24 at 21:31






This was down-voted within 30 seconds of posting - if I can improve the question please let me know!
– EdwardHaigh
Nov 24 at 21:31






1




1




In the first line there is a typo because $(((λb.λt.λe((b t) e) λx.λt.λx) 4) 5)$ is not a $lambda$-term. In particular, what does $lambda x ) 4$ mean?
– Taroccoesbrocco
Nov 24 at 22:25




In the first line there is a typo because $(((λb.λt.λe((b t) e) λx.λt.λx) 4) 5)$ is not a $lambda$-term. In particular, what does $lambda x ) 4$ mean?
– Taroccoesbrocco
Nov 24 at 22:25




1




1




@Taroccoesbrocco - You're quite right, I made two typos. It should now be correct.
– EdwardHaigh
Nov 24 at 22:31




@Taroccoesbrocco - You're quite right, I made two typos. It should now be correct.
– EdwardHaigh
Nov 24 at 22:31










2 Answers
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In line 4. there is a $beta$-redex $(lambda x lambda y. x 4)$, where $lambda x lambda y. x$ is applied to $4$.
This means that $(lambda x lambda y. x 4)$ rewrites$-$via a $beta$-step$-$to $(lambda y. x){4 / x} $ (which is $(lambda y. x)$ where the free occurrence of $x$ is replaced by $4$), i.e. $ lambda y. 4$.



Since the $beta$-redex $(lambda x lambda y. x 4)$ is inside a context (i.e. it is a sub-term of a bigger term), then from line 4., if we put the sub-term $(lambda x lambda y. x 4)$ into its context, we have:
begin{align}
((lambda x lambda y. x 4) 5) &to_beta (lambda y. x 5){4 / x} = (lambda y. 4 5)
\ &to_beta 4 {5/y} = 5
end{align}
where, again, $(lambda y. 4 5) $ is a $beta$-redex and then rewrites$-$via a $beta$-step$-$to $4 {5/y}$ (which is $4$ where the free occurrences of $y$ are replaced by $5$), i.e. $4$ because there are not free occurrences of $y$ in $4$.






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    When going from line 4 to line 5, we substitute $x=4$ into $lambda y. x$, because the inner argument is bound to the $lambda x$. The 5 is then passed to the result and bound to the $lambda y$, so that we substitute $y=5$ into $lambda y. 4$. Since $y$ does not appear free in $4$, this does not alter the expression, so we end up with $4$.






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      In line 4. there is a $beta$-redex $(lambda x lambda y. x 4)$, where $lambda x lambda y. x$ is applied to $4$.
      This means that $(lambda x lambda y. x 4)$ rewrites$-$via a $beta$-step$-$to $(lambda y. x){4 / x} $ (which is $(lambda y. x)$ where the free occurrence of $x$ is replaced by $4$), i.e. $ lambda y. 4$.



      Since the $beta$-redex $(lambda x lambda y. x 4)$ is inside a context (i.e. it is a sub-term of a bigger term), then from line 4., if we put the sub-term $(lambda x lambda y. x 4)$ into its context, we have:
      begin{align}
      ((lambda x lambda y. x 4) 5) &to_beta (lambda y. x 5){4 / x} = (lambda y. 4 5)
      \ &to_beta 4 {5/y} = 5
      end{align}
      where, again, $(lambda y. 4 5) $ is a $beta$-redex and then rewrites$-$via a $beta$-step$-$to $4 {5/y}$ (which is $4$ where the free occurrences of $y$ are replaced by $5$), i.e. $4$ because there are not free occurrences of $y$ in $4$.






      share|cite|improve this answer


























        2














        In line 4. there is a $beta$-redex $(lambda x lambda y. x 4)$, where $lambda x lambda y. x$ is applied to $4$.
        This means that $(lambda x lambda y. x 4)$ rewrites$-$via a $beta$-step$-$to $(lambda y. x){4 / x} $ (which is $(lambda y. x)$ where the free occurrence of $x$ is replaced by $4$), i.e. $ lambda y. 4$.



        Since the $beta$-redex $(lambda x lambda y. x 4)$ is inside a context (i.e. it is a sub-term of a bigger term), then from line 4., if we put the sub-term $(lambda x lambda y. x 4)$ into its context, we have:
        begin{align}
        ((lambda x lambda y. x 4) 5) &to_beta (lambda y. x 5){4 / x} = (lambda y. 4 5)
        \ &to_beta 4 {5/y} = 5
        end{align}
        where, again, $(lambda y. 4 5) $ is a $beta$-redex and then rewrites$-$via a $beta$-step$-$to $4 {5/y}$ (which is $4$ where the free occurrences of $y$ are replaced by $5$), i.e. $4$ because there are not free occurrences of $y$ in $4$.






        share|cite|improve this answer
























          2












          2








          2






          In line 4. there is a $beta$-redex $(lambda x lambda y. x 4)$, where $lambda x lambda y. x$ is applied to $4$.
          This means that $(lambda x lambda y. x 4)$ rewrites$-$via a $beta$-step$-$to $(lambda y. x){4 / x} $ (which is $(lambda y. x)$ where the free occurrence of $x$ is replaced by $4$), i.e. $ lambda y. 4$.



          Since the $beta$-redex $(lambda x lambda y. x 4)$ is inside a context (i.e. it is a sub-term of a bigger term), then from line 4., if we put the sub-term $(lambda x lambda y. x 4)$ into its context, we have:
          begin{align}
          ((lambda x lambda y. x 4) 5) &to_beta (lambda y. x 5){4 / x} = (lambda y. 4 5)
          \ &to_beta 4 {5/y} = 5
          end{align}
          where, again, $(lambda y. 4 5) $ is a $beta$-redex and then rewrites$-$via a $beta$-step$-$to $4 {5/y}$ (which is $4$ where the free occurrences of $y$ are replaced by $5$), i.e. $4$ because there are not free occurrences of $y$ in $4$.






          share|cite|improve this answer












          In line 4. there is a $beta$-redex $(lambda x lambda y. x 4)$, where $lambda x lambda y. x$ is applied to $4$.
          This means that $(lambda x lambda y. x 4)$ rewrites$-$via a $beta$-step$-$to $(lambda y. x){4 / x} $ (which is $(lambda y. x)$ where the free occurrence of $x$ is replaced by $4$), i.e. $ lambda y. 4$.



          Since the $beta$-redex $(lambda x lambda y. x 4)$ is inside a context (i.e. it is a sub-term of a bigger term), then from line 4., if we put the sub-term $(lambda x lambda y. x 4)$ into its context, we have:
          begin{align}
          ((lambda x lambda y. x 4) 5) &to_beta (lambda y. x 5){4 / x} = (lambda y. 4 5)
          \ &to_beta 4 {5/y} = 5
          end{align}
          where, again, $(lambda y. 4 5) $ is a $beta$-redex and then rewrites$-$via a $beta$-step$-$to $4 {5/y}$ (which is $4$ where the free occurrences of $y$ are replaced by $5$), i.e. $4$ because there are not free occurrences of $y$ in $4$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 24 at 23:03









          Taroccoesbrocco

          4,92761838




          4,92761838























              2














              When going from line 4 to line 5, we substitute $x=4$ into $lambda y. x$, because the inner argument is bound to the $lambda x$. The 5 is then passed to the result and bound to the $lambda y$, so that we substitute $y=5$ into $lambda y. 4$. Since $y$ does not appear free in $4$, this does not alter the expression, so we end up with $4$.






              share|cite|improve this answer


























                2














                When going from line 4 to line 5, we substitute $x=4$ into $lambda y. x$, because the inner argument is bound to the $lambda x$. The 5 is then passed to the result and bound to the $lambda y$, so that we substitute $y=5$ into $lambda y. 4$. Since $y$ does not appear free in $4$, this does not alter the expression, so we end up with $4$.






                share|cite|improve this answer
























                  2












                  2








                  2






                  When going from line 4 to line 5, we substitute $x=4$ into $lambda y. x$, because the inner argument is bound to the $lambda x$. The 5 is then passed to the result and bound to the $lambda y$, so that we substitute $y=5$ into $lambda y. 4$. Since $y$ does not appear free in $4$, this does not alter the expression, so we end up with $4$.






                  share|cite|improve this answer












                  When going from line 4 to line 5, we substitute $x=4$ into $lambda y. x$, because the inner argument is bound to the $lambda x$. The 5 is then passed to the result and bound to the $lambda y$, so that we substitute $y=5$ into $lambda y. 4$. Since $y$ does not appear free in $4$, this does not alter the expression, so we end up with $4$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 24 at 23:01









                  Kevin

                  1,597722




                  1,597722






























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