Thinning lemma in simply typed lambda calculus












0












$begingroup$


From "Type Theory and Formal Proof" by Rob Nederpelt and Herman Geuvers:




Definition 2.4.2



(1) A statement is of the form $M : alpha$, where $M in
Lambda_{mathbb{T}}$
and $sigma in mathbb{T}$. In such a
statement, $M$ is called the subject and $alpha$ the type.



(2) A declaration is a statement with a variable as subject.



(3) A context is a list of declarations with different subjects.



Definition 2.4.5 (Derivation rules for $lambda to$)



(var) $Gamma vdash x : alpha$ if $x : alpha in Gamma$



(appl) $Gamma vdash M : alpha to tau quad Gamma vdash N
: alpha implies Gamma vdash M N : tau$



(abst) $Gamma, x : alpha vdash M : tau implies Gamma vdash
lambda x : alpha . M : alpha to tau$



Definition 2.10.1



(2) Context $Gamma'$ is a subcontext of context $Gamma$, or $Gamma'
subseteq Gamma$
, if all declarations occurring in $Gamma'$ also
occur in $Gamma$, in the same order.



Lemma 2.10.5



(1) (Thinning) Let $Gamma'$ and $Gamma''$ be contexts such that
$Gamma' subseteq Gamma''$. If $Gamma' vdash M : alpha$, then
also $Gamma'' vdash M : alpha$.




Note: I replaced the horizontal bar between the premisses and conclusion in the derivation rules with $implies$ since I could not get the bar to typeset as intended.



Suppose I assign the following in Lemma 2.10.5:



$Gamma' = y : B$



$Gamma'' = x : C, y : B$



$M = lambda x : A . y$



$alpha = A to B$



Then



$Gamma' vdash M : alpha = y : B vdash lambda x : A . y : A to B$



$Gamma'' vdash M : alpha = x : C, y : B vdash lambda x : A . y : A to B$.



For a derivation of the first I have:



(i) $y : B, x : A vdash y : B$ (var)



(ii) $y : B vdash lambda x : A . y : A to B$ (abst on i)



I am unable to find a derivation for the second as the lemma implies. Is there a derivation, or am I missing something else?










share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    From "Type Theory and Formal Proof" by Rob Nederpelt and Herman Geuvers:




    Definition 2.4.2



    (1) A statement is of the form $M : alpha$, where $M in
    Lambda_{mathbb{T}}$
    and $sigma in mathbb{T}$. In such a
    statement, $M$ is called the subject and $alpha$ the type.



    (2) A declaration is a statement with a variable as subject.



    (3) A context is a list of declarations with different subjects.



    Definition 2.4.5 (Derivation rules for $lambda to$)



    (var) $Gamma vdash x : alpha$ if $x : alpha in Gamma$



    (appl) $Gamma vdash M : alpha to tau quad Gamma vdash N
    : alpha implies Gamma vdash M N : tau$



    (abst) $Gamma, x : alpha vdash M : tau implies Gamma vdash
    lambda x : alpha . M : alpha to tau$



    Definition 2.10.1



    (2) Context $Gamma'$ is a subcontext of context $Gamma$, or $Gamma'
    subseteq Gamma$
    , if all declarations occurring in $Gamma'$ also
    occur in $Gamma$, in the same order.



    Lemma 2.10.5



    (1) (Thinning) Let $Gamma'$ and $Gamma''$ be contexts such that
    $Gamma' subseteq Gamma''$. If $Gamma' vdash M : alpha$, then
    also $Gamma'' vdash M : alpha$.




    Note: I replaced the horizontal bar between the premisses and conclusion in the derivation rules with $implies$ since I could not get the bar to typeset as intended.



    Suppose I assign the following in Lemma 2.10.5:



    $Gamma' = y : B$



    $Gamma'' = x : C, y : B$



    $M = lambda x : A . y$



    $alpha = A to B$



    Then



    $Gamma' vdash M : alpha = y : B vdash lambda x : A . y : A to B$



    $Gamma'' vdash M : alpha = x : C, y : B vdash lambda x : A . y : A to B$.



    For a derivation of the first I have:



    (i) $y : B, x : A vdash y : B$ (var)



    (ii) $y : B vdash lambda x : A . y : A to B$ (abst on i)



    I am unable to find a derivation for the second as the lemma implies. Is there a derivation, or am I missing something else?










    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      From "Type Theory and Formal Proof" by Rob Nederpelt and Herman Geuvers:




      Definition 2.4.2



      (1) A statement is of the form $M : alpha$, where $M in
      Lambda_{mathbb{T}}$
      and $sigma in mathbb{T}$. In such a
      statement, $M$ is called the subject and $alpha$ the type.



      (2) A declaration is a statement with a variable as subject.



      (3) A context is a list of declarations with different subjects.



      Definition 2.4.5 (Derivation rules for $lambda to$)



      (var) $Gamma vdash x : alpha$ if $x : alpha in Gamma$



      (appl) $Gamma vdash M : alpha to tau quad Gamma vdash N
      : alpha implies Gamma vdash M N : tau$



      (abst) $Gamma, x : alpha vdash M : tau implies Gamma vdash
      lambda x : alpha . M : alpha to tau$



      Definition 2.10.1



      (2) Context $Gamma'$ is a subcontext of context $Gamma$, or $Gamma'
      subseteq Gamma$
      , if all declarations occurring in $Gamma'$ also
      occur in $Gamma$, in the same order.



      Lemma 2.10.5



      (1) (Thinning) Let $Gamma'$ and $Gamma''$ be contexts such that
      $Gamma' subseteq Gamma''$. If $Gamma' vdash M : alpha$, then
      also $Gamma'' vdash M : alpha$.




      Note: I replaced the horizontal bar between the premisses and conclusion in the derivation rules with $implies$ since I could not get the bar to typeset as intended.



      Suppose I assign the following in Lemma 2.10.5:



      $Gamma' = y : B$



      $Gamma'' = x : C, y : B$



      $M = lambda x : A . y$



      $alpha = A to B$



      Then



      $Gamma' vdash M : alpha = y : B vdash lambda x : A . y : A to B$



      $Gamma'' vdash M : alpha = x : C, y : B vdash lambda x : A . y : A to B$.



      For a derivation of the first I have:



      (i) $y : B, x : A vdash y : B$ (var)



      (ii) $y : B vdash lambda x : A . y : A to B$ (abst on i)



      I am unable to find a derivation for the second as the lemma implies. Is there a derivation, or am I missing something else?










      share|cite|improve this question









      $endgroup$




      From "Type Theory and Formal Proof" by Rob Nederpelt and Herman Geuvers:




      Definition 2.4.2



      (1) A statement is of the form $M : alpha$, where $M in
      Lambda_{mathbb{T}}$
      and $sigma in mathbb{T}$. In such a
      statement, $M$ is called the subject and $alpha$ the type.



      (2) A declaration is a statement with a variable as subject.



      (3) A context is a list of declarations with different subjects.



      Definition 2.4.5 (Derivation rules for $lambda to$)



      (var) $Gamma vdash x : alpha$ if $x : alpha in Gamma$



      (appl) $Gamma vdash M : alpha to tau quad Gamma vdash N
      : alpha implies Gamma vdash M N : tau$



      (abst) $Gamma, x : alpha vdash M : tau implies Gamma vdash
      lambda x : alpha . M : alpha to tau$



      Definition 2.10.1



      (2) Context $Gamma'$ is a subcontext of context $Gamma$, or $Gamma'
      subseteq Gamma$
      , if all declarations occurring in $Gamma'$ also
      occur in $Gamma$, in the same order.



      Lemma 2.10.5



      (1) (Thinning) Let $Gamma'$ and $Gamma''$ be contexts such that
      $Gamma' subseteq Gamma''$. If $Gamma' vdash M : alpha$, then
      also $Gamma'' vdash M : alpha$.




      Note: I replaced the horizontal bar between the premisses and conclusion in the derivation rules with $implies$ since I could not get the bar to typeset as intended.



      Suppose I assign the following in Lemma 2.10.5:



      $Gamma' = y : B$



      $Gamma'' = x : C, y : B$



      $M = lambda x : A . y$



      $alpha = A to B$



      Then



      $Gamma' vdash M : alpha = y : B vdash lambda x : A . y : A to B$



      $Gamma'' vdash M : alpha = x : C, y : B vdash lambda x : A . y : A to B$.



      For a derivation of the first I have:



      (i) $y : B, x : A vdash y : B$ (var)



      (ii) $y : B vdash lambda x : A . y : A to B$ (abst on i)



      I am unable to find a derivation for the second as the lemma implies. Is there a derivation, or am I missing something else?







      lambda-calculus






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      asked Jan 14 at 4:13









      user695931user695931

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          This may just be a consequence of Convention 1.7.2 From now on, we identify α-convertible λ-terms.






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            $begingroup$

            This may just be a consequence of Convention 1.7.2 From now on, we identify α-convertible λ-terms.






            share|cite|improve this answer









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              This may just be a consequence of Convention 1.7.2 From now on, we identify α-convertible λ-terms.






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                $begingroup$

                This may just be a consequence of Convention 1.7.2 From now on, we identify α-convertible λ-terms.






                share|cite|improve this answer









                $endgroup$



                This may just be a consequence of Convention 1.7.2 From now on, we identify α-convertible λ-terms.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 14 at 4:29









                user695931user695931

                17511




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