Thinning lemma in simply typed lambda calculus
$begingroup$
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?
lambda-calculus
asked Jan 14 at 4:13
user695931user695931
16811
$endgroup$
add a comment |
$begingroup$
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?
lambda-calculus
asked Jan 14 at 4:13
user695931user695931
16811
$endgroup$
add a comment |
$begingroup$
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?
lambda-calculus
asked Jan 14 at 4:13
user695931user695931
16811
$endgroup$
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
lambda-calculus
asked Jan 14 at 4:13
user695931user695931
16811
asked Jan 14 at 4:13
user695931user695931
16811
asked Jan 14 at 4:13
user695931user695931
16811
asked Jan 14 at 4:13
user695931user695931
16811
asked Jan 14 at 4:13
user695931user695931
16811
16811
add a comment |
add a comment |
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This may just be a consequence of Convention 1.7.2 From now on, we identify α-convertible λ-terms.
answered Jan 14 at 4:29
user695931user695931
16811
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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.
answered Jan 14 at 4:29
user695931user695931
16811
$endgroup$
add a comment |
$begingroup$
This may just be a consequence of Convention 1.7.2 From now on, we identify α-convertible λ-terms.
answered Jan 14 at 4:29
user695931user695931
16811
$endgroup$
add a comment |
$begingroup$
This may just be a consequence of Convention 1.7.2 From now on, we identify α-convertible λ-terms.
answered Jan 14 at 4:29
user695931user695931
16811
$endgroup$
This may just be a consequence of Convention 1.7.2 From now on, we identify α-convertible λ-terms.
answered Jan 14 at 4:29
user695931user695931
16811
answered Jan 14 at 4:29
user695931user695931
16811
answered Jan 14 at 4:29
user695931user695931
16811
answered Jan 14 at 4:29
user695931user695931
16811
16811
add a comment |
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