Lemma relating to alpha equivalence of lambda terms
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From Type Theory and Formal Proof, An Introduction by Rob Nederpelt and Herman Geuvers:
Lemma 1.7.1 Let $M_{1} =_{alpha} N_{1}$ and $M_{2} =_{alpha} N_{2}$. Then also:
(1) $M_{1}N_{1} =_{alpha} M_{2}N_{2}$,
(2) $lambda x . M_{1} =_{alpha} lambda x . M_{2}$,
(3) $M_{1}[x := N_{1}] =_{alpha} M_{2}[x := N_{2}]$.
Why does this hold? There does not seem to be a stated connection between $M_{1}$ and $M_{2}$ or $N_{1}$ and $N_{2}$.
lambda-calculus
asked Dec 21 '18 at 2:31
user695931user695931
17511
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add a comment |
$begingroup$
From Type Theory and Formal Proof, An Introduction by Rob Nederpelt and Herman Geuvers:
Lemma 1.7.1 Let $M_{1} =_{alpha} N_{1}$ and $M_{2} =_{alpha} N_{2}$. Then also:
(1) $M_{1}N_{1} =_{alpha} M_{2}N_{2}$,
(2) $lambda x . M_{1} =_{alpha} lambda x . M_{2}$,
(3) $M_{1}[x := N_{1}] =_{alpha} M_{2}[x := N_{2}]$.
Why does this hold? There does not seem to be a stated connection between $M_{1}$ and $M_{2}$ or $N_{1}$ and $N_{2}$.
lambda-calculus
asked Dec 21 '18 at 2:31
user695931user695931
17511
$endgroup$
1
$begingroup$
It is just a typo. The hypothesis of lemma 1.7.1 should be $M_1 =_alpha M_2$ and $N_1 =_alpha N_2$.
$endgroup$
– Taroccoesbrocco
Dec 21 '18 at 3:56
$begingroup$
I see. Thank you!
$endgroup$
– user695931
Dec 21 '18 at 4:04
add a comment |
$begingroup$
From Type Theory and Formal Proof, An Introduction by Rob Nederpelt and Herman Geuvers:
Lemma 1.7.1 Let $M_{1} =_{alpha} N_{1}$ and $M_{2} =_{alpha} N_{2}$. Then also:
(1) $M_{1}N_{1} =_{alpha} M_{2}N_{2}$,
(2) $lambda x . M_{1} =_{alpha} lambda x . M_{2}$,
(3) $M_{1}[x := N_{1}] =_{alpha} M_{2}[x := N_{2}]$.
Why does this hold? There does not seem to be a stated connection between $M_{1}$ and $M_{2}$ or $N_{1}$ and $N_{2}$.
lambda-calculus
asked Dec 21 '18 at 2:31
user695931user695931
17511
$endgroup$
From Type Theory and Formal Proof, An Introduction by Rob Nederpelt and Herman Geuvers:
Lemma 1.7.1 Let $M_{1} =_{alpha} N_{1}$ and $M_{2} =_{alpha} N_{2}$. Then also:
(1) $M_{1}N_{1} =_{alpha} M_{2}N_{2}$,
(2) $lambda x . M_{1} =_{alpha} lambda x . M_{2}$,
(3) $M_{1}[x := N_{1}] =_{alpha} M_{2}[x := N_{2}]$.
Why does this hold? There does not seem to be a stated connection between $M_{1}$ and $M_{2}$ or $N_{1}$ and $N_{2}$.
lambda-calculus
lambda-calculus
asked Dec 21 '18 at 2:31
user695931user695931
17511
asked Dec 21 '18 at 2:31
user695931user695931
17511
asked Dec 21 '18 at 2:31
user695931user695931
17511
asked Dec 21 '18 at 2:31
user695931user695931
17511
asked Dec 21 '18 at 2:31
user695931user695931
17511
17511
1
$begingroup$
It is just a typo. The hypothesis of lemma 1.7.1 should be $M_1 =_alpha M_2$ and $N_1 =_alpha N_2$.
$endgroup$
– Taroccoesbrocco
Dec 21 '18 at 3:56
$begingroup$
I see. Thank you!
$endgroup$
– user695931
Dec 21 '18 at 4:04
add a comment |
1
$begingroup$
It is just a typo. The hypothesis of lemma 1.7.1 should be $M_1 =_alpha M_2$ and $N_1 =_alpha N_2$.
$endgroup$
– Taroccoesbrocco
Dec 21 '18 at 3:56
$begingroup$
I see. Thank you!
$endgroup$
– user695931
Dec 21 '18 at 4:04
1
1
$begingroup$
It is just a typo. The hypothesis of lemma 1.7.1 should be $M_1 =_alpha M_2$ and $N_1 =_alpha N_2$.
$endgroup$
– Taroccoesbrocco
Dec 21 '18 at 3:56
$begingroup$
It is just a typo. The hypothesis of lemma 1.7.1 should be $M_1 =_alpha M_2$ and $N_1 =_alpha N_2$.
$endgroup$
– Taroccoesbrocco
Dec 21 '18 at 3:56
$begingroup$
I see. Thank you!
$endgroup$
– user695931
Dec 21 '18 at 4:04
$begingroup$
I see. Thank you!
$endgroup$
– user695931
Dec 21 '18 at 4:04
add a comment |
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$begingroup$
It is just a typo. The hypothesis of lemma 1.7.1 should be $M_1 =_alpha M_2$ and $N_1 =_alpha N_2$.
$endgroup$
– Taroccoesbrocco
Dec 21 '18 at 3:56
$begingroup$
I see. Thank you!
$endgroup$
– user695931
Dec 21 '18 at 4:04