Does this have a name? $vDash(a_1to(a_2to(cdots(a_{n-1}to a_{n})cdots)))leftrightarrow((a_1wedge...
$begingroup$
For all formula $alpha_1, alpha_2, cdots, alpha_n$,
$$vDash (alpha_1to(alpha_2to(cdots(alpha_{n-1}toalpha_{n})cdots))) leftrightarrow ((alpha_1wedge alpha_2wedgecdotswedgealpha_{n-1})to alpha_n) $$
In fact, I proved this theorem when $n = 3$, But I think this is generally true. Does this theorem have a name? And I would be grateful if you give me a hint of a general case proof.
logic propositional-calculus
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add a comment |
$begingroup$
For all formula $alpha_1, alpha_2, cdots, alpha_n$,
$$vDash (alpha_1to(alpha_2to(cdots(alpha_{n-1}toalpha_{n})cdots))) leftrightarrow ((alpha_1wedge alpha_2wedgecdotswedgealpha_{n-1})to alpha_n) $$
In fact, I proved this theorem when $n = 3$, But I think this is generally true. Does this theorem have a name? And I would be grateful if you give me a hint of a general case proof.
logic propositional-calculus
$endgroup$
4
$begingroup$
Exportation is tha "base" case. Do we need a name for the "general" case ? Maybe Iterated Exportation ...
$endgroup$
– Mauro ALLEGRANZA
Jan 22 at 14:29
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@MauroALLEGRANZA Thank you very much!
$endgroup$
– amoogae
Jan 22 at 15:46
3
$begingroup$
This also corresponds to uncurrying from a propositions-as-types perspective.
$endgroup$
– Derek Elkins
Jan 22 at 20:30
add a comment |
$begingroup$
For all formula $alpha_1, alpha_2, cdots, alpha_n$,
$$vDash (alpha_1to(alpha_2to(cdots(alpha_{n-1}toalpha_{n})cdots))) leftrightarrow ((alpha_1wedge alpha_2wedgecdotswedgealpha_{n-1})to alpha_n) $$
In fact, I proved this theorem when $n = 3$, But I think this is generally true. Does this theorem have a name? And I would be grateful if you give me a hint of a general case proof.
logic propositional-calculus
$endgroup$
For all formula $alpha_1, alpha_2, cdots, alpha_n$,
$$vDash (alpha_1to(alpha_2to(cdots(alpha_{n-1}toalpha_{n})cdots))) leftrightarrow ((alpha_1wedge alpha_2wedgecdotswedgealpha_{n-1})to alpha_n) $$
In fact, I proved this theorem when $n = 3$, But I think this is generally true. Does this theorem have a name? And I would be grateful if you give me a hint of a general case proof.
logic propositional-calculus
logic propositional-calculus
edited Jan 22 at 14:31
Blue
49k870156
49k870156
asked Jan 22 at 14:24
amoogaeamoogae
487
487
4
$begingroup$
Exportation is tha "base" case. Do we need a name for the "general" case ? Maybe Iterated Exportation ...
$endgroup$
– Mauro ALLEGRANZA
Jan 22 at 14:29
$begingroup$
@MauroALLEGRANZA Thank you very much!
$endgroup$
– amoogae
Jan 22 at 15:46
3
$begingroup$
This also corresponds to uncurrying from a propositions-as-types perspective.
$endgroup$
– Derek Elkins
Jan 22 at 20:30
add a comment |
4
$begingroup$
Exportation is tha "base" case. Do we need a name for the "general" case ? Maybe Iterated Exportation ...
$endgroup$
– Mauro ALLEGRANZA
Jan 22 at 14:29
$begingroup$
@MauroALLEGRANZA Thank you very much!
$endgroup$
– amoogae
Jan 22 at 15:46
3
$begingroup$
This also corresponds to uncurrying from a propositions-as-types perspective.
$endgroup$
– Derek Elkins
Jan 22 at 20:30
4
4
$begingroup$
Exportation is tha "base" case. Do we need a name for the "general" case ? Maybe Iterated Exportation ...
$endgroup$
– Mauro ALLEGRANZA
Jan 22 at 14:29
$begingroup$
Exportation is tha "base" case. Do we need a name for the "general" case ? Maybe Iterated Exportation ...
$endgroup$
– Mauro ALLEGRANZA
Jan 22 at 14:29
$begingroup$
@MauroALLEGRANZA Thank you very much!
$endgroup$
– amoogae
Jan 22 at 15:46
$begingroup$
@MauroALLEGRANZA Thank you very much!
$endgroup$
– amoogae
Jan 22 at 15:46
3
3
$begingroup$
This also corresponds to uncurrying from a propositions-as-types perspective.
$endgroup$
– Derek Elkins
Jan 22 at 20:30
$begingroup$
This also corresponds to uncurrying from a propositions-as-types perspective.
$endgroup$
– Derek Elkins
Jan 22 at 20:30
add a comment |
0
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4
$begingroup$
Exportation is tha "base" case. Do we need a name for the "general" case ? Maybe Iterated Exportation ...
$endgroup$
– Mauro ALLEGRANZA
Jan 22 at 14:29
$begingroup$
@MauroALLEGRANZA Thank you very much!
$endgroup$
– amoogae
Jan 22 at 15:46
3
$begingroup$
This also corresponds to uncurrying from a propositions-as-types perspective.
$endgroup$
– Derek Elkins
Jan 22 at 20:30