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Directory:Jon Awbrey/Papers/Propositions As Types
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Revision as of 14:54, 24 March 2009
468 bytes added
,
14:54, 24 March 2009
→Step 1
:
markup
Line 193:
Line 193:
We are asked to find an explication of <math>\operatorname{P}</math> in terms of primitive combinators.
We are asked to find an explication of <math>\operatorname{P}</math> in terms of primitive combinators.
−
<pre>
Proceed as follows:
Proceed as follows:
−
(xy)z
=
(xy)(x(
zK
))
=
x(y((
zK
)S))
+
{| align="center" cellpadding="8" width="90%"
−
+
|
−
(
zK
)S
=
(
zK
)(z(
SK
))
=
z(K((
SK
)S))
+
<math>\begin{array}{ccccc}
−
+
(xy)z
−
=>
+
&
=
&
−
+
(xy)(x(
z\operatorname{K}
))
−
x(y(
zP
))
=
(xy)z
=
x(y(z(K((
SK
)S))))
+
&
=
&
−
+
x(y((
z\operatorname{K}
)
\operatorname{
S
}
))
−
=>
+
\\[8pt]
−
+
(
z\operatorname{K}
)
\operatorname{
S
}
−
P
=
(K((
SK
)S))
+
&
=
&
−
</
pre
>
+
(
z\operatorname{K}
)(z(
\operatorname{S}\operatorname{K}
))
+
&
=
&
+
z(
\operatorname{
K
}
((
\operatorname{S}\operatorname{K}
)
\operatorname{
S
}
))
+
\\[6pt]
+
& & \Downarrow
+
\\[8pt]
+
x(y(
z\operatorname{P}
))
+
&
=
&
+
(xy)z
+
&
=
&
+
x(y(z(
\operatorname{
K
}
((
\operatorname{S}\operatorname{K}
)
\operatorname{
S
}
))))
+
\\[8pt]
+
& & \Downarrow
+
\\[8pt]
+
\operatorname{
P
}
+
&
=
&
+
(
\operatorname{
K
}
((
\operatorname{S}\operatorname{K}
)
\operatorname{
S
}
))
+
\end{array}
</
math
>
+
|}
===Step 2===
===Step 2===
Jon Awbrey
12,089
edits