Problem 3.24:

pair:
 <pair.Qrest, V1.V2.S> => <Qrest, <v2,V1>.S>

Because there is an automatic transition between the stuck state and
the error state, we do not need to do any extra handling for errors.  

left:
<left.Qrest, <Vleft, Vright>.S> => <Qrest, Vleft.S>

right:
<right.Qrest, <Vleft, Vright>.S> => <Qrest, Vright.S>


Problem 3.27:

a.i) 

F = {[]command} x Pfstackval+ x Pdictval*
Here I use the same notation as was presented in the Notes, using
Pdictval to represent a (Identifier x Value) pair.

I : Program -> Cposttext = Lambda(Q) . <Q, []value, [](Identifier X Value)

O: F -> Answer = Lambda<[]command, V.S', Pdictval*> .
			(Value -> Answer V)


a.ii)

<I.Q, S, D> => <Q, I.S, D>

<def.Qrest, V.I.S, D> => <Qrest, S, <I, V>.D>

<get.Qrest, S, <I, V>.D> => <Qrest, V.S, <I, V>.D>
Note that this representation assumes <I, V> is on the top of the
Dictionary list, which is not neccessarily true.  The full definition
requires representation of cases where the pair is in the middle and
the end of the Dictionary.



b.i>

F = {[]command} x Pfstackval+ x Pdictval*

I : Program -> Cposttext = Lambda(Q) . <Q, []value, {unbound}>

O: F -> Answer = Lambda<[]command, V.S', Dictionary> .
			(Value -> Answer V)

a.ii)

<I.Q, S, D> => <Q, I.S, D>

<def.Qrest, V.I.S, D> => <Qrest, S, bind(I, V, D)>

<get.Qrest, S, <I, V>.D> => <Qrest, V.S, I->(V+I1->(V1..))>
Note that this representation assumes <I, V> is on the top of the
Dictionary list, which is not neccessarily true.  The full definition
requires representation of cases where the pair is in the middle and
the end of the Dictionary.