By Frank Pfenning
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V is derivable, then for any such that . e : is derivable, . v : is also derivable. Proof: By induction on the structure of the deduction D of e ,! v. 2. More directly, from the form of the judgment established by a derivation we draw conclusions about the possible forms of the premiss, which, of course, must also derivable. ev z: Case: D = z ,! z Then we have to show that for any type such that . z : is derivable, . z : is derivable. This is obvious. D1 e1 ,! v1 Case: D = ev s: Then s e1 ,! s v1 .
16). 2 Intuitively the type nat can be interpreted by the set of natural numbers. We write vnat for values v such that . v : nat. It can easily be seen by induction on the structure of the derivation of vnat Value that vnat could be de ned inductively by vnat ::= z j s vnat : The meaning or denotation of a value vnat , vnat ] , can be de ned almost trivially as z] = 0 s vnat] = vnat] + 1: It is immediate that this is a bijection between closed values of type nat and the natural numbers. The meaning of an arbitrary closed expression enat of type nat can then be de ned by v] if enat ,!
For example, `EV ev case z (ev s ev z) ev z * eval would be derivable, but the object above does not represent a valid evaluation. The problem is that the rst premiss of the rule ev case z must be an evaluation yielding z, while the corresponding argument to ev case z, namely (ev s ev z), represents an evaluation yielding s z. One solution to this representation problem is to introduce a validity predicate and de ne when a given object of type eval represents a valid deduction. This is, for example, the solution one would take in a framework such as higher-order Horn clauses or hereditary Harrop formulas.