If R and S are reducibility candidates of types U and V , we can de ne a set R ! S of terms of type U!V by:

By 6.2.3, we know that R ! S is a reducibility candidate of type U!V .

14.2Reducibility with parameters

sequence of reducibility candidates of corresponding types; then we can de ne a set REDT [R=X] (parametric reducibility) of terms of type T [U=X] as follows:

1.If T = Xi, then REDT [R=X] = Ri;

2.If T = V !W , then REDT [R=X] = REDV [R=X] ! REDW [R=X];

3. If T = Y: W then REDT [R=X] is the set of terms t of type T [U=X] such that, for every type V and reducibility candidate S of this type, then t V 2 REDW [R=X; S=Y ].

Lemma REDT [R=X] is a reducibility candidate of type T [U=X].

Proof By induction on T : the only case to consider is T = Y: W .

(CR 1) If t 2 REDT [R=X], take an arbitrary type V and an arbitrary candidate S of type V (for example, the strongly normalisable terms of type V ). Then t V 2 REDW [R=X; S=Y ], and so, by induction hypothesis (CR 1), we know that t V is strongly normalisable. But (t) (t V ), so t is strongly normalisable.

(CR 3) Let t be neutral and suppose all the t0 one step from t are in REDT [R=X]. Take V and S: applying a conversion inside t V , the result is a t0 V since t is neutral, and t0 V is in REDW [R=X; S=Y ] since t0 is. By induction hypothesis (CR 3) we see that t V 2 REDW [R=X; S=Y ], and so t 2 REDT [R=X].

14.2.1Substitution

The following lemma says that parametric reducibility behaves well with respect to substitution:

Lemma REDT [V=Y ][R=X] = REDT [R=X; REDV [R=X]=Y ]

Here we make hidden use of the comprehension scheme, since, in order to be able to use the predicate REDV [R=X] as a parameter, it is necessary to know that it is a set.

This lemma is proved by a straightforward induction on T . The only di culty was to formulate it precisely!

14.2.2Universal abstraction

Lemma If for every type V and candidate S, w[V=Y ] 2 REDW [R=X; S=Y ], thenY: w 2 RED Y: W [R=X].

Proof We have to show that ( Y: w) V 2 REDW [R=X; S=Y ] for every type V and candidate S of type V . We argue by induction on (w). Converting a redex of ( Y: w) V gives:

( Y: w0) V with (w0) < (w), which is in REDW [R=X; S=Y ] by the induction hypothesis.

w[V=Y ] which is in REDW [R=X; S=Y ] by assumption.

14.2.3Universal application

Lemma If t 2 RED Y: W [R=X], then t V 2 REDW [V=Y ][R=X] for every type V .