Step of Proof: decidable-filter 11,40

Inference at * 1 
Iof proof for Lemma decidable-filter:



1. T : Type
2. T List
3. P : {x:T| (x  [])} 
4. x[]. Dec(P(x))
  L':T List. (L'  [] & (x:T. (x  L' ((x  []) & P(x)))) 
latex

 by ((InstConcl [[]]) 
CollapseTHEN (MaAuto)) 
latex


C1

C1:   []  []
C.


Definitions[], suptype(ST), #$n, {i..j}, i  j < k, P  Q, P  Q, P & Q, A c B, S  T, |g|, l[i], , , A  B, A, False, Void, ||as||, x(s), a < b, increasing(f;k), x:AB(x), x:A  B(x), Dec(P), P  Q, P  Q, left + right, L1  L2, xLP(x), type List, f(a), {x:AB(x)} , x:AB(x), x:AB(x), (x  l), s = t, t  T, , Type
Lemmasnil member, increasing wf, length wf1, int seg wf, le wf, sublist wf, rev implies wf, member wf, subtype rel wf, iff wf, nat wf, length wf2, select wf, l member wf

origin