G:Sequent.
L:Sequent List.
s
L.(
(s) = 0)
& (sL.|= s 
|= G )
& (a:Assignment. sL.a |
s a | G)
normalization_lemma
xL.P(x)
== (letrec list_exists L = (Case of L; nil 
False ; h.t P(h) 
list_exists(t)) ) (L)
Thm* G:Sequent.
L:Sequent List.
sL.((s) = 0)
& (sL.|= s |= G )
& (a:Assignment. sL.a | s a | G)
normalization_lemma
In prior sections: list 3 jlc sequent rank