let s be non empty typealg ; for z being type of s
for p being Proof of s
for v being Element of dom p st (p . v) `2 = 5 holds
ex w being Element of dom p ex X being FinSequence of s ex x, y being type of s ex Y being FinSequence of s st
( w = v ^ <*0*> & (p . v) `1 = [((X ^ <*(x * y)*>) ^ Y),z] & (p . w) `1 = [(((X ^ <*x*>) ^ <*y*>) ^ Y),z] )
let z be type of s; for p being Proof of s
for v being Element of dom p st (p . v) `2 = 5 holds
ex w being Element of dom p ex X being FinSequence of s ex x, y being type of s ex Y being FinSequence of s st
( w = v ^ <*0*> & (p . v) `1 = [((X ^ <*(x * y)*>) ^ Y),z] & (p . w) `1 = [(((X ^ <*x*>) ^ <*y*>) ^ Y),z] )
let p be Proof of s; for v being Element of dom p st (p . v) `2 = 5 holds
ex w being Element of dom p ex X being FinSequence of s ex x, y being type of s ex Y being FinSequence of s st
( w = v ^ <*0*> & (p . v) `1 = [((X ^ <*(x * y)*>) ^ Y),z] & (p . w) `1 = [(((X ^ <*x*>) ^ <*y*>) ^ Y),z] )
let v be Element of dom p; ( (p . v) `2 = 5 implies ex w being Element of dom p ex X being FinSequence of s ex x, y being type of s ex Y being FinSequence of s st
( w = v ^ <*0*> & (p . v) `1 = [((X ^ <*(x * y)*>) ^ Y),z] & (p . w) `1 = [(((X ^ <*x*>) ^ <*y*>) ^ Y),z] ) )
A1:
v is correct
by Def12;
assume A2:
(p . v) `2 = 5
; ex w being Element of dom p ex X being FinSequence of s ex x, y being type of s ex Y being FinSequence of s st
( w = v ^ <*0*> & (p . v) `1 = [((X ^ <*(x * y)*>) ^ Y),z] & (p . w) `1 = [(((X ^ <*x*>) ^ <*y*>) ^ Y),z] )
then A3:
ex X being FinSequence of s ex x, y being type of s ex Y being FinSequence of s st
( (p . v) `1 = [((X ^ <*(x * y)*>) ^ Y),z] & (p . (v ^ <*0*>)) `1 = [(((X ^ <*x*>) ^ <*y*>) ^ Y),z] )
by A1, Def4;
branchdeg v = 1
by A1, A2, Def4;
then
v ^ <*0*> in dom p
by Th1;
hence
ex w being Element of dom p ex X being FinSequence of s ex x, y being type of s ex Y being FinSequence of s st
( w = v ^ <*0*> & (p . v) `1 = [((X ^ <*(x * y)*>) ^ Y),z] & (p . w) `1 = [(((X ^ <*x*>) ^ <*y*>) ^ Y),z] )
by A3; verum