reconsider P9 = P as Subset of T by TREES_1:def 11;

let x be object ; :: thesis:
assume A2: x in P9 ; :: thesis:
then reconsider x9 = x as FinSequence by TREES_1:def 10;
reconsider x9 = x9 as Element of T by A2;

let p be FinSequence of NAT ; :: thesis:
assume A3: p in P ; :: thesis:

per cases ;

suppose p <> x9 ; :: thesis:

then not p,x9 are_c=-comparable by A2, A3, TREES_1:def 10;
hence not p is_a_proper_prefix_of x9 ; :: thesis:

end;

suppose p = x9 ; :: thesis:

hence not p is_a_proper_prefix_of x9 ; :: thesis:

end;

hence x in { t where t is Element of T : for p being FinSequence of NAT st p in P holds
not p is_a_proper_prefix_of t } ; :: thesis:

end;

then P c= { t where t is Element of T : for p being FinSequence of NAT st p in P holds
not p is_a_proper_prefix_of t } ;
then reconsider Y = { t where t is Element of T : for p being FinSequence of NAT st p in P holds
not p is_a_proper_prefix_of t } as non empty set by A1, XBOOLE_1:3;
consider Z being set such that
A4: Z = { (p ^ s) where p is Element of T, s is Element of T1 : p in P } ;
reconsider X = Y \/ Z as non empty set ;
A5: for x being set st x in { t where t is Element of T : for p being FinSequence of NAT st p in P holds
not p is_a_proper_prefix_of t } holds
( x is FinSequence of NAT & x in NAT * & x in T )

proof

let x be set ; :: thesis:
assume x in { t where t is Element of T : for p being FinSequence of NAT st p in P holds
not p is_a_proper_prefix_of t } ; :: thesis:
then A6: ex t being Element of T st
( x = t & ( for p being FinSequence of NAT st p in P holds
not p is_a_proper_prefix_of t ) ) ;
hence x is FinSequence of NAT ; :: thesis:
thus ( x in NAT * & x in T ) by A6, FINSEQ_1:def 11; :: thesis:

end;

X is Tree-like

proof

thus X c= NAT * :: according to TREES_1:def 3 :: thesis:

proof

let x be object ; :: according to TARSKI:def 3 :: thesis:
assume x in X ; :: thesis:
then A7: ( x in { t where t is Element of T : for p being FinSequence of NAT st p in P holds
not p is_a_proper_prefix_of t } or x in { (p ^ s) where p is Element of T, s is Element of T1 : p in P } ) by A4, XBOOLE_0:def 3;

now :: thesis:

assume x in { (p ^ s) where p is Element of T, s is Element of T1 : p in P } ; :: thesis:
then ex p being Element of T ex s being Element of T1 st
( x = p ^ s & p in P ) ;
hence x is FinSequence of NAT ; :: thesis:

end;

hence x in NAT * by A5, A7, FINSEQ_1:def 11; :: thesis:

end;

thus for q being FinSequence of NAT st q in X holds
ProperPrefixes q c= X :: thesis:

proof

let q be FinSequence of NAT ; :: thesis:
assume A8: q in X ; :: thesis:

A9: now :: thesis:

assume A10: q in { t where t is Element of T : for p being FinSequence of NAT st p in P holds
not p is_a_proper_prefix_of t } ; :: thesis:
then ex t being Element of T st
( q = t & ( for p being FinSequence of NAT st p in P holds
not p is_a_proper_prefix_of t ) ) ;
then A11: ProperPrefixes q c= T by TREES_1:def 3;
thus ProperPrefixes q c= X :: thesis:

proof

let x be object ; :: according to TARSKI:def 3 :: thesis:
assume A12: x in ProperPrefixes q ; :: thesis:
then consider p1 being FinSequence such that
A13: x = p1 and
A14: p1 is_a_proper_prefix_of q by TREES_1:def 2;
reconsider p1 = p1 as Element of T by A11, A12, A13;
for p being FinSequence of NAT st p in P holds
not p is_a_proper_prefix_of p1

proof

let p be FinSequence of NAT ; :: thesis:
assume A15: p in P ; :: thesis:
ex t being Element of T st
( q = t & ( for p being FinSequence of NAT st p in P holds
not p is_a_proper_prefix_of t ) ) by A10;
hence not p is_a_proper_prefix_of p1 by A14, A15, XBOOLE_1:56; :: thesis:

end;

then x in { s1 where s1 is Element of T : for p being FinSequence of NAT st p in P holds
not p is_a_proper_prefix_of s1 } by A13;
hence x in X by XBOOLE_0:def 3; :: thesis:

end;

now :: thesis:

assume q in Z ; :: thesis:
then consider p being Element of T, s being Element of T1 such that
A16: q = p ^ s and
A17: p in P by A4;
thus ProperPrefixes q c= X :: thesis:

proof

let x be object ; :: according to TARSKI:def 3 :: thesis:
assume A18: x in ProperPrefixes q ; :: thesis:
then reconsider r = x as FinSequence by TREES_1:11;
r is_a_proper_prefix_of p ^ s by A16, A18, TREES_1:12;
then r is_a_prefix_of p ^ s ;
then consider r1 being FinSequence such that
A19: p ^ s = r ^ r1 by TREES_1:1;

A20: now :: thesis:

assume len p <= len r ; :: thesis:
then consider r2 being FinSequence such that
A21: p ^ r2 = r by A19, FINSEQ_1:47;
p ^ s = p ^ (r2 ^ r1) by A19, A21, FINSEQ_1:32;
then s = r2 ^ r1 by FINSEQ_1:33;
then s | (dom r2) = r2 by FINSEQ_1:21;
then A22: s | (Seg (len r2)) = r2 by FINSEQ_1:def 3;
then reconsider r2 = r2 as FinSequence of NAT by FINSEQ_1:18;
r2 is_a_prefix_of s by A22;
then reconsider r2 = r2 as Element of T1 by TREES_1:20;
r = p ^ r2 by A21;
then r in { (w ^ v) where w is Element of T, v is Element of T1 : w in P } by A17;
hence r in X by A4, XBOOLE_0:def 3; :: thesis:

end;

now :: thesis:

assume len r <= len p ; :: thesis:
then ex r2 being FinSequence st r ^ r2 = p by A19, FINSEQ_1:47;
then p | (dom r) = r by FINSEQ_1:21;
then A23: p | (Seg (len r)) = r by FINSEQ_1:def 3;
then reconsider r3 = r as FinSequence of NAT by FINSEQ_1:18;
A24: r3 is_a_prefix_of p by A23;
then reconsider r3 = r3 as Element of T by TREES_1:20;
for p9 being FinSequence of NAT st p9 in P holds
not p9 is_a_proper_prefix_of r3

proof

let p9 be FinSequence of NAT ; :: thesis:
assume A25: p9 in P ; :: thesis:
assume A26: p9 is_a_proper_prefix_of r3 ; :: thesis:
then p9 is_a_prefix_of r3 ;
then p9 is_a_prefix_of p by A24;
then A27: p,p9 are_c=-comparable ;

per cases ;

suppose p <> p9 ; :: thesis:

hence contradiction by A17, A25, A27, TREES_1:def 10; :: thesis:

end;

suppose p = p9 ; :: thesis:

hence contradiction by A24, A26, XBOOLE_0:def 8; :: thesis:

end;

then r3 in { t where t is Element of T : for p9 being FinSequence of NAT st p9 in P holds
not p9 is_a_proper_prefix_of t } ;
hence r in X by XBOOLE_0:def 3; :: thesis:

end;

hence x in X by A20; :: thesis:

end;

hence ProperPrefixes q c= X by A8, A9, XBOOLE_0:def 3; :: thesis:

end;

let q be FinSequence of NAT ; :: thesis:
let k, n be Nat; :: thesis:
assume that
A28: q ^ <*k*> in X and
A29: n <= k ; :: thesis:

A30: now :: thesis:

assume A31: q ^ <*k*> in { t where t is Element of T : for p being FinSequence of NAT st p in P holds
not p is_a_proper_prefix_of t } ; :: thesis:
then ex s being Element of T st
( q ^ <*k*> = s & ( for p being FinSequence of NAT st p in P holds
not p is_a_proper_prefix_of s ) ) ;
then reconsider u = q ^ <*n*> as Element of T by A29, TREES_1:def 3;

now :: thesis:

let p be FinSequence of NAT ; :: thesis:
assume A32: p in P ; :: thesis:
assume p is_a_proper_prefix_of u ; :: thesis:
then A33: p is_a_prefix_of q by TREES_1:9;
ex s being Element of T st
( q ^ <*k*> = s & ( for p being FinSequence of NAT st p in P holds
not p is_a_proper_prefix_of s ) ) by A31;
hence contradiction by A32, A33, TREES_1:8; :: thesis:

end;

then q ^ <*n*> in { t where t is Element of T : for p being FinSequence of NAT st p in P holds
not p is_a_proper_prefix_of t } ;
hence q ^ <*n*> in X by XBOOLE_0:def 3; :: thesis:

end;

now :: thesis:

assume q ^ <*k*> in Z ; :: thesis:
then consider p being Element of T, s being Element of T1 such that
A34: q ^ <*k*> = p ^ s and
A35: p in P by A4;

A36: now :: thesis:

assume len q <= len p ; :: thesis:
then consider r being FinSequence such that
A37: q ^ r = p by A34, FINSEQ_1:47;
q ^ <*k*> = q ^ (r ^ s) by A34, A37, FINSEQ_1:32;
then A38: <*k*> = r ^ s by FINSEQ_1:33;

A39: now :: thesis:

assume A40: r = <*k*> ; :: thesis:
then reconsider s9 = q ^ <*n*> as Element of T by A29, A37, TREES_1:def 3;

now :: thesis:

let p9 be FinSequence of NAT ; :: thesis:
assume A41: p9 in P ; :: thesis:
assume A42: p9 is_a_proper_prefix_of s9 ; :: thesis:
A43: len p = (len q) + (len <*k*>) by A37, A40, FINSEQ_1:22
.= (len q) + 1 by FINSEQ_1:40
.= (len q) + (len <*n*>) by FINSEQ_1:40
.= len s9 by FINSEQ_1:22 ;

per cases ;

suppose p9 = p ; :: thesis:

hence contradiction by A42, A43, CARD_2:102; :: thesis:

end;

suppose A44: p9 <> p ; :: thesis:

( q is_a_prefix_of p & p9 is_a_prefix_of q ) by A37, A42, TREES_1:1, TREES_1:9;
then p9 is_a_prefix_of p ;
then p,p9 are_c=-comparable ;
hence contradiction by A35, A41, A44, TREES_1:def 10; :: thesis:

end;

hence q ^ <*n*> in { t where t is Element of T : for p being FinSequence of NAT st p in P holds
not p is_a_proper_prefix_of t } ; :: thesis:

end;

now :: thesis:

assume that
A45: s = <*k*> and
A46: r = {} ; :: thesis:
s = (<*> NAT) ^ s by FINSEQ_1:34;
then (<*> NAT) ^ <*n*> in T1 by A29, A45, TREES_1:def 3;
then reconsider t = <*n*> as Element of T1 by FINSEQ_1:34;
q ^ <*n*> = p ^ t by A37, A46, FINSEQ_1:34;
hence q ^ <*n*> in Z by A4, A35; :: thesis:

end;

hence q ^ <*n*> in X by A38, A39, FINSEQ_1:88, XBOOLE_0:def 3; :: thesis:

end;

now :: thesis:

assume len p <= len q ; :: thesis:
then consider r being FinSequence such that
A47: p ^ r = q by A34, FINSEQ_1:47;
p ^ (r ^ <*k*>) = p ^ s by A34, A47, FINSEQ_1:32;
then A48: r ^ <*k*> = s by FINSEQ_1:33;
then s | (dom r) = r by FINSEQ_1:21;
then s | (Seg (len r)) = r by FINSEQ_1:def 3;
then reconsider r = r as FinSequence of NAT by FINSEQ_1:18;
reconsider t = r ^ <*n*> as Element of T1 by A29, A48, TREES_1:def 3;
q ^ <*n*> = p ^ t by A47, FINSEQ_1:32;
then q ^ <*n*> in { (p9 ^ v) where p9 is Element of T, v is Element of T1 : p9 in P } by A35;
hence q ^ <*n*> in X by A4, XBOOLE_0:def 3; :: thesis:

end;

hence q ^ <*n*> in X by A36; :: thesis:

end;

hence q ^ <*n*> in X by A28, A30, XBOOLE_0:def 3; :: thesis:

end;

then reconsider X = X as Tree ;
take X ; :: thesis:
let q be FinSequence of NAT ; :: thesis:
thus ( not q in X or ( q in T & ( for p being FinSequence of NAT st p in P holds
not p is_a_proper_prefix_of q ) ) or ex p, r being FinSequence of NAT st
( p in P & r in T1 & q = p ^ r ) ) :: thesis:

proof

assume A49: q in X ; :: thesis:

A50: now :: thesis:

assume q in Y ; :: thesis:
then ex s being Element of T st
( q = s & ( for p being FinSequence of NAT st p in P holds
not p is_a_proper_prefix_of s ) ) ;
hence ( ( q in T & ( for p being FinSequence of NAT st p in P holds
not p is_a_proper_prefix_of q ) ) or ex p, r being FinSequence of NAT st
( p in P & r in T1 & q = p ^ r ) ) ; :: thesis:

end;

now :: thesis:

assume q in Z ; :: thesis:
then ex p being Element of T ex s being Element of T1 st
( q = p ^ s & p in P ) by A4;
hence ex p, r being FinSequence of NAT st
( p in P & r in T1 & q = p ^ r ) ; :: thesis:

end;

hence ( ( q in T & ( for p being FinSequence of NAT st p in P holds
not p is_a_proper_prefix_of q ) ) or ex p, r being FinSequence of NAT st
( p in P & r in T1 & q = p ^ r ) ) by A49, A50, XBOOLE_0:def 3; :: thesis:

end;

assume A51: ( ( q in T & ( for p being FinSequence of NAT st p in P holds
not p is_a_proper_prefix_of q ) ) or ex p, r being FinSequence of NAT st
( p in P & r in T1 & q = p ^ r ) ) ; :: thesis:
A52: ( q in T & ( for p being FinSequence of NAT st p in P holds
not p is_a_proper_prefix_of q ) implies q in { t where t is Element of T : for p being FinSequence of NAT st p in P holds
not p is_a_proper_prefix_of t } ) ;

now :: thesis:

given p, r being FinSequence of NAT such that A53: ( p in P & r in T1 & q = p ^ r ) ; :: thesis:
P c= T by TREES_1:def 11;
hence q in Z by A4, A53; :: thesis:

end;

hence q in X by A51, A52, XBOOLE_0:def 3; :: thesis: