let T1, T2 be Tree; ( ( T1 is binary & T2 is binary ) iff tree T1,T2 is binary )
set RT = tree T1,T2;
hereby ( tree T1,T2 is binary implies ( T1 is binary & T2 is binary ) )
assume that A1:
T1 is
binary
and A2:
T2 is
binary
;
tree T1,T2 is binary now let t be
Element of
tree T1,
T2;
( not t in Leaves (tree T1,T2) implies succ b1 = {(b1 ^ <*0 *>),(b1 ^ <*1*>)} )assume A3:
not
t in Leaves (tree T1,T2)
;
succ b1 = {(b1 ^ <*0 *>),(b1 ^ <*1*>)}per cases
( t = {} or ex p being FinSequence st
( ( p in T1 & t = <*0 *> ^ p ) or ( p in T2 & t = <*1*> ^ p ) ) )
by TREES_3:71;
suppose
ex
p being
FinSequence st
( (
p in T1 &
t = <*0 *> ^ p ) or (
p in T2 &
t = <*1*> ^ p ) )
;
succ b1 = {(b1 ^ <*0 *>),(b1 ^ <*1*>)}then consider p being
FinSequence such that A4:
( (
p in T1 &
t = <*0 *> ^ p ) or (
p in T2 &
t = <*1*> ^ p ) )
;
A5:
now assume that A6:
p in T2
and A7:
t = <*1*> ^ p
;
succ t = {(t ^ <*0 *>),(t ^ <*1*>)}reconsider p =
p as
Element of
T2 by A6;
per cases
( p in Leaves T2 or not p in Leaves T2 )
;
suppose
not
p in Leaves T2
;
succ t = {(t ^ <*0 *>),(t ^ <*1*>)}then A8:
succ p = {(p ^ <*0 *>),(p ^ <*1*>)}
by A2, Def2;
now let x be
set ;
( ( x in succ t implies x in {(t ^ <*0 *>),(t ^ <*1*>)} ) & ( x in {(t ^ <*0 *>),(t ^ <*1*>)} implies x in succ t ) )hereby ( x in {(t ^ <*0 *>),(t ^ <*1*>)} implies x in succ t )
assume A9:
x in succ t
;
x in {(t ^ <*0 *>),(t ^ <*1*>)}then
x in { (t ^ <*n*>) where n is Element of NAT : t ^ <*n*> in tree T1,T2 }
by TREES_2:def 5;
then consider n being
Element of
NAT such that A10:
x = t ^ <*n*>
and A11:
t ^ <*n*> in tree T1,
T2
;
A12:
x = <*1*> ^ (p ^ <*n*>)
by A7, A10, FINSEQ_1:45;
then reconsider pn =
p ^ <*n*> as
Element of
T2 by A10, A11, TREES_3:73;
pn in succ p
by A7, A9, A12, Th9;
then
(
pn = p ^ <*0 *> or
pn = p ^ <*1*> )
by A8, TARSKI:def 2;
then
(
x = t ^ <*0 *> or
x = t ^ <*1*> )
by A10, FINSEQ_1:46;
hence
x in {(t ^ <*0 *>),(t ^ <*1*>)}
by TARSKI:def 2;
verum
end; assume
x in {(t ^ <*0 *>),(t ^ <*1*>)}
;
x in succ tthen
(
x = (<*1*> ^ p) ^ <*0 *> or
x = (<*1*> ^ p) ^ <*1*> )
by A7, TARSKI:def 2;
then A13:
(
x = <*1*> ^ (p ^ <*0 *>) or
x = <*1*> ^ (p ^ <*1*>) )
by FINSEQ_1:45;
(
p ^ <*0 *> in succ p &
p ^ <*1*> in succ p )
by A8, TARSKI:def 2;
hence
x in succ t
by A7, A13, Th9;
verum end; hence
succ t = {(t ^ <*0 *>),(t ^ <*1*>)}
by TARSKI:2;
verum end; end; end; now assume that A14:
p in T1
and A15:
t = <*0 *> ^ p
;
succ t = {(t ^ <*0 *>),(t ^ <*1*>)}reconsider p =
p as
Element of
T1 by A14;
per cases
( p in Leaves T1 or not p in Leaves T1 )
;
suppose
not
p in Leaves T1
;
succ t = {(t ^ <*0 *>),(t ^ <*1*>)}then A16:
succ p = {(p ^ <*0 *>),(p ^ <*1*>)}
by A1, Def2;
now let x be
set ;
( ( x in succ t implies x in {(t ^ <*0 *>),(t ^ <*1*>)} ) & ( x in {(t ^ <*0 *>),(t ^ <*1*>)} implies x in succ t ) )hereby ( x in {(t ^ <*0 *>),(t ^ <*1*>)} implies x in succ t )
assume A17:
x in succ t
;
x in {(t ^ <*0 *>),(t ^ <*1*>)}then
x in { (t ^ <*n*>) where n is Element of NAT : t ^ <*n*> in tree T1,T2 }
by TREES_2:def 5;
then consider n being
Element of
NAT such that A18:
x = t ^ <*n*>
and A19:
t ^ <*n*> in tree T1,
T2
;
A20:
x = <*0 *> ^ (p ^ <*n*>)
by A15, A18, FINSEQ_1:45;
then reconsider pn =
p ^ <*n*> as
Element of
T1 by A18, A19, TREES_3:72;
pn in succ p
by A15, A17, A20, Th9;
then
(
pn = p ^ <*0 *> or
pn = p ^ <*1*> )
by A16, TARSKI:def 2;
then
(
x = t ^ <*0 *> or
x = t ^ <*1*> )
by A18, FINSEQ_1:46;
hence
x in {(t ^ <*0 *>),(t ^ <*1*>)}
by TARSKI:def 2;
verum
end; assume
x in {(t ^ <*0 *>),(t ^ <*1*>)}
;
x in succ tthen
(
x = (<*0 *> ^ p) ^ <*0 *> or
x = (<*0 *> ^ p) ^ <*1*> )
by A15, TARSKI:def 2;
then A21:
(
x = <*0 *> ^ (p ^ <*0 *>) or
x = <*0 *> ^ (p ^ <*1*>) )
by FINSEQ_1:45;
(
p ^ <*0 *> in succ p &
p ^ <*1*> in succ p )
by A16, TARSKI:def 2;
hence
x in succ t
by A15, A21, Th9;
verum end; hence
succ t = {(t ^ <*0 *>),(t ^ <*1*>)}
by TARSKI:2;
verum end; end; end; hence
succ t = {(t ^ <*0 *>),(t ^ <*1*>)}
by A4, A5;
verum end; end; end; hence
tree T1,
T2 is
binary
by Def2;
verum
end;
assume A22:
tree T1,T2 is binary
; ( T1 is binary & T2 is binary )
hence
T1 is binary
by Def2; T2 is binary
hence
T2 is binary
by Def2; verum