let x be Surreal; x - x == 0_No
set y = 0_No ;
defpred S1[ Ordinal] means for x being Surreal st born x = $1 holds
x + (- x) == 0_No ;
A1:
for D being Ordinal st ( for C being Ordinal st C in D holds
S1[C] ) holds
S1[D]
proof
let D be
Ordinal;
( ( for C being Ordinal st C in D holds
S1[C] ) implies S1[D] )
assume A2:
for
C being
Ordinal st
C in D holds
S1[
C]
;
S1[D]
let x be
Surreal;
( born x = D implies x + (- x) == 0_No )
assume A3:
born x = D
;
x + (- x) == 0_No
set X =
x + (- x);
A4:
- x = [(-- (R_ x)),(-- (L_ x))]
by Th7;
then
(
L_ (- x) = -- (R_ x) &
R_ (- x) = -- (L_ x) )
;
then A5:
x + (- x) = [(((L_ x) ++ {(- x)}) \/ ({x} ++ (-- (R_ x)))),(((R_ x) ++ {(- x)}) \/ ({x} ++ (-- (L_ x))))]
by Th28;
A6:
L_ (x + (- x)) << {0_No}
proof
let l,
r be
Surreal;
SURREAL0:def 20 ( not l in L_ (x + (- x)) or not r in {0_No} or not r <= l )
assume A7:
(
l in L_ (x + (- x)) &
r in {0_No} &
r <= l )
;
contradiction
A8:
r = 0_No
by A7, TARSKI:def 1;
A9:
{r} << R_ l
by A7, SURREAL0:43;
per cases
( l in {x} ++ (-- (R_ x)) or l in (L_ x) ++ {(- x)} )
by A5, A7, XBOOLE_0:def 3;
suppose
l in {x} ++ (-- (R_ x))
;
contradictionthen consider ll,
lr being
Surreal such that A10:
(
ll in {x} &
lr in -- (R_ x) &
l = ll + lr )
by Def8;
consider xr being
Surreal such that A11:
(
xr in R_ x &
- xr = lr )
by A10, Def4;
x = ll
by A10, TARSKI:def 1;
then A12:
l = [(((L_ x) ++ {(- xr)}) \/ ({x} ++ (L_ (- xr)))),(((R_ x) ++ {(- xr)}) \/ ({x} ++ (R_ (- xr))))]
by Th28, A10, A11;
xr in (L_ x) \/ (R_ x)
by A11, XBOOLE_0:def 3;
then A13:
xr + (- xr) == 0_No
by A3, A2, SURREALO:1;
- xr in {(- xr)}
by TARSKI:def 1;
then
xr + (- xr) in (R_ x) ++ {(- xr)}
by A11, Def8;
then
xr + (- xr) in R_ l
by A12, XBOOLE_0:def 3;
hence
contradiction
by A13, A9, A7, A8;
verum end; suppose
l in (L_ x) ++ {(- x)}
;
contradictionthen consider xl,
lr being
Surreal such that A14:
(
xl in L_ x &
lr in {(- x)} &
l = xl + lr )
by Def8;
lr = - x
by A14, TARSKI:def 1;
then A15:
l = [(((L_ xl) ++ {(- x)}) \/ ({xl} ++ (L_ (- x)))),(((R_ xl) ++ {(- x)}) \/ ({xl} ++ (R_ (- x))))]
by A14, Th28;
xl in (L_ x) \/ (R_ x)
by A14, XBOOLE_0:def 3;
then A16:
xl + (- xl) == 0_No
by A3, A2, SURREALO:1;
(
- xl in -- (L_ x) &
xl in {xl} )
by A14, Def4, TARSKI:def 1;
then
xl + (- xl) in {xl} ++ (-- (L_ x))
by Def8;
then
xl + (- xl) in R_ l
by A4, A15, XBOOLE_0:def 3;
hence
contradiction
by A16, A9, A7, A8;
verum end; end;
end;
{(x + (- x))} << R_ 0_No
;
hence
x + (- x) <= 0_No
by A6, SURREAL0:43;
SURREALO:def 2 0_No <= x + (- x)
A17:
{0_No} << R_ (x + (- x))
proof
let l,
r be
Surreal;
SURREAL0:def 20 ( not l in {0_No} or not r in R_ (x + (- x)) or not r <= l )
assume A18:
(
l in {0_No} &
r in R_ (x + (- x)) &
r <= l )
;
contradiction
A19:
l = 0_No
by A18, TARSKI:def 1;
A20:
L_ r << {l}
by SURREAL0:43, A18;
per cases
( r in (R_ x) ++ {(- x)} or r in {x} ++ (-- (L_ x)) )
by A5, A18, XBOOLE_0:def 3;
suppose
r in (R_ x) ++ {(- x)}
;
contradictionthen consider ll,
lr being
Surreal such that A21:
(
ll in R_ x &
lr in {(- x)} &
r = ll + lr )
by Def8;
lr = - x
by A21, TARSKI:def 1;
then A22:
r = [(((L_ ll) ++ {(- x)}) \/ ({ll} ++ (L_ (- x)))),(((R_ ll) ++ {(- x)}) \/ ({ll} ++ (R_ (- x))))]
by A21, Th28;
ll in (L_ x) \/ (R_ x)
by A21, XBOOLE_0:def 3;
then A23:
ll + (- ll) == 0_No
by A3, A2, SURREALO:1;
(
- ll in L_ (- x) &
ll in {ll} )
by A21, Def4, A4, TARSKI:def 1;
then
ll + (- ll) in {ll} ++ (L_ (- x))
by Def8;
then
ll + (- ll) in L_ r
by A22, XBOOLE_0:def 3;
hence
contradiction
by A23, A20, A19, A18;
verum end; suppose
r in {x} ++ (-- (L_ x))
;
contradictionthen consider ll,
lr being
Surreal such that A24:
(
ll in {x} &
lr in -- (L_ x) &
r = ll + lr )
by Def8;
consider xr being
Surreal such that A25:
(
xr in L_ x &
- xr = lr )
by A24, Def4;
x = ll
by A24, TARSKI:def 1;
then A26:
r = [(((L_ x) ++ {(- xr)}) \/ ({x} ++ (L_ (- xr)))),(((R_ x) ++ {(- xr)}) \/ ({x} ++ (R_ (- xr))))]
by Th28, A24, A25;
xr in (L_ x) \/ (R_ x)
by A25, XBOOLE_0:def 3;
then A27:
xr + (- xr) == 0_No
by A3, A2, SURREALO:1;
- xr in {(- xr)}
by TARSKI:def 1;
then
xr + (- xr) in (L_ x) ++ {(- xr)}
by A25, Def8;
then
xr + (- xr) in L_ r
by A26, XBOOLE_0:def 3;
hence
contradiction
by A27, A20, A18, A19;
verum end; end;
end;
L_ 0_No << {(x + (- x))}
;
hence
0_No <= x + (- x)
by SURREAL0:43, A17;
verum
end;
for D being Ordinal holds S1[D]
from ORDINAL1:sch 2(A1);
hence
x - x == 0_No
; verum