:: AFF_1 semantic presentation
:: deftheorem Def1 defines LIN AFF_1:def 1 :
theorem Th1: :: AFF_1:1
canceled;
theorem Th2: :: AFF_1:2
canceled;
theorem Th3: :: AFF_1:3
canceled;
theorem Th4: :: AFF_1:4
canceled;
theorem Th5: :: AFF_1:5
canceled;
theorem Th6: :: AFF_1:6
canceled;
theorem Th7: :: AFF_1:7
canceled;
theorem Th8: :: AFF_1:8
canceled;
theorem Th9: :: AFF_1:9
canceled;
theorem Th10: :: AFF_1:10
theorem Th11: :: AFF_1:11
Lemma4:
for b1 being AffinSpace
for b2, b3, b4, b5 being Element of b1 holds
( b2,b3 // b4,b5 implies b4,b5 // b2,b3 )
theorem Th12: :: AFF_1:12
Lemma6:
for b1 being AffinSpace
for b2, b3, b4, b5 being Element of b1 holds
( b2,b3 // b4,b5 implies b3,b2 // b4,b5 )
proof
let c
1 be
AffinSpace;
let c
2, c
3, c
4, c
5 be
Element of c
1;
assume
c
2,c
3 // c
4,c
5
;
then
( c
2,c
3 // c
4,c
5 & c
2,c
3 // c
3,c
2 )
by Th11;
then
( c
3,c
2 // c
4,c
5 or c
2 = c
3 )
by DIRAF:47;
hence
c
3,c
2 // c
4,c
5
by Th12;
end;
Lemma7:
for b1 being AffinSpace
for b2, b3, b4, b5 being Element of b1 holds
( b2,b3 // b4,b5 implies b2,b3 // b5,b4 )
proof
let c
1 be
AffinSpace;
let c
2, c
3, c
4, c
5 be
Element of c
1;
assume
c
2,c
3 // c
4,c
5
;
then
c
4,c
5 // c
2,c
3
by Lemma4;
then
c
5,c
4 // c
2,c
3
by Lemma6;
hence
c
2,c
3 // c
5,c
4
by Lemma4;
end;
theorem Th13: :: AFF_1:13
for b
1 being
AffinSpacefor b
2, b
3, b
4, b
5 being
Element of b
1 holds
( b
2,b
3 // b
4,b
5 implies ( b
2,b
3 // b
5,b
4 & b
3,b
2 // b
4,b
5 & b
3,b
2 // b
5,b
4 & b
4,b
5 // b
2,b
3 & b
4,b
5 // b
3,b
2 & b
5,b
4 // b
2,b
3 & b
5,b
4 // b
3,b
2 ) )
proof
let c
1 be
AffinSpace;
let c
2, c
3, c
4, c
5 be
Element of c
1;
assume E9:
c
2,c
3 // c
4,c
5
;
hence
( c
2,c
3 // c
5,c
4 & c
3,c
2 // c
4,c
5 )
by Lemma6, Lemma7;
hence
c
3,c
2 // c
5,c
4
by Lemma6;
thus
c
4,c
5 // c
2,c
3
by E9, Lemma4;
hence
( c
4,c
5 // c
3,c
2 & c
5,c
4 // c
2,c
3 )
by Lemma6, Lemma7;
hence
c
5,c
4 // c
3,c
2
by Lemma7;
end;
theorem Th14: :: AFF_1:14
for b
1 being
AffinSpacefor b
2, b
3, b
4, b
5, b
6, b
7 being
Element of b
1 holds
( b
2 <> b
3 & not ( not ( b
2,b
3 // b
4,b
5 & b
2,b
3 // b
6,b
7 ) & not ( b
2,b
3 // b
4,b
5 & b
6,b
7 // b
2,b
3 ) & not ( b
4,b
5 // b
2,b
3 & b
6,b
7 // b
2,b
3 ) & not ( b
4,b
5 // b
2,b
3 & b
2,b
3 // b
6,b
7 ) ) implies b
4,b
5 // b
6,b
7 )
proof
let c
1 be
AffinSpace;
let c
2, c
3, c
4, c
5, c
6, c
7 be
Element of c
1;
assume
( c
2 <> c
3 & not ( not ( c
2,c
3 // c
4,c
5 & c
2,c
3 // c
6,c
7 ) & not ( c
2,c
3 // c
4,c
5 & c
6,c
7 // c
2,c
3 ) & not ( c
4,c
5 // c
2,c
3 & c
6,c
7 // c
2,c
3 ) & not ( c
4,c
5 // c
2,c
3 & c
2,c
3 // c
6,c
7 ) ) )
;
then
( c
2 <> c
3 & c
2,c
3 // c
4,c
5 & c
2,c
3 // c
6,c
7 )
by Th13;
hence
c
4,c
5 // c
6,c
7
by DIRAF:47;
end;
Lemma10:
for b1 being AffinSpace
for b2, b3, b4 being Element of b1 holds
( LIN b2,b3,b4 implies ( LIN b2,b4,b3 & LIN b3,b2,b4 ) )
proof
let c
1 be
AffinSpace;
let c
2, c
3, c
4 be
Element of c
1;
assume
LIN c
2,c
3,c
4
;
then
c
2,c
3 // c
2,c
4
by Def1;
then
( c
2,c
4 // c
2,c
3 & c
3,c
2 // c
3,c
4 )
by Th13, DIRAF:47;
hence
(
LIN c
2,c
4,c
3 &
LIN c
3,c
2,c
4 )
by Def1;
end;
theorem Th15: :: AFF_1:15
for b
1 being
AffinSpacefor b
2, b
3, b
4 being
Element of b
1 holds
(
LIN b
2,b
3,b
4 implies (
LIN b
2,b
4,b
3 &
LIN b
3,b
2,b
4 &
LIN b
3,b
4,b
2 &
LIN b
4,b
2,b
3 &
LIN b
4,b
3,b
2 ) )
proof
let c
1 be
AffinSpace;
let c
2, c
3, c
4 be
Element of c
1;
assume
LIN c
2,c
3,c
4
;
hence
(
LIN c
2,c
4,c
3 &
LIN c
3,c
2,c
4 )
by Lemma10;
hence
(
LIN c
3,c
4,c
2 &
LIN c
4,c
2,c
3 )
by Lemma10;
hence
LIN c
4,c
3,c
2
by Lemma10;
end;
theorem Th16: :: AFF_1:16
proof
let c
1 be
AffinSpace;
let c
2, c
3 be
Element of c
1;
( c
2,c
2 // c
2,c
3 & c
2,c
3 // c
2,c
3 & c
2,c
3 // c
2,c
2 )
by Th11, Th12;
hence
(
LIN c
2,c
2,c
3 &
LIN c
2,c
3,c
3 &
LIN c
2,c
3,c
2 )
by Def1;
end;
theorem Th17: :: AFF_1:17
for b
1 being
AffinSpacefor b
2, b
3, b
4, b
5, b
6 being
Element of b
1 holds
( b
2 <> b
3 &
LIN b
2,b
3,b
4 &
LIN b
2,b
3,b
5 &
LIN b
2,b
3,b
6 implies
LIN b
4,b
5,b
6 )
proof
let c
1 be
AffinSpace;
let c
2, c
3, c
4, c
5, c
6 be
Element of c
1;
assume E14:
( c
2 <> c
3 &
LIN c
2,c
3,c
4 &
LIN c
2,c
3,c
5 &
LIN c
2,c
3,c
6 )
;
E15:
now
assume
c
2 = c
4
;
then
( c
4 <> c
3 & c
4,c
3 // c
4,c
5 & c
4,c
3 // c
4,c
6 )
by E14, Def1;
then
c
4,c
5 // c
4,c
6
by Th14;
hence
LIN c
4,c
5,c
6
by Def1;
end;
now
assume E16:
c
2 <> c
4
;
( c
2,c
3 // c
2,c
4 & c
2,c
3 // c
2,c
5 & c
2,c
3 // c
2,c
6 )
by E14, Def1;
then
( c
2,c
4 // c
2,c
5 & c
2,c
4 // c
2,c
6 )
by E14, Th14;
then
( c
4,c
2 // c
4,c
5 & c
4,c
2 // c
4,c
6 )
by DIRAF:47;
then
c
4,c
5 // c
4,c
6
by E16, Th14;
hence
LIN c
4,c
5,c
6
by Def1;
end;
hence
LIN c
4,c
5,c
6
by E15;
end;
theorem Th18: :: AFF_1:18
proof
let c
1 be
AffinSpace;
let c
2, c
3, c
4, c
5 be
Element of c
1;
assume E15:
( c
2 <> c
3 &
LIN c
2,c
3,c
4 & c
2,c
3 // c
4,c
5 )
;
now
assume E16:
c
4 <> c
2
;
c
2,c
3 // c
2,c
4
by E15, Def1;
then
c
2,c
4 // c
4,c
5
by E15, Th14;
then
c
4,c
2 // c
4,c
5
by Th13;
then
LIN c
4,c
2,c
5
by Def1;
then
(
LIN c
2,c
4,c
5 &
LIN c
2,c
4,c
3 &
LIN c
2,c
4,c
2 )
by E15, Th15, Th16;
hence
LIN c
2,c
3,c
5
by E16, Th17;
end;
hence
LIN c
2,c
3,c
5
by E15, Def1;
end;
theorem Th19: :: AFF_1:19
proof
let c
1 be
AffinSpace;
let c
2, c
3, c
4, c
5 be
Element of c
1;
assume E16:
(
LIN c
2,c
3,c
4 &
LIN c
2,c
3,c
5 )
;
now
assume E17:
c
2 <> c
3
;
now
E18:
( c
2,c
3 // c
2,c
4 & c
2,c
3 // c
2,c
5 )
by E16, Def1;
then
c
2,c
4 // c
2,c
5
by E17, Th14;
then
c
4,c
2 // c
4,c
5
by DIRAF:47;
then
c
2,c
4 // c
4,c
5
by Th13;
hence
c
2,c
3 // c
4,c
5
by E18, Th14;
end;
hence
c
2,c
3 // c
4,c
5
;
end;
hence
c
2,c
3 // c
4,c
5
by Th12;
end;
theorem Th20: :: AFF_1:20
for b
1 being
AffinSpacefor b
2, b
3, b
4, b
5, b
6 being
Element of b
1 holds
( b
2 <> b
3 &
LIN b
4,b
5,b
2 &
LIN b
4,b
5,b
3 &
LIN b
2,b
3,b
6 implies
LIN b
4,b
5,b
6 )
proof
let c
1 be
AffinSpace;
let c
2, c
3, c
4, c
5, c
6 be
Element of c
1;
assume E17:
( c
2 <> c
3 &
LIN c
4,c
5,c
2 &
LIN c
4,c
5,c
3 &
LIN c
2,c
3,c
6 )
;
now
assume E18:
c
4 <> c
5
;
(
LIN c
4,c
5,c
5 &
LIN c
4,c
5,c
4 )
by Th16;
then
(
LIN c
3,c
2,c
5 &
LIN c
3,c
2,c
4 &
LIN c
3,c
2,c
6 )
by E17, E18, Th15, Th17;
hence
LIN c
4,c
5,c
6
by E17, Th17;
end;
hence
LIN c
4,c
5,c
6
by Th16;
end;
theorem Th21: :: AFF_1:21
proof
let c
1 be
AffinSpace;
consider c
2, c
3, c
4 being
Element of c
1 such that E18:
not c
2,c
3 // c
2,c
4
by DIRAF:47;
not
LIN c
2,c
3,c
4
by E18, Def1;
hence
not for b
1, b
2, b
3 being
Element of c
1 holds
LIN b
1,b
2,b
3
;
end;
theorem Th22: :: AFF_1:22
proof
let c
1 be
AffinSpace;
let c
2, c
3 be
Element of c
1;
assume E18:
c
2 <> c
3
;
assume E19:
for b
1 being
Element of c
1 holds
LIN c
2,c
3,b
1
;
consider c
4, c
5, c
6 being
Element of c
1 such that E20:
not
LIN c
4,c
5,c
6
by Th21;
(
LIN c
2,c
3,c
4 &
LIN c
2,c
3,c
5 &
LIN c
2,c
3,c
6 )
by E19;
hence
not verum
by E18, E20, Th17;
end;
theorem Th23: :: AFF_1:23
for b
1 being
AffinSpacefor b
2, b
3, b
4, b
5 being
Element of b
1 holds
( not
LIN b
2,b
3,b
4 &
LIN b
2,b
4,b
5 & b
3,b
4 // b
3,b
5 implies b
4 = b
5 )
proof
let c
1 be
AffinSpace;
let c
2, c
3, c
4, c
5 be
Element of c
1;
assume E18:
( not
LIN c
2,c
3,c
4 &
LIN c
2,c
4,c
5 & c
3,c
4 // c
3,c
5 )
;
assume E19:
c
4 <> c
5
;
LIN c
3,c
4,c
5
by E18, Def1;
then
(
LIN c
4,c
5,c
3 &
LIN c
4,c
5,c
2 &
LIN c
4,c
5,c
4 )
by E18, Th15, Th16;
hence
not verum
by E18, E19, Th17;
end;
definition
let c
1 be
AffinSpace;
let c
2, c
3 be
Element of c
1;
func Line c
2,c
3 -> Subset of a
1 means :
Def2:
:: AFF_1:def 2
for b
1 being
Element of a
1 holds
( b
1 in a
4 iff
LIN a
2,a
3,b
1 );
existence
ex b1 being Subset of c1 st
for b2 being Element of c1 holds
( b2 in b1 iff LIN c2,c3,b2 )
proof
defpred S
1[
set ] means for b
1 being
Element of c
1 holds
( b
1 = a
1 implies
LIN c
2,c
3,b
1 );
consider c
4 being
Subset of c
1 such that E18:
for b
1 being
set holds
( b
1 in c
4 iff ( b
1 in the
carrier of c
1 & S
1[b
1] ) )
from SUBSET_1:sch 1();
take
c
4
;
let c
5 be
Element of c
1;
thus
( c
5 in c
4 implies
LIN c
2,c
3,c
5 )
by E18;
assume
LIN c
2,c
3,c
5
;
then
( c
5 in the
carrier of c
1 & ( for b
1 being
Element of c
1 holds
( b
1 = c
5 implies
LIN c
2,c
3,b
1 ) ) )
;
hence
c
5 in c
4
by E18;
end;
uniqueness
for b1, b2 being Subset of c1 holds
( ( for b3 being Element of c1 holds
( b3 in b1 iff LIN c2,c3,b3 ) ) & ( for b3 being Element of c1 holds
( b3 in b2 iff LIN c2,c3,b3 ) ) implies b1 = b2 )
proof
let c
4, c
5 be
Subset of c
1;
assume that E18:
for b
1 being
Element of c
1 holds
( b
1 in c
4 iff
LIN c
2,c
3,b
1 )
and E19:
for b
1 being
Element of c
1 holds
( b
1 in c
5 iff
LIN c
2,c
3,b
1 )
;
for b
1 being
set holds
( b
1 in c
4 iff b
1 in c
5 )
proof
let c
6 be
set ;
thus
( c
6 in c
4 implies c
6 in c
5 )
assume E20:
c
6 in c
5
;
then reconsider c
7 = c
6 as
Element of c
1 ;
LIN c
2,c
3,c
7
by E19, E20;
hence
c
6 in c
4
by E18;
end;
hence
c
4 = c
5
by TARSKI:2;
end;
end;
:: deftheorem Def2 defines Line AFF_1:def 2 :
Lemma19:
for b1 being AffinSpace
for b2, b3 being Element of b1 holds Line b2,b3 c= Line b3,b2
theorem Th24: :: AFF_1:24
canceled;
theorem Th25: :: AFF_1:25
theorem Th26: :: AFF_1:26
theorem Th27: :: AFF_1:27
proof
let c
1 be
AffinSpace;
let c
2, c
3, c
4, c
5 be
Element of c
1;
assume E23:
( c
2 in Line c
3,c
4 & c
5 in Line c
3,c
4 & c
2 <> c
5 )
;
then E24:
(
LIN c
3,c
4,c
2 &
LIN c
3,c
4,c
5 )
by Def2;
hence
Line c
2,c
5 c= Line c
3,c
4
by TARSKI:def 3;
end;
theorem Th28: :: AFF_1:28
proof
let c
1 be
AffinSpace;
let c
2, c
3, c
4, c
5 be
Element of c
1;
assume E24:
( c
2 in Line c
3,c
4 & c
5 in Line c
3,c
4 & c
3 <> c
4 )
;
then E25:
(
LIN c
3,c
4,c
2 &
LIN c
3,c
4,c
5 )
by Def2;
hence
Line c
3,c
4 c= Line c
2,c
5
by TARSKI:def 3;
end;
:: deftheorem Def3 defines being_line AFF_1:def 3 :
Lemma25:
for b1 being AffinSpace
for b2, b3 being Element of b1
for b4 being Subset of b1 holds
( b4 is_line & b2 in b4 & b3 in b4 & b2 <> b3 implies b4 = Line b2,b3 )
proof
let c
1 be
AffinSpace;
let c
2, c
3 be
Element of c
1;
let c
4 be
Subset of c
1;
assume that E26:
c
4 is_line
and E27:
( c
2 in c
4 & c
3 in c
4 & c
2 <> c
3 )
;
E28:
ex b
1, b
2 being
Element of c
1 st
( b
1 <> b
2 & c
4 = Line b
1,b
2 )
by E26, Def3;
then E29:
Line c
2,c
3 c= c
4
by E27, Th27;
c
4 c= Line c
2,c
3
by E27, E28, Th28;
hence
c
4 = Line c
2,c
3
by E29, XBOOLE_0:def 10;
end;
theorem Th29: :: AFF_1:29
canceled;
theorem Th30: :: AFF_1:30
theorem Th31: :: AFF_1:31
theorem Th32: :: AFF_1:32
theorem Th33: :: AFF_1:33
proof
let c
1 be
AffinSpace;
let c
2, c
3, c
4 be
Element of c
1;
E30:
not (
LIN c
2,c
3,c
4 & ( for b
1 being
Subset of c
1 holds
not ( b
1 is_line & c
2 in b
1 & c
3 in b
1 & c
4 in b
1 ) ) )
proof
assume E31:
LIN c
2,c
3,c
4
;
now
assume E34:
c
2 <> c
4
;
E35:
LIN c
2,c
4,c
3
by E31, Th15;
set c
5 =
Line c
2,c
4;
(
Line c
2,c
4 is_line & c
2 in Line c
2,c
4 & c
3 in Line c
2,c
4 & c
4 in Line c
2,c
4 )
by E34, E35, Def2, Def3, Th26;
hence
ex b
1 being
Subset of c
1 st
( b
1 is_line & c
2 in b
1 & c
3 in b
1 & c
4 in b
1 )
;
end;
hence
ex b
1 being
Subset of c
1 st
( b
1 is_line & c
2 in b
1 & c
3 in b
1 & c
4 in b
1 )
by E32, E33;
end;
( ex b
1 being
Subset of c
1 st
( b
1 is_line & c
2 in b
1 & c
3 in b
1 & c
4 in b
1 ) implies
LIN c
2,c
3,c
4 )
proof
given c
5 being
Subset of c
1 such that E31:
c
5 is_line
and E32:
( c
2 in c
5 & c
3 in c
5 & c
4 in c
5 )
;
consider c
6, c
7 being
Element of c
1 such that E33:
( c
6 <> c
7 & c
5 = Line c
6,c
7 )
by E31, Def3;
(
LIN c
6,c
7,c
2 &
LIN c
6,c
7,c
3 &
LIN c
6,c
7,c
4 )
by E32, E33, Def2;
hence
LIN c
2,c
3,c
4
by E33, Th17;
end;
hence
(
LIN c
2,c
3,c
4 iff ex b
1 being
Subset of c
1 st
( b
1 is_line & c
2 in b
1 & c
3 in b
1 & c
4 in b
1 ) )
by E30;
end;
:: deftheorem Def4 defines // AFF_1:def 4 :
:: deftheorem Def5 defines // AFF_1:def 5 :
theorem Th34: :: AFF_1:34
canceled;
theorem Th35: :: AFF_1:35
canceled;
theorem Th36: :: AFF_1:36
proof
let c
1 be
AffinSpace;
let c
2, c
3, c
4, c
5 be
Element of c
1;
assume E33:
( c
2 in Line c
3,c
4 & c
3 <> c
4 )
;
then E34:
LIN c
3,c
4,c
2
by Def2;
E35:
( c
5 in Line c
3,c
4 implies c
3,c
4 // c
2,c
5 )
( c
3,c
4 // c
2,c
5 implies c
5 in Line c
3,c
4 )
hence
( c
5 in Line c
3,c
4 iff c
3,c
4 // c
2,c
5 )
by E35;
end;
theorem Th37: :: AFF_1:37
proof
let c
1 be
AffinSpace;
let c
2, c
3 be
Element of c
1;
let c
4 be
Subset of c
1;
assume E34:
( c
4 is_line & c
2 in c
4 )
;
E35:
now
assume E36:
c
3 in c
4
;
consider c
5, c
6 being
Element of c
1 such that E37:
( c
5 <> c
6 & c
4 = Line c
5,c
6 )
by E34, Def3;
c
5,c
6 // c
2,c
3
by E34, E36, E37, Th36;
then
c
2,c
3 // c
5,c
6
by Th13;
hence
c
2,c
3 // c
4
by E37, Def4;
end;
now
assume
c
2,c
3 // c
4
;
then consider c
5, c
6 being
Element of c
1 such that E36:
( c
5 <> c
6 & c
4 = Line c
5,c
6 & c
2,c
3 // c
5,c
6 )
by Def4;
c
5,c
6 // c
2,c
3
by E36, Th13;
hence
c
3 in c
4
by E34, E36, Th36;
end;
hence
( c
3 in c
4 iff c
2,c
3 // c
4 )
by E35;
end;
theorem Th38: :: AFF_1:38
theorem Th39: :: AFF_1:39
theorem Th40: :: AFF_1:40
theorem Th41: :: AFF_1:41
proof
let c
1 be
AffinSpace;
let c
2, c
3, c
4, c
5 be
Element of c
1;
let c
6 be
Subset of c
1;
assume that E37:
( c
2 in c
6 & c
3 in c
6 )
and E38:
c
6 is_line
and E39:
c
2 <> c
3
;
E40:
( c
4,c
5 // c
6 implies c
4,c
5 // c
2,c
3 )
proof
assume
c
4,c
5 // c
6
;
then consider c
7, c
8 being
Element of c
1 such that E41:
( c
7 <> c
8 & c
6 = Line c
7,c
8 & c
4,c
5 // c
7,c
8 )
by Def4;
c
7,c
8 // c
2,c
3
by E37, E41, Th36;
hence
c
4,c
5 // c
2,c
3
by E41, Th14;
end;
( c
4,c
5 // c
2,c
3 implies c
4,c
5 // c
6 )
hence
( c
4,c
5 // c
6 iff c
4,c
5 // c
2,c
3 )
by E40;
end;
theorem Th42: :: AFF_1:42
canceled;
theorem Th43: :: AFF_1:43
theorem Th44: :: AFF_1:44
proof
let c
1 be
AffinSpace;
let c
2, c
3 be
Element of c
1;
let c
4 be
Subset of c
1;
assume E39:
c
4 is_line
;
E40:
not ( c
2,c
3 // c
4 & ( for b
1, b
2 being
Element of c
1 holds
not ( b
1 <> b
2 & b
1 in c
4 & b
2 in c
4 & c
2,c
3 // b
1,b
2 ) ) )
proof
assume
c
2,c
3 // c
4
;
then consider c
5, c
6 being
Element of c
1 such that E41:
( c
5 <> c
6 & c
4 = Line c
5,c
6 & c
2,c
3 // c
5,c
6 )
by Def4;
( c
5 <> c
6 & c
5 in c
4 & c
6 in c
4 & c
2,c
3 // c
5,c
6 )
by E41, Th26;
hence
ex b
1, b
2 being
Element of c
1 st
( b
1 <> b
2 & b
1 in c
4 & b
2 in c
4 & c
2,c
3 // b
1,b
2 )
;
end;
( ex b
1, b
2 being
Element of c
1 st
( b
1 <> b
2 & b
1 in c
4 & b
2 in c
4 & c
2,c
3 // b
1,b
2 ) implies c
2,c
3 // c
4 )
proof
assume
ex b
1, b
2 being
Element of c
1 st
( b
1 <> b
2 & b
1 in c
4 & b
2 in c
4 & c
2,c
3 // b
1,b
2 )
;
then consider c
5, c
6 being
Element of c
1 such that E41:
( c
5 <> c
6 & c
5 in c
4 & c
6 in c
4 & c
2,c
3 // c
5,c
6 )
;
c
4 = Line c
5,c
6
by E39, E41, Lemma25;
hence
c
2,c
3 // c
4
by E41, Def4;
end;
hence
( c
2,c
3 // c
4 iff ex b
1, b
2 being
Element of c
1 st
( b
1 <> b
2 & b
1 in c
4 & b
2 in c
4 & c
2,c
3 // b
1,b
2 ) )
by E40;
end;
theorem Th45: :: AFF_1:45
proof
let c
1 be
AffinSpace;
let c
2, c
3, c
4, c
5 be
Element of c
1;
let c
6 be
Subset of c
1;
assume that E39:
c
6 is_line
and E40:
( c
2,c
3 // c
6 & c
4,c
5 // c
6 )
;
consider c
7, c
8 being
Element of c
1 such that E41:
( c
7 <> c
8 & c
6 = Line c
7,c
8 & c
2,c
3 // c
7,c
8 )
by E40, Def4;
( c
7 in c
6 & c
8 in c
6 )
by E41, Th26;
then
c
4,c
5 // c
7,c
8
by E39, E40, E41, Th41;
hence
c
2,c
3 // c
4,c
5
by E41, Th14;
end;
theorem Th46: :: AFF_1:46
for b
1 being
AffinSpacefor b
2, b
3, b
4, b
5 being
Element of b
1for b
6 being
Subset of b
1 holds
( b
2,b
3 // b
6 & b
2,b
3 // b
4,b
5 & b
2 <> b
3 implies b
4,b
5 // b
6 )
proof
let c
1 be
AffinSpace;
let c
2, c
3, c
4, c
5 be
Element of c
1;
let c
6 be
Subset of c
1;
assume E40:
( c
2,c
3 // c
6 & c
2,c
3 // c
4,c
5 & c
2 <> c
3 )
;
then E41:
c
6 is_line
by Th40;
then consider c
7, c
8 being
Element of c
1 such that E42:
( c
7 <> c
8 & c
7 in c
6 & c
8 in c
6 & c
2,c
3 // c
7,c
8 )
by E40, Th44;
( c
7 <> c
8 & c
7 in c
6 & c
8 in c
6 & c
4,c
5 // c
7,c
8 )
by E40, E42, Th14;
hence
c
4,c
5 // c
6
by E41, Th44;
end;
theorem Th47: :: AFF_1:47