let SAS be AffinPlane; for o, a being Element of SAS ex p being Element of SAS st
for b, c being Element of SAS holds
( o,a // o,p & ex d being Element of SAS st
( o,p // o,b implies ( o,c // o,d & p,c // b,d ) ) )
let o, a be Element of SAS; ex p being Element of SAS st
for b, c being Element of SAS holds
( o,a // o,p & ex d being Element of SAS st
( o,p // o,b implies ( o,c // o,d & p,c // b,d ) ) )
ex p being Element of SAS st
( o <> p & o,a // o,p )
proof
consider x,
y being
Element of
SAS such that A1:
x <> y
by DIRAF:40;
now ( a = o implies ex p being Element of SAS st
( o <> p & o,a // o,p ) )assume
a = o
;
ex p being Element of SAS st
( o <> p & o,a // o,p )then A2:
(
o,
a // o,
x &
o,
a // o,
y )
by AFF_1:3;
(
o <> x or
o <> y )
by A1;
hence
ex
p being
Element of
SAS st
(
o <> p &
o,
a // o,
p )
by A2;
verum end;
hence
ex
p being
Element of
SAS st
(
o <> p &
o,
a // o,
p )
by AFF_1:2;
verum
end;
then consider p being Element of SAS such that
A3:
o <> p
and
A4:
o,a // o,p
;
take
p
; for b, c being Element of SAS holds
( o,a // o,p & ex d being Element of SAS st
( o,p // o,b implies ( o,c // o,d & p,c // b,d ) ) )
thus
for b, c being Element of SAS holds
( o,a // o,p & ex d being Element of SAS st
( o,p // o,b implies ( o,c // o,d & p,c // b,d ) ) )
verumproof
let b,
c be
Element of
SAS;
( o,a // o,p & ex d being Element of SAS st
( o,p // o,b implies ( o,c // o,d & p,c // b,d ) ) )
ex
d being
Element of
SAS st
(
o,
p // o,
b implies (
o,
c // o,
d &
p,
c // b,
d ) )
proof
now ( o,p // o,b implies ex d being Element of SAS st
( o,p // o,b implies ( o,c // o,d & p,c // b,d ) ) )assume
o,
p // o,
b
;
ex d being Element of SAS st
( o,p // o,b implies ( o,c // o,d & p,c // b,d ) )then
p,
o // o,
b
by AFF_1:4;
then consider d being
Element of
SAS such that A5:
(
c,
o // o,
d &
c,
p // b,
d )
by A3, DIRAF:40;
(
o,
c // o,
d &
p,
c // b,
d )
by A5, AFF_1:4;
hence
ex
d being
Element of
SAS st
(
o,
p // o,
b implies (
o,
c // o,
d &
p,
c // b,
d ) )
;
verum end;
hence
ex
d being
Element of
SAS st
(
o,
p // o,
b implies (
o,
c // o,
d &
p,
c // b,
d ) )
;
verum
end;
hence
(
o,
a // o,
p & ex
d being
Element of
SAS st
(
o,
p // o,
b implies (
o,
c // o,
d &
p,
c // b,
d ) ) )
by A4;
verum
end;