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 )
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 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_1:13;
then consider d being
Element of
SAS such that A5:
(
c,
o // o,
d &
c,
p // b,
d )
by A3, DIRAF:47;
(
o,
c // o,
d &
p,
c // b,
d )
by A5, AFF_1:13;
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;