let SAS be AffinPlane; :: thesis: 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; :: thesis: 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
A6:
( o <> p & o,a // o,p )
;
take
p
; :: thesis: 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 ) ) )
:: thesis: verumproof
let b,
c be
Element of
SAS;
:: thesis: ( 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
;
:: thesis: 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 A7:
(
c,
o // o,
d &
c,
p // b,
d )
by A6, DIRAF:47;
(
o,
c // o,
d &
p,
c // b,
d )
by A7, AFF_1:13;
hence
ex
d being
Element of
SAS st
(
o,
p // o,
b implies (
o,
c // o,
d &
p,
c // b,
d ) )
;
:: thesis: verum end;
hence
ex
d being
Element of
SAS st
(
o,
p // o,
b implies (
o,
c // o,
d &
p,
c // b,
d ) )
;
:: thesis: 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 A6;
:: thesis: verum
end;