let G be Group; :: thesis: for A being non empty Subset of G
for x being Element of G holds card A = card (A * x)
let A be non empty Subset of G; :: thesis: for x being Element of G holds card A = card (A * x)
let x be Element of G; :: thesis: card A = card (A * x)
set B = A * x;
( not A is empty & not A * x is empty )

proof

consider a being Element of A;
a * x in A * x by GROUP_2:34;
hence ( not A is empty & not A * x is empty ) ; :: thesis: verum

end;

then reconsider B = A * x as non empty Subset of G ;
deffunc H₁( Element of G) -> Element of the carrier of G = $1 * x;
consider f being Function of A,the carrier of G such that
A1: for a being Element of A holds f . a = H₁(a) from FUNCT_2:sch 4();
A2: dom f = A by FUNCT_2:def 1;
A3: rng f = B

proof

A4: rng f c= B

proof

let s be set ; :: according to TARSKI:def 3 :: thesis: ( not s in rng f or s in B )
assume s in rng f ; :: thesis: s in B
then consider a being set such that
A5: ( a in A & s = f . a ) by A2, FUNCT_1:def 5;
reconsider a = a as Element of A by A5;
s = H₁(a) by A1, A5;
hence s in B by GROUP_2:34; :: thesis: verum

end;

B c= rng f

proof

let s be set ; :: according to TARSKI:def 3 :: thesis: ( not s in B or s in rng f )
assume s in B ; :: thesis: s in rng f
then consider a being Element of G such that
A6: ( s = a * x & a in A ) by GROUP_2:34;
ex x being set st
( x in dom f & s = f . x )

proof

take x = a; :: thesis: ( x in dom f & s = f . x )
thus ( x in dom f & s = f . x ) by A1, A6, FUNCT_2:def 1; :: thesis: verum

end;

hence s in rng f by FUNCT_1:def 5; :: thesis: verum

end;

hence rng f = B by A4, XBOOLE_0:def 10; :: thesis: verum

end;

then reconsider f = f as Function of A,B by A2, FUNCT_2:4;
A7: for a, b being Element of A st f . a = f . b holds
a = b

proof

let a, b be Element of A; :: thesis: ( f . a = f . b implies a = b )
assume f . a = f . b ; :: thesis: a = b
then H₁(a) = f . b by A1;
then a * x = b * x by A1;
hence a = b by GROUP_1:14; :: thesis: verum

end;

A,B are_equipotent

proof

take f ; :: according to WELLORD2:def 4 :: thesis: ( f is one-to-one & dom f = A & rng f = B )
thus ( f is one-to-one & dom f = A & rng f = B ) by A3, A7, FUNCT_2:def 1, GROUP_6:1; :: thesis: verum

end;

hence card A = card (A * x) by CARD_1:21; :: thesis: verum