let a, b be real number ; :: thesis: for n being Element of NAT st a < b & ( ( a >= 0 & n >= 1 ) or ex m being Element of NAT st n = (2 * m) + 1 ) holds
n -root a < n -root b

let n be Element of NAT ; :: thesis: ( a < b & ( ( a >= 0 & n >= 1 ) or ex m being Element of NAT st n = (2 * m) + 1 ) implies n -root a < n -root b )
assume that
A1: a < b and
A2: ( ( 0 <= a & n >= 1 ) or ex m being Element of NAT st n = (2 * m) + 1 ) ; :: thesis: n -root a < n -root b
A3: now
let a, b be real number ; :: thesis: for n being Element of NAT st 0 <= a & n >= 1 & a < b holds
n -root a < n -root b

let n be Element of NAT ; :: thesis: ( 0 <= a & n >= 1 & a < b implies n -root a < n -root b )
assume A4: ( 0 <= a & n >= 1 & a < b ) ; :: thesis: n -root a < n -root b
then n -Root a < n -Root b by PREPOWER:37;
then n -Root a < n -root b by A4, Def1;
hence n -root a < n -root b by A4, Def1; :: thesis: verum
end;
now
assume A5: ex m being Element of NAT st n = (2 * m) + 1 ; :: thesis: n -root a < n -root b
then consider m being Element of NAT such that
A6: n = (2 * m) + 1 ;
A7: n >= 0 + 1 by A6, XREAL_1:8;
now
per cases ( a >= 0 or a < 0 ) ;
end;
end;
hence n -root a < n -root b ; :: thesis: verum
end;
hence n -root a < n -root b by A1, A2, A3; :: thesis: verum