let k be Nat; :: thesis: is PI * (k + 1) -periodic
defpred S1[ Nat] means is PI * (\$1 + 1) -periodic ;
A1: S1[ 0 ] by Lm14;
A2: for k being Nat st S1[k] holds
S1[k + 1]
proof
let k be Nat; :: thesis: ( S1[k] implies S1[k + 1] )
assume A3: |.sin.| is PI * (k + 1) -periodic ; :: thesis: S1[k + 1]
|.sin.| is PI * ((k + 1) + 1) -periodic
proof
for x being Real st x in dom holds
( x + (PI * ((k + 1) + 1)) in dom & x - (PI * ((k + 1) + 1)) in dom & . x = . (x + (PI * ((k + 1) + 1))) )
proof
let x be Real; :: thesis: ( x in dom implies ( x + (PI * ((k + 1) + 1)) in dom & x - (PI * ((k + 1) + 1)) in dom & . x = . (x + (PI * ((k + 1) + 1))) ) )
assume A4: x in dom ; :: thesis: ( x + (PI * ((k + 1) + 1)) in dom & x - (PI * ((k + 1) + 1)) in dom & . x = . (x + (PI * ((k + 1) + 1))) )
then A5: ( x + (PI * (k + 1)) in dom & x - (PI * (k + 1)) in dom ) by ;
A6: ( x + (PI * ((k + 1) + 1)) in dom sin & x - (PI * ((k + 1) + 1)) in dom sin ) by ;
then ( x + (PI * ((k + 1) + 1)) in dom & x - (PI * ((k + 1) + 1)) in dom ) by VALUED_1:def 11;
then . (x + (PI * ((k + 1) + 1))) = |.(sin . ((x + (PI * (k + 1))) + PI)).| by VALUED_1:def 11
.= |.(- (sin . (x + (PI * (k + 1))))).| by SIN_COS:78
.= |.(sin . (x + (PI * (k + 1)))).| by COMPLEX1:52
.= . (x + (PI * (k + 1))) by ;
hence ( x + (PI * ((k + 1) + 1)) in dom & x - (PI * ((k + 1) + 1)) in dom & . x = . (x + (PI * ((k + 1) + 1))) ) by ; :: thesis: verum
end;
hence |.sin.| is PI * ((k + 1) + 1) -periodic by Th1; :: thesis: verum
end;
hence S1[k + 1] ; :: thesis: verum
end;
for n being Nat holds S1[n] from NAT_1:sch 2(A1, A2);
hence |.sin.| is PI * (k + 1) -periodic ; :: thesis: verum