let k be Nat; :: thesis: sin ^2 is PI * (k + 1) -periodic
defpred S1[ Nat] means sin ^2 is PI * (\$1 + 1) -periodic ;
A1: S1[ 0 ] by Lm20;
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 ^2 is PI * (k + 1) -periodic ; :: thesis: S1[k + 1]
sin ^2 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))) )
A5: ( x + (PI * ((k + 1) + 1)) in dom sin & x - (PI * ((k + 1) + 1)) in dom sin ) by ;
() . (x + (PI * ((k + 1) + 1))) = (sin . ((x + (PI * (k + 1))) + PI)) ^2 by VALUED_1:11
.= (- (sin . (x + (PI * (k + 1))))) ^2 by SIN_COS:78
.= (sin . (x + (PI * (k + 1)))) ^2
.= () . (x + (PI * (k + 1))) by VALUED_1:11 ;
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 ^2 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 ^2 is PI * (k + 1) -periodic ; :: thesis: verum