let k be Nat; :: thesis: sin is (2 * PI) * (k + 1) -periodic
defpred S₁[ Nat] means sin is (2 * PI) * ($1 + 1) -periodic ;
A1: S₁[ 0 ] by Lm2;
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 (2 * PI) * (k + 1) -periodic ; :: thesis: S₁[k + 1]
sin is (2 * PI) * ((k + 1) + 1) -periodic

proof

for x being Real st x in dom sin holds
sin . x = sin . (x + ((2 * PI) * ((k + 1) + 1)))

proof

let x be Real; :: thesis: ( x in dom sin implies sin . x = sin . (x + ((2 * PI) * ((k + 1) + 1))) )
assume A4: x in dom sin ; :: thesis: sin . x = sin . (x + ((2 * PI) * ((k + 1) + 1)))
sin . (x + ((2 * PI) * (k + 1))) = sin . ((x + ((2 * PI) * (k + 1))) + (2 * PI)) by SIN_COS:78;
hence sin . x = sin . (x + ((2 * PI) * ((k + 1) + 1))) by A3, A4; :: thesis: verum

end;

then ( (2 * PI) * ((k + 1) + 1) <> 0 & ( for x being Real st x in dom sin holds
( x + ((2 * PI) * ((k + 1) + 1)) in dom sin & x - ((2 * PI) * ((k + 1) + 1)) in dom sin & sin . x = sin . (x + ((2 * PI) * ((k + 1) + 1))) ) ) ) by SIN_COS:24, XREAL_0:def 1;
hence sin is (2 * 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 (2 * PI) * (k + 1) -periodic ; :: thesis: verum