let f, g be Function of I[01],(Tunit_circle 2); :: thesis: ( ( for x being Point of I[01] holds f . x = |[(cos (((2 * PI) * r) * x)),(sin (((2 * PI) * r) * x))]| ) & ( for x being Point of I[01] holds g . x = |[(cos (((2 * PI) * r) * x)),(sin (((2 * PI) * r) * x))]| ) implies f = g )

assume that

A2: for x being Point of I[01] holds f . x = |[(cos (((2 * PI) * r) * x)),(sin (((2 * PI) * r) * x))]| and

A3: for x being Point of I[01] holds g . x = |[(cos (((2 * PI) * r) * x)),(sin (((2 * PI) * r) * x))]| ; :: thesis: f = g

for x being Point of I[01] holds f . x = g . x

assume that

A2: for x being Point of I[01] holds f . x = |[(cos (((2 * PI) * r) * x)),(sin (((2 * PI) * r) * x))]| and

A3: for x being Point of I[01] holds g . x = |[(cos (((2 * PI) * r) * x)),(sin (((2 * PI) * r) * x))]| ; :: thesis: f = g

for x being Point of I[01] holds f . x = g . x

proof

hence
f = g
; :: thesis: verum
let x be Point of I[01]; :: thesis: f . x = g . x

thus f . x = |[(cos (((2 * PI) * r) * x)),(sin (((2 * PI) * r) * x))]| by A2

.= g . x by A3 ; :: thesis: verum

end;thus f . x = |[(cos (((2 * PI) * r) * x)),(sin (((2 * PI) * r) * x))]| by A2

.= g . x by A3 ; :: thesis: verum