- #1

TomMe

- 51

- 0

Can anyone show me how to prove exactly that the composition of 2 function is again a function, by using the following 3 formulas?

Suppose f: A -> B and g: B -> C are functions, then

(1)[tex]\forall \ a \ \epsilon \ A : \exists ! \ b \ \epsilon \ B : (a,b) \ \epsilon \ f[/tex]

(2)[tex]\forall \ b \ \epsilon \ B : \exists ! \ c \ \epsilon \ C : (b,c) \ \epsilon \ g[/tex].

And by definition the composition of relations f and g is

(3)[tex] g \ o \ f = \{(a,c) \ | \ \exists \ b \ \epsilon \ B : (a,b) \ \epsilon \ f \ and \ (b,c) \ \epsilon \ g \}[/tex].

I should be getting [tex]\forall \ a \ \epsilon \ A : \exists ! \ c \ \epsilon \ C : (a,c) \ \epsilon \ g \ o \ f[/tex] but I'm not sure how to combine the givens. I can do it in words, no problem, but I'm not that good in the use of quantifiers.

Thanks in advance.

Suppose f: A -> B and g: B -> C are functions, then

(1)[tex]\forall \ a \ \epsilon \ A : \exists ! \ b \ \epsilon \ B : (a,b) \ \epsilon \ f[/tex]

(2)[tex]\forall \ b \ \epsilon \ B : \exists ! \ c \ \epsilon \ C : (b,c) \ \epsilon \ g[/tex].

And by definition the composition of relations f and g is

(3)[tex] g \ o \ f = \{(a,c) \ | \ \exists \ b \ \epsilon \ B : (a,b) \ \epsilon \ f \ and \ (b,c) \ \epsilon \ g \}[/tex].

I should be getting [tex]\forall \ a \ \epsilon \ A : \exists ! \ c \ \epsilon \ C : (a,c) \ \epsilon \ g \ o \ f[/tex] but I'm not sure how to combine the givens. I can do it in words, no problem, but I'm not that good in the use of quantifiers.

Thanks in advance.

Last edited: