# Given x o y = xy-x-y+2 and f(x) = x+1 , prove that f(xy) = f(x) o f(y).

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We have x o y = xy - x - y + 2 and f(x) = x + 1.

f(xy) = xy + 1

f(x) o f(y) = (x + 1) o (y + 1)

=> (x + 1)(y + 1) - (x + 1) - (y + 1) + 2

=> xy + x + y + 1 - x - 1 - y - 1 + 2

=> xy + 1

We see that both f(xy) and f(x) o f(y) give xy + 1.

**This proves that f(xy) = f(x) o f(y).**

We'll write f(xy) = xy+1

Now, we'll manage the right side:

f(x) o f(y)=(x+1)o(y+1)

We'll apply the rule of composition of 2 terms :

(x+1)o(y+1)=(x+1)(y+1)-(x+1)-(y+1)+2

(x+1)o(y+1)=xy+x+y+1-x-1-y-1+2

(x+1)o(y+1)=xy+1=f(xy)

We notice that managing both sides, we've get the same result.

**There fore, the identity f(xy)=f(x)of(y) is verified.**