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For all non-negative integers $x$ and $y$, $f(x,y)$ is defined as below.

$f\left ( 0,y \right )=y+1f\left ( x+1,0 \right )= f\left ( x, 1\right )f\left ( x+1, y+1 \right )=f\left ( x,f\left ( x+1,y \right ) \right ).$

Then what is the value of $f(1,2)$?________

Lakshman

For all non-negative integers $x$ and $y$, $f(x,y)$ is defined as below.

$f( 0,y) = y+1$

$f(x+1,0) = f( x, 1)$

$f( x+1, y+1) = f\left( x,f( x+1,y) \right)$

Then what is the value of $f(1,2)?$________

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Given that$:f( 0,y) = y+1\rightarrow(1)$

$f(x+1,0) = f( x, 1)\rightarrow(2)$

$f( x+1, y+1) = f\left( x,f( x+1,y) \right)\rightarrow(3)$

Now, put $x = 0,y = 1,$ in equation $(3)$ we get,

$f(1,2) = f(0,f(1,1))\rightarrow(4)$

And, put $x = 0,y = 0$ in equation $(3)$ we get,

$f(1,1) = f(0,f(1,0))\rightarrow(5)$

And, put $x = 0$ in equation $(2)$ we get,

$f(1,0) = f( 0, 1)\rightarrow(6)$

And, put $y = 1,$ in equation $(1)$ we get,

$f( 0,1) = 1+1 = 2$

Now, put $f(0,1) = 2,$ in equation $(6),$ we get,

$f(1,0) = 2$

Now, Put $f(0,1) = 2,$ in equation $(5),$ we get,

$f(1,1) = f(0,2)\rightarrow(7)$

Put $y = 2$ in equation $(1)$ we get, $f(0,2) = 2+1 = 3$

$\therefore f(1,1) = 3$

Now, put $f(1,1) = 3,$ in equation $(4),$ we get,

$f(1,2) = f(0,3)\rightarrow(8)$

Put $y = 3$ in equation $(1)$ we get, $f(0,3) = 3+1 = 4$

$\therefore f(1,2) = 4.$

So, the correct answer is $4.$
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