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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 $f(1,2)$ ________

Given that,

• $f( 0,y) = y+1 \quad \longrightarrow (1)$
• $f(x+1,0) = f( x, 1) \quad \longrightarrow (2)$
• $f( x+1, y+1) = f\left( x,f( x+1,y) \right) \quad \longrightarrow (3)$

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

• $f(1,2) = f(0,f(1,1)) \quad \longrightarrow (4)$

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

• $f(1,1) = f(0,f(1,0)) \quad \longrightarrow (5)$

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

• $f(1,0) = f( 0, 1) \quad \longrightarrow (6)$

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,

• $f(1,1) = f(0,2) \quad \longrightarrow (7)$

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

• $f(1,1) = 3$

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

• $f(1,2) = f(0,3) \quad \longrightarrow (8)$

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

$\therefore \boxed{f(1,2) = 4}$

Correct Answer $:4$

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