I have a statement $P(x,y,z)$ in natural numbers and I want to prove using mathematical induction and I managed to prove the following steps
1) $P(1,1,1),P(1,2,1)$
2) $P(x,1,z)\land P(x,2,z)\land P(x,3,z)\land \cdots \land P(x,y,z) \implies P(x+1,y,z)$
3) $P(x,y,z) \land P(x -1,y+1,z) \implies P(x,y+1,z)$
4) $P(x,y,z) \land P(x+1,y -1,z) \implies P(x,y,z+1)$
Is my statement proved? My doubt is on step 2 because the premise also includes $y+1$. If $P$ is not proved by above 4 steps, then which steps I need to work on more to prove my statement $P$?
I assume the statement is of the form $$\forall x\forall y\forall z\,P(x,y,z). $$ Step 2 with some adjustments is essentially what you are after to prove the whole thing. Let $x,z$ be arbitrarily chosen, it suffices to show $$P(x,0,z)\qquad \mbox{and}\qquad \forall y \left (P(x,y,z)\implies P(x,s(y),z)\right ), $$ where $s(y)$ is the successor of $y$. If $0\notin\mathbb N$ then the base case to be proved is $P(x,1,z)$.
This works in general for statements about natural numbers such as $$\forall x_1\forall x_2\ldots \forall x_n\ Q(x_1,x_2,\ldots ,x_n). $$ You fix all but one variable, say $x_1$, and conduct induction on $x_1$. Since the other variables are arbitrarily chosen the statement $Q$ must be true for all possible choices if you prove the inductive step for $x_1$.