Can we expand "induction principle" to a partial order $(X, \leq)$? We know that every infinite can be made well-ordered with an unknown order.
Also  we can expand the induction principle on any infinite set  in
the sense that
it can made well ordered.
Now partially ordered set may not be a well ordered set with respect
to the partial order.
Let a partially ordered set $(X, \leq)$  with respect to this particular
order $'\leq'$ and suppose that this partial order $'\leq '$ does not make the
set $X$ well-ordered.
My question is-
Can we expand "induction principle" to the partially order set $(X,\leq)$  keeping
in mind that $(X,\leq)$ is not well ordered?
I have great confusion here.
 A: We can't even do induction on a total order if it's not well-ordered. Like on $\Bbb Q$ or $\Bbb R$ with the standard order. So in general a partial order is out of the question.
One could impose a well-ordering-like requirement on the partial order (every non-empty subset of $X$ has a minimal element), and then it is called a well-founded partial order. In that case one can indeed induct on a partial order.
A: One way to view the principle of induction is a manner to show there can be no counterexamples to the proposition being proved. In general the set of values for which a proposition fails can be any subset. So if you want to be able to show a proposition is valid by establishing that there are no minimal counterexamples, which is what the principle of strong induction does, then one needs an ordering with the property that every non-empty set has a minimal element. For total orderings, that is the property of being a well ordering. For partial orderings, it is the property of being well-founded. So if you want to employ an inductive argument of strong induction type (showing that validity of a proposition for all $y<x$ implies validity for $x$), then nothing weaker than having a well founded partial ordering will allow for such an induction principle to be valid.
To see why, take a partial ordering for which some non-empty set $S$ has no minimal element. Then for the proposition $P$ for which $P(x)$ says $x\notin S$ it is true that $\forall x: (\forall y: (y<x\to P(y))\to P(x)$ (because it is equivalent to the negation of "$S$ has a minimal element"), but the conclusion of the induction principle, namely $\forall x:P(x)$, is false. Therefore the strong induction principle is not valid in such a situation.
A: Induction can be performed over any relation $R$ over a set $X$, provided the relation is well-founded: any subset $S \subseteq X$ must have a minimal element with respect to $R$. A minimal element of $S$ is an element $m \in S$ such that there is no $x \in S$ with $x R m$. 
For total orders, being well-founded is equivalent to being well-ordered.
Note: Assuming the axiom of dependent choice (a weaker form of the axiom of choice), one can show that a relation is well-founded if and only if there is no infinite descending chain of elements in $X$ (with respect to the relation $R$).
Note 2: The Axiom of Choice is equivalent to stating that any set can be well-ordered. Thus, one could in principle do (transfinite) induction over any set. 
The problem is that the Axiom of Choice is not constructive, which means that no one knows anything about what the well-ordering given by the axiom looks like. Therefore, it is in practice impossible to use that well-ordering.
