You seem to be asking about well-founded induction. It generalizes many forms of induction, including the usual induction on numbers and transfinite induction on ordinals.
Consider a relation $<$ on a set $A$. Say that $S \subseteq A$ is $<$-progressive when, for all $x \in A$, $$(\forall y < x \,.\, y \in S) \Rightarrow x \in S.$$ In words, an element is in $S$ as soon as all of its predecessors are. There is a logical counter-part: say that $\phi$ is a property of elements of $A$, then $\phi$ is $<$-progressive when, for all $x \in A$, $$(\forall y < x \,.\, \phi(y)) \Rightarrow \phi(x).$$
A well-founded order is a relation $<$ on a set $A$ such that, if $S \subseteq A$ is $<$-progressive then $S = A$. A well-founded order enjoys the following induction principle: if $\phi$ is a $<$-progressive property then $\phi(x)$ holds for all $x \in A$.
A strict order is a relation $<$ which is irreflexive and transitive. We may associate with it a partial order $\leq$ by $$x \leq y \iff x = y \lor x < y.$$
We have the following characterization:
Theorem. Let $<$ be an irreflexive, transitive relation on $A$. The following are equivalent:
- $<$ is well-founded,
- every nonempty subset $S \subseteq A$ has a $\leq$-minimal element,
- there are no infinite descending chains $\cdots < x_3 < x_2 < x_1$ in $A$.
To summarize, an irreflexive and transitive relation $<$ without infinite descending chains gives us the following induction principle: Suppose that for every $x \in A$ we have $(\forall y < x . \phi(y)) \Rightarrow \phi(x)$. Then $\forall z \in A. \phi(z)$.
A final remark: a linearly ordered well-founded relation is just a well-ordered relation. Induction on well-ordered relations is a bit more familiar, as it is just ordinal induction.