Has by well-foundedness every non-empty class an $R$-minimal element? Also if axiom REG is not assumed? I asked this question here on Math.SE but uptil now it was not answered. So I decided to give it a try. Thank you in advance.
Working in $\mathbf{ZF}$ let $R$ be a proper class of ordered pairs that is well-founded. This means that for every non-empty set $a$ there is a set $b\in a$ such that $cRb\implies c\notin a$. Here $cRb$ is a notation for $\langle c,b\rangle\in R$ and $b$ is a so-called $R$-minimal element of $a$. If $R$ is local (i.e. collections $\{x\mid xRb\}$ are all sets) then it can be shown that also non-empty proper classes have $R$-minimal elements. 
I encountered a proof that made the condition of being local redundant. It made use of an operation on classes that adds to each class a set that is contained in it (Bottom-operation) but the definition of this operation relied on the regularity axiom. 
My question:

Is there a proof that every non-empty class has an $R$-minimal element that does not make use of the regularity axiom?

Another formulation:

If $R$ is a class of ordered pairs that is well-founded, then is it legal to apply $R$-induction if all axioms of $\mathbf{ZF}$ are accepted with exception of the axiom of regularity?

Edit:
Maybe a bounty will help. If the question will not be answered then I will start cherishing the fact that I asked a good question ($6$ upvotes) that 'nobody' could answer :).
 A: Here's a counterexample: it is consistent with $\mathbf{ZFC}^-$ that $\mathbf{U} := \{x : x = \{x\}\}$ is a proper class with no infinite subsets [1]. Once you have this, consider the class $\mathbf{R} := \mathcal{P}(\mathbf{U})$ of all finite subsets of $\mathbf{U}$, ordered with respect to $\supsetneqq$. Then given any subset $A \subseteq \mathbf{R}$, $\bigcup A$ is finite and as $A \subseteq \mathcal{P}(\bigcup A)$, $A$ too is finite and thus has a $\supsetneqq$-minimal element. Therefore $\mathbf{R}$ is well-founded. However, given an $A \in \mathbf{R}$ there exists an $x \in \mathbf{U} \backslash A$ and $A \cup \{x\} \supsetneqq A$, proving that $\mathbf{R}$ has no $\supsetneqq$-minimal element.
[1] Exercise II.9.11 from Set Theory, Kenneth Kunen, 2011.
A: Let me mention another counterexample. In [1, Thm. 11], we construct a model of $\mathrm{ZFC}^-$ with the collection schema which contains a definable class relation $\langle A,<\rangle$ such that


*

*$<$ is a dense linear order on $A$ with no least element;

*every subset of $A$ is well-ordered by $<$.
The second condition says that $<$ is well-founded in the sense used in the question, nevertheless it has no minimal element by 1.
[1] A. S. Daghighi, M. Golshani, J. D. Hamkins, E. Jeřábek, The foundation axiom and elementary self-embeddings of the universe, in: Infinity, Computability, and Metamathematics: Festschrift celebrating the 60th birthdays of Peter Koepke and Philip Welch (S. Geschke, B. Löwe, and P. Schlicht, eds.), College Publications, London, 2014, pp. 89–112.
