2

A is probable because of B, C, and D

D is probable because of E, F and A

Initially it looks like it is circular reasoning, however in the first statement D is not required for A to be probable, B and C do that on their own. D just acts as support to the notion.

In the second statement, A is not required for D to be probable, but A just acts as a support to the notion.

2

Yes, literally that is circular reasoning.

However, it may be the case that one of the following is true:

  • A is probable because of B and C. D is probable because of E, F and A.
  • A is probable because of B, C and D. D is probable because of E and F.

And obviously, these don't use circular reasoning.

1

It is circular reasoning if you infer A is likely, and then infer D is more likely because you've inferred that A is likely and so on. That being said, the two statements in themselves are not inconsistent with each other when considered independently. The problem is when you reason that one statement reinforces the other.

A real world situation where this could come up is one research lab does an experiment and finds that A depends on B,C,D, but doesn't control for E,F. A second group does an independent experiment and finds that D depends on E,F,A (but doesn't control for B,C). Both results would be valid, but there is no valid way to just combine these two studies into one, more powerful (in the statistical sense), result because each one fails to consider variables involved in the other.

The mathematical/formal approach for dealing with these types of situations are Bayesian networks; the systematic use of this approach will avoid circular reasoning.

In this language, map onto statements about P(A | B, C, D) [read as the probability of A depends on variables B,C,D] and P(D | E, F, A) and these, by themselves, don't fully constrain the possible models for the joint relationship between all of the variables. Some of the important information that is missing are the conditional independence statements, i.e. which variables are (conditionally) independent of which other variables, in particular we don't know if A depends on E,F or D depends on A,B (or whether E,F depend on B,C).

0

If one says A implies B, and that B implies A then, at least formally we can say that this is a circular argument.

If one alternatively has A implies B to a certain degree, ie it's plausible; and also the reverse, B implies A to a certain degree; then one may not, even formally say that we have circular logic - for the degree of plausibility may tend more in one direction than another.

-1

You can visualize the probabilistic dependencies by drawing a directed graph with

  • the propositions A, B, C, D, E, F as vertices and
  • a directed arc (x,y) fom x to y in case y is probable because of x. You get arcs (B,A), (C,A), (D,A), (A,D), (E,D) and (F,D).

The resulting directed graph has several cycle-free directed paths containing the vertices A and D and starting with a vertex different from A and D.

This property shows that A and D are probable because of other propositions, hence the probability of their respective truth is not circular.

  • I'm not familiar with this stuff, but isn't it clear simply from the term "probable" and more than one condition that it can't be strictly "circular" in a tautological sense? And apart from that "circular" would have very little meaning? Just curious. I am probably just misconstruing "circular." – Nelson Alexander Oct 29 '15 at 13:21
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    That there is something not circular in that reasoning doesn't mean it isn't still circular. – user2953 Oct 29 '15 at 14:27
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    @Nelson Alexander The graph method is quite general and applies also for large systems of dependencies where one does not see the solution at first sight. Needless to say, you can implement it by a computer program using graph traversal. – Jo Wehler Oct 29 '15 at 15:04
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    If you're trying to do a Bayesian network type of analysis, then there's a problem since there is a A->D->A loop in your network. If not you should indicate what formalism allows you to draw the conclusion you have drawn. – Dave Oct 29 '15 at 15:20
  • @Dave Which point of my argumentation seems problematic to you? - I do not consider the problem in question to be of Bayesian type, because there is no backward reasoning. – Jo Wehler Oct 29 '15 at 16:44

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