You can answer such questions
to the negative via Fuzzy Testing:

However such relation induced above is transitive ?

Typical counter example, showing
the native compare/3 is not transitive:

/* SWI-Prolog 9.3.25 */
?- repeat, fuzzy(X), fuzzy(Y), compare(<,X,Y), fuzzy(Z),
compare(<,Y,Z), compare(>,X,Z).
X = f(f(f(X, 1), 1), 1),
Y = f(f(Y, 1), 0),
Z = f(f(f(Z, 1), 0), 1) .
Etc..

Works also for Trealla Prolog, so I
assume compare_expensive/3 also fails:

/* Trealla Prolog 2.78.5 */
?- repeat, fuzzy(X), fuzzy(Y), compare(<,X,Y), fuzzy(Z),
compare(<,Y,Z), compare(>,X,Z).
X = f(f(f(...,0),0),1), Y = f(f(...,1),0), Z = f(f(...,0),1)
; ... .
Etc..

In essence, my nasty fuzzer uses
variants of Matt Carlsons example.

P.S.: From a logic viewpoint you could view
Fuzzy Testing as a form of randomized model
checking. A little famous Fuzzy Tester was Sandsifter:

"The tool discovered undocumented instructions
in all major processors, shared bugs in nearly
every major assembler and disassembler, flaws in
enterprise hypervisors, and critical x86 hardware" https://github.com/xoreaxeaxeax/sandsifter/blob/master/references/domas_breaking_the_x86_isa_wp.pdf

So we are essentially all using Trojan Horses 🐎 daily?

Mild Shock schrieb:

Hi,

The same example values also create fishy 🐟
sorting using native sorting in Scryer Prolog:

/* Scryer Prolog 0.9.4-417 */
?- values([z,x,y], A), sort(A, B),
   values([x,y,z], C), sort(C, D), B == D.
   false. /* fishy 🐟 */

Or using native sorting in SWI-Prolog:

/* SWI-Prolog 9.3.25 */
?- values([z,x,y], A), sort(A, B),
   values([x,y,z], C), sort(C, D), B == D.
false. /* fishy 🐟 */

Bye

Mild Shock schrieb:

I checked that your examples are not counter
examples for my compare_with_stack/3.

What makes you think the values I show, X, Y
and Z, are possible in a total linear ordering?
The values also break predsort/3, you can easily
verify that sort([x,y,z]) =\= sort([y,x,z]):

value(x, X) :- X = X-0-9-7-6-5-4-3-2-1.
value(y, Y) :- Y = Y-7-5-8-2-4-1.
value(z, Z) :- H = H-9-7-6-5-4-3-2-1-0, Z = H-9-7-6-5-4-3-2-1.

values(L, R) :- maplist(value, L, R).

?- values([x,y,z], A), predsort(compare_with_stack, A, B),
    values([y,x,z], C), predsort(compare_with_stack, C, D),
    B == D.
false.

But expectation would be sort([x,y,z]) ==
sort([y,x,z]) since sort/2 should be immune
to permutation. If this isn’t enough proof that
there is something fishy in compare_with_stack/3 ,

well then I don’t know, maybe the earth is indeed flat?

Mild Shock schrieb:

Hi,

Now somebody was so friendly to spear head
a new Don Quixote attempt in fighting the
windmills of compare/3. Interestingly my

favorite counter example still goes through:

?- X = X-0-9-7-6-5-4-3-2-1, Y = Y-7-5-8-2-4-1,
    compare_with_stack(C, X, Y).
X = X-0-9-7-6-5-4-3-2-1,
Y = Y-7-5-8-2-4-1,
C = (<).

?- H = H-9-7-6-5-4-3-2-1-0, Z = H-9-7-6-5-4-3-2-1, Y = Y-7-5-8-2-4-1,
    compare_with_stack(C, Z, Y).
H = H-9-7-6-5-4-3-2-1-0,
Z = H-9-7-6-5-4-3-2-1,
Y = Y-7-5-8-2-4-1,
C = (>).

?- H = H-9-7-6-5-4-3-2-1-0, Z = H-9-7-6-5-4-3-2-1, X =
X-0-9-7-6-5-4-3-2-1,
    compare_with_stack(C, Z, X).
H = H-9-7-6-5-4-3-2-1-0,
Z = X, X = X-0-9-7-6-5-4-3-2-1,
C = (=).

I posted it here in March 2023:

Careful with compare/3 and Brent algorithm
https://swi-prolog.discourse.group/t/careful-with-compare-3-and-brent-algorithm/6413


Its based that rational terms are indeed in
some relation to rational numbers. The above
terms are related to:

10/81 = 0.(123456790) = 0.12345679(012345679)

Bye

Mild Shock schrieb:

Hi,

That false/0 and not fail/0 is now all over the place,
I don't mean in person but for example here:

?- X=f(f(X), X), Y=f(Y, f(Y)), X = Y.
false.

Is a little didactical nightmare.

Syntactic unification has mathematical axioms (1978),
to fully formalize unifcation you would need to
formalize both (=)/2 and (≠)/2 (sic!), otherwise you
rely on some negation as failure concept.

Keith L. Clark, Negation as Failure
https://link.springer.com/chapter/10.1007/978-1-4684-3384-5_11

You can realize a subset of a mixture of (=)/2
and (≠)/2 in the form of a vanilla unify Prolog
predicate using some of the meta programming
facilities of Prolog, like var/1 and having some

negation as failure reading:

/* Vanilla Unify */
unify(V, W) :- var(V), var(W), !, (V \== W -> V = W; true).
unify(V, T) :- var(V), !, V = T.
unify(S, W) :- var(W), !, W = S.
unify(S, T) :- functor(S, F, N), functor(T, F, N),
      S =.. [F|L], T =.. [F|R], maplist(unify, L, R).

I indeed get:

?- X=f(f(X), X), Y=f(Y, f(Y)), unify(X,Y).
false.

If the vanilla unify/2 already fails then unify
with and without subject to occurs check, will also
fail, and unify with and without ability to
handle rational terms, will also fail:

Bye

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