Humans are just overwhelmed by computers (Re: Wait till they find out about compare/3)

 > 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