From Newsgroup: sci.physics

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 (rea)/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 (rea)/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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From Newsgroup: sci.physics

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 (rea)/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 (rea)/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),
-a-a-a-a 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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