Some challenge for a top-level with cyclic terms

General
1

Currently I get these results:

/* SWI-Prolog 9.1.7 */
?- A=(B,0), B=(A,1), T=f(A,B).
A = ((A, 1), 0),
B = (A, 1),
T = f(A, (A, 1)).

?- A=(B,0), B=(A,1), T=f(B,A).
A = ((A, 1), 0),
B = (A, 1),
T = f((A, 1), A).

But I wonder, wouldn’t it be shorter and more intuitive to show:

?- A=(B,0), B=(A,1), T=f(A,B).
A = (B, 0),
B = (A, 1),
T = f(A, B).

?- A=(B,0), B=(A,1), T=f(B,A).
A = (B, 0),
B = (A, 1),
T = f(B, A).

Edit 13.04.2023
Amazingly, Trealla Prolog just does it:

/* Trealla Prolog 2.14.17 */
?- A=(B,0), B=(A,1), T=f(A,B).
   A = (B,0), B = (A,1), T = f(A,B).

?- A=(B,0), B=(A,1), T=f(B,A).
   A = (B,0), B = (A,1), T = f(B,A).

It clearly raises the bar, Scryer-Prolog isn’t very convincing:

/* Scryer Prolog 0.9.1-228 */
?- A=(B,0), B=(A,1), T=f(A,B).
   A = ((A,1),0), B = ((B,0),1), T = f(((A,1),0),((B,0),1)).

?- A=(B,0), B=(A,1), T=f(B,A).
   A = ((A,1),0), B = ((B,0),1), T = f(((B,0),1),((A,1),0)).
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