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Asked: 2026-06-14T11:16:11+00:00

I have the following prolog rules. b(b(true)) –> [true]. b(b(false)) –> [false]. b(b(E,[=],E)) –>

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I have the following prolog rules.

b(b(true)) --> [true].
b(b(false)) --> [false].
b(b(E,[=],E)) --> e(E),[=],e(E).
b(b([not],B)) --> [not],b(B).
e(e(I)) --> i(I).
e(e(N)) --> n(N).
e(e(N,O,E)) --> n(N),o(O),e(E).
e(e(I,O,E)) --> i(I),o(O),e(E).
o(o(+)) --> [+].
o(o(-))--> [-].
o(o(*))--> [*].
o(o(/)) --> [/].
i(i(x)) --> [x].
i(i(y)) --> [y].
i(i(z)) --> [z].
i(i(u)) --> [u].
i(i(v)) --> [v].
n(n(0)) --> [0].
n(n(1)) --> [1].
n(n(2)) --> [2].
n(n(3)) --> [3].
n(n(4)) --> [4].
n(n(5)) --> [5].
n(n(6)) --> [6].
n(n(7)) --> [7].
n(n(8)) --> [8].
n(n(9)) --> [9].

But I dunno why 

[6] 26 ?- b(A,[x,=,4],[]).
false

fails. I tried to debug the code. 4 is not getting matched with n(n(4)). I couldn’t understand wat the problem is.

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1 Answer

  1. Editorial Team

    (You would have received an answer much earlier if you would have used the prolog tag instead)

    So, how can we localize the error? Let’s start with your query:

    ?- phrase(b(A),[x,=,4]).
   false.

    Bad!
    Shall I really fire up a trace/debugger?
    What you are currently asking is: What are the solutions for A for a well-formed sentence.
    Alas, there is none.

    Can’t we ask Prolog a more general question?
    So, dear Prolog – at least – do you know any sentence?
    Please say a word!

    ?- phrase(b(A),L).
   A = b(true), L = [true]
;  A = b(false), L = [false]
;  ... .

    So there is something – I will remove the A, because we want to see the sentences, first.

    ?- phrase(b(_),L).
   L = [true]
;  L = [false]
;  L = [x,=,x]
;  L = [y,=,y]
;  L = [z,=,z]
;  L = [u,=,u]
;  L = [v,=,v]
;  L = [0,=,0]
;  L = [1,=,1]
;  L = [2,=,2]
;  L = [3,=,3]
;  L = [4,=,4]
;  L = [5,=,5]
;  L = [6,=,6]
;  L = [7,=,7]
;  L = [8,=,8]
;  L = [9,=,9]
;  L = [0,+,x,=,0,+,x]
;  ... .

    Do you see a pattern here?
    Maybe we can refine this.
    What are the sentences of length 3 you know of?

    ?- L = [_,_,_], phrase(b(_),L).
  L = [x,=,x]
;  L = [y,=,y]
;  L = [z,=,z]
;  L = [u,=,u]
;  L = [v,=,v]
;  L = [0,=,0]
;  L = [1,=,1]
;  L = [2,=,2]
;  L = [3,=,3]
;  L = [4,=,4]
;  L = [5,=,5]
;  L = [6,=,6]
;  L = [7,=,7]
;  L = [8,=,8]
;  L = [9,=,9]
;  L = [not,not,true]
;  L = [not,not,false]
;  false.

    In other words: There are no other three word sentences than those.
    Now, do you get it?
    There is a rule for sentences with =:

    b(b(E,[=],E)) -->
   e(E),[=],e(E).

    It requires that the expression on the left-hand side is the same as the expression on the right-hand side. Therefore only those sentences above.

    When debugging a Prolog program, there is much less need to type (on the keyboard) test data than in other languages. Instead, use variables to let Prolog fill-in-the-variables. After all, Prolog is much faster on that than we are ; and it does not suffer CTS.

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