Left Termination of the query pattern gopher_in_2(g, a) w.r.t. the given Prolog program could successfully be proven:



↳ Prolog
  ↳ PrologToPiTRSProof

Clauses:

gopher(nil, nil).
gopher(cons(nil, Y), cons(nil, Y)).
gopher(cons(cons(U, V), W), X) :- gopher(cons(U, cons(V, W)), X).

Queries:

gopher(g,a).

We use the technique of [30].Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

gopher_in(cons(cons(U, V), W), X) → U1(U, V, W, X, gopher_in(cons(U, cons(V, W)), X))
gopher_in(cons(nil, Y), cons(nil, Y)) → gopher_out(cons(nil, Y), cons(nil, Y))
gopher_in(nil, nil) → gopher_out(nil, nil)
U1(U, V, W, X, gopher_out(cons(U, cons(V, W)), X)) → gopher_out(cons(cons(U, V), W), X)

The argument filtering Pi contains the following mapping:
gopher_in(x1, x2)  =  gopher_in(x1)
cons(x1, x2)  =  cons(x1, x2)
U1(x1, x2, x3, x4, x5)  =  U1(x5)
nil  =  nil
gopher_out(x1, x2)  =  gopher_out(x2)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog



↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof

Pi-finite rewrite system:
The TRS R consists of the following rules:

gopher_in(cons(cons(U, V), W), X) → U1(U, V, W, X, gopher_in(cons(U, cons(V, W)), X))
gopher_in(cons(nil, Y), cons(nil, Y)) → gopher_out(cons(nil, Y), cons(nil, Y))
gopher_in(nil, nil) → gopher_out(nil, nil)
U1(U, V, W, X, gopher_out(cons(U, cons(V, W)), X)) → gopher_out(cons(cons(U, V), W), X)

The argument filtering Pi contains the following mapping:
gopher_in(x1, x2)  =  gopher_in(x1)
cons(x1, x2)  =  cons(x1, x2)
U1(x1, x2, x3, x4, x5)  =  U1(x5)
nil  =  nil
gopher_out(x1, x2)  =  gopher_out(x2)


Using Dependency Pairs [1,30] we result in the following initial DP problem:
Pi DP problem:
The TRS P consists of the following rules:

GOPHER_IN(cons(cons(U, V), W), X) → U11(U, V, W, X, gopher_in(cons(U, cons(V, W)), X))
GOPHER_IN(cons(cons(U, V), W), X) → GOPHER_IN(cons(U, cons(V, W)), X)

The TRS R consists of the following rules:

gopher_in(cons(cons(U, V), W), X) → U1(U, V, W, X, gopher_in(cons(U, cons(V, W)), X))
gopher_in(cons(nil, Y), cons(nil, Y)) → gopher_out(cons(nil, Y), cons(nil, Y))
gopher_in(nil, nil) → gopher_out(nil, nil)
U1(U, V, W, X, gopher_out(cons(U, cons(V, W)), X)) → gopher_out(cons(cons(U, V), W), X)

The argument filtering Pi contains the following mapping:
gopher_in(x1, x2)  =  gopher_in(x1)
cons(x1, x2)  =  cons(x1, x2)
U1(x1, x2, x3, x4, x5)  =  U1(x5)
nil  =  nil
gopher_out(x1, x2)  =  gopher_out(x2)
U11(x1, x2, x3, x4, x5)  =  U11(x5)
GOPHER_IN(x1, x2)  =  GOPHER_IN(x1)

We have to consider all (P,R,Pi)-chains

↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof

Pi DP problem:
The TRS P consists of the following rules:

GOPHER_IN(cons(cons(U, V), W), X) → U11(U, V, W, X, gopher_in(cons(U, cons(V, W)), X))
GOPHER_IN(cons(cons(U, V), W), X) → GOPHER_IN(cons(U, cons(V, W)), X)

The TRS R consists of the following rules:

gopher_in(cons(cons(U, V), W), X) → U1(U, V, W, X, gopher_in(cons(U, cons(V, W)), X))
gopher_in(cons(nil, Y), cons(nil, Y)) → gopher_out(cons(nil, Y), cons(nil, Y))
gopher_in(nil, nil) → gopher_out(nil, nil)
U1(U, V, W, X, gopher_out(cons(U, cons(V, W)), X)) → gopher_out(cons(cons(U, V), W), X)

The argument filtering Pi contains the following mapping:
gopher_in(x1, x2)  =  gopher_in(x1)
cons(x1, x2)  =  cons(x1, x2)
U1(x1, x2, x3, x4, x5)  =  U1(x5)
nil  =  nil
gopher_out(x1, x2)  =  gopher_out(x2)
U11(x1, x2, x3, x4, x5)  =  U11(x5)
GOPHER_IN(x1, x2)  =  GOPHER_IN(x1)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 1 SCC with 1 less node.

↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ PiDP
              ↳ UsableRulesProof

Pi DP problem:
The TRS P consists of the following rules:

GOPHER_IN(cons(cons(U, V), W), X) → GOPHER_IN(cons(U, cons(V, W)), X)

The TRS R consists of the following rules:

gopher_in(cons(cons(U, V), W), X) → U1(U, V, W, X, gopher_in(cons(U, cons(V, W)), X))
gopher_in(cons(nil, Y), cons(nil, Y)) → gopher_out(cons(nil, Y), cons(nil, Y))
gopher_in(nil, nil) → gopher_out(nil, nil)
U1(U, V, W, X, gopher_out(cons(U, cons(V, W)), X)) → gopher_out(cons(cons(U, V), W), X)

The argument filtering Pi contains the following mapping:
gopher_in(x1, x2)  =  gopher_in(x1)
cons(x1, x2)  =  cons(x1, x2)
U1(x1, x2, x3, x4, x5)  =  U1(x5)
nil  =  nil
gopher_out(x1, x2)  =  gopher_out(x2)
GOPHER_IN(x1, x2)  =  GOPHER_IN(x1)

We have to consider all (P,R,Pi)-chains
For (infinitary) constructor rewriting [30] we can delete all non-usable rules from R.

↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ PiDP
              ↳ UsableRulesProof
                ↳ PiDP
                  ↳ PiDPToQDPProof

Pi DP problem:
The TRS P consists of the following rules:

GOPHER_IN(cons(cons(U, V), W), X) → GOPHER_IN(cons(U, cons(V, W)), X)

R is empty.
The argument filtering Pi contains the following mapping:
cons(x1, x2)  =  cons(x1, x2)
GOPHER_IN(x1, x2)  =  GOPHER_IN(x1)

We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem [30] into ordinary QDP problem [15] by application of Pi.

↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ PiDP
              ↳ UsableRulesProof
                ↳ PiDP
                  ↳ PiDPToQDPProof
                    ↳ QDP
                      ↳ RuleRemovalProof

Q DP problem:
The TRS P consists of the following rules:

GOPHER_IN(cons(cons(U, V), W)) → GOPHER_IN(cons(U, cons(V, W)))

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

GOPHER_IN(cons(cons(U, V), W)) → GOPHER_IN(cons(U, cons(V, W)))


Used ordering: POLO with Polynomial interpretation [25]:

POL(GOPHER_IN(x1)) = 2·x1   
POL(cons(x1, x2)) = 2 + 2·x1 + x2   



↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ PiDP
              ↳ UsableRulesProof
                ↳ PiDP
                  ↳ PiDPToQDPProof
                    ↳ QDP
                      ↳ RuleRemovalProof
                        ↳ QDP
                          ↳ PisEmptyProof

Q DP problem:
P is empty.
R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.