Left Termination of the query pattern samefringe_in_2(g, g) 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).
samefringe(nil, nil).
samefringe(cons(U, V), cons(X, Y)) :- ','(gopher(cons(U, V), cons(U1, V1)), ','(gopher(cons(X, Y), cons(X1, Y1)), samefringe(V1, Y1))).

Queries:

samefringe(g,g).

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:

samefringe_in(cons(U, V), cons(X, Y)) → U2(U, V, X, Y, gopher_in(cons(U, V), cons(U1, V1)))
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)
U2(U, V, X, Y, gopher_out(cons(U, V), cons(U1, V1))) → U3(U, V, X, Y, U1, V1, gopher_in(cons(X, Y), cons(X1, Y1)))
U3(U, V, X, Y, U1, V1, gopher_out(cons(X, Y), cons(X1, Y1))) → U4(U, V, X, Y, samefringe_in(V1, Y1))
samefringe_in(nil, nil) → samefringe_out(nil, nil)
U4(U, V, X, Y, samefringe_out(V1, Y1)) → samefringe_out(cons(U, V), cons(X, Y))

The argument filtering Pi contains the following mapping:
samefringe_in(x1, x2)  =  samefringe_in(x1, x2)
cons(x1, x2)  =  cons(x1, x2)
U2(x1, x2, x3, x4, x5)  =  U2(x3, x4, x5)
gopher_in(x1, x2)  =  gopher_in(x1)
U1(x1, x2, x3, x4, x5)  =  U1(x5)
nil  =  nil
gopher_out(x1, x2)  =  gopher_out(x2)
U3(x1, x2, x3, x4, x5, x6, x7)  =  U3(x6, x7)
U4(x1, x2, x3, x4, x5)  =  U4(x5)
samefringe_out(x1, x2)  =  samefringe_out

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:

samefringe_in(cons(U, V), cons(X, Y)) → U2(U, V, X, Y, gopher_in(cons(U, V), cons(U1, V1)))
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)
U2(U, V, X, Y, gopher_out(cons(U, V), cons(U1, V1))) → U3(U, V, X, Y, U1, V1, gopher_in(cons(X, Y), cons(X1, Y1)))
U3(U, V, X, Y, U1, V1, gopher_out(cons(X, Y), cons(X1, Y1))) → U4(U, V, X, Y, samefringe_in(V1, Y1))
samefringe_in(nil, nil) → samefringe_out(nil, nil)
U4(U, V, X, Y, samefringe_out(V1, Y1)) → samefringe_out(cons(U, V), cons(X, Y))

The argument filtering Pi contains the following mapping:
samefringe_in(x1, x2)  =  samefringe_in(x1, x2)
cons(x1, x2)  =  cons(x1, x2)
U2(x1, x2, x3, x4, x5)  =  U2(x3, x4, x5)
gopher_in(x1, x2)  =  gopher_in(x1)
U1(x1, x2, x3, x4, x5)  =  U1(x5)
nil  =  nil
gopher_out(x1, x2)  =  gopher_out(x2)
U3(x1, x2, x3, x4, x5, x6, x7)  =  U3(x6, x7)
U4(x1, x2, x3, x4, x5)  =  U4(x5)
samefringe_out(x1, x2)  =  samefringe_out


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

SAMEFRINGE_IN(cons(U, V), cons(X, Y)) → U21(U, V, X, Y, gopher_in(cons(U, V), cons(U1, V1)))
SAMEFRINGE_IN(cons(U, V), cons(X, Y)) → GOPHER_IN(cons(U, V), cons(U1, V1))
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)
U21(U, V, X, Y, gopher_out(cons(U, V), cons(U1, V1))) → U31(U, V, X, Y, U1, V1, gopher_in(cons(X, Y), cons(X1, Y1)))
U21(U, V, X, Y, gopher_out(cons(U, V), cons(U1, V1))) → GOPHER_IN(cons(X, Y), cons(X1, Y1))
U31(U, V, X, Y, U1, V1, gopher_out(cons(X, Y), cons(X1, Y1))) → U41(U, V, X, Y, samefringe_in(V1, Y1))
U31(U, V, X, Y, U1, V1, gopher_out(cons(X, Y), cons(X1, Y1))) → SAMEFRINGE_IN(V1, Y1)

The TRS R consists of the following rules:

samefringe_in(cons(U, V), cons(X, Y)) → U2(U, V, X, Y, gopher_in(cons(U, V), cons(U1, V1)))
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)
U2(U, V, X, Y, gopher_out(cons(U, V), cons(U1, V1))) → U3(U, V, X, Y, U1, V1, gopher_in(cons(X, Y), cons(X1, Y1)))
U3(U, V, X, Y, U1, V1, gopher_out(cons(X, Y), cons(X1, Y1))) → U4(U, V, X, Y, samefringe_in(V1, Y1))
samefringe_in(nil, nil) → samefringe_out(nil, nil)
U4(U, V, X, Y, samefringe_out(V1, Y1)) → samefringe_out(cons(U, V), cons(X, Y))

The argument filtering Pi contains the following mapping:
samefringe_in(x1, x2)  =  samefringe_in(x1, x2)
cons(x1, x2)  =  cons(x1, x2)
U2(x1, x2, x3, x4, x5)  =  U2(x3, x4, x5)
gopher_in(x1, x2)  =  gopher_in(x1)
U1(x1, x2, x3, x4, x5)  =  U1(x5)
nil  =  nil
gopher_out(x1, x2)  =  gopher_out(x2)
U3(x1, x2, x3, x4, x5, x6, x7)  =  U3(x6, x7)
U4(x1, x2, x3, x4, x5)  =  U4(x5)
samefringe_out(x1, x2)  =  samefringe_out
U41(x1, x2, x3, x4, x5)  =  U41(x5)
U31(x1, x2, x3, x4, x5, x6, x7)  =  U31(x6, x7)
SAMEFRINGE_IN(x1, x2)  =  SAMEFRINGE_IN(x1, x2)
U21(x1, x2, x3, x4, x5)  =  U21(x3, x4, x5)
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:

SAMEFRINGE_IN(cons(U, V), cons(X, Y)) → U21(U, V, X, Y, gopher_in(cons(U, V), cons(U1, V1)))
SAMEFRINGE_IN(cons(U, V), cons(X, Y)) → GOPHER_IN(cons(U, V), cons(U1, V1))
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)
U21(U, V, X, Y, gopher_out(cons(U, V), cons(U1, V1))) → U31(U, V, X, Y, U1, V1, gopher_in(cons(X, Y), cons(X1, Y1)))
U21(U, V, X, Y, gopher_out(cons(U, V), cons(U1, V1))) → GOPHER_IN(cons(X, Y), cons(X1, Y1))
U31(U, V, X, Y, U1, V1, gopher_out(cons(X, Y), cons(X1, Y1))) → U41(U, V, X, Y, samefringe_in(V1, Y1))
U31(U, V, X, Y, U1, V1, gopher_out(cons(X, Y), cons(X1, Y1))) → SAMEFRINGE_IN(V1, Y1)

The TRS R consists of the following rules:

samefringe_in(cons(U, V), cons(X, Y)) → U2(U, V, X, Y, gopher_in(cons(U, V), cons(U1, V1)))
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)
U2(U, V, X, Y, gopher_out(cons(U, V), cons(U1, V1))) → U3(U, V, X, Y, U1, V1, gopher_in(cons(X, Y), cons(X1, Y1)))
U3(U, V, X, Y, U1, V1, gopher_out(cons(X, Y), cons(X1, Y1))) → U4(U, V, X, Y, samefringe_in(V1, Y1))
samefringe_in(nil, nil) → samefringe_out(nil, nil)
U4(U, V, X, Y, samefringe_out(V1, Y1)) → samefringe_out(cons(U, V), cons(X, Y))

The argument filtering Pi contains the following mapping:
samefringe_in(x1, x2)  =  samefringe_in(x1, x2)
cons(x1, x2)  =  cons(x1, x2)
U2(x1, x2, x3, x4, x5)  =  U2(x3, x4, x5)
gopher_in(x1, x2)  =  gopher_in(x1)
U1(x1, x2, x3, x4, x5)  =  U1(x5)
nil  =  nil
gopher_out(x1, x2)  =  gopher_out(x2)
U3(x1, x2, x3, x4, x5, x6, x7)  =  U3(x6, x7)
U4(x1, x2, x3, x4, x5)  =  U4(x5)
samefringe_out(x1, x2)  =  samefringe_out
U41(x1, x2, x3, x4, x5)  =  U41(x5)
U31(x1, x2, x3, x4, x5, x6, x7)  =  U31(x6, x7)
SAMEFRINGE_IN(x1, x2)  =  SAMEFRINGE_IN(x1, x2)
U21(x1, x2, x3, x4, x5)  =  U21(x3, x4, x5)
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 2 SCCs with 4 less nodes.

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

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:

samefringe_in(cons(U, V), cons(X, Y)) → U2(U, V, X, Y, gopher_in(cons(U, V), cons(U1, V1)))
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)
U2(U, V, X, Y, gopher_out(cons(U, V), cons(U1, V1))) → U3(U, V, X, Y, U1, V1, gopher_in(cons(X, Y), cons(X1, Y1)))
U3(U, V, X, Y, U1, V1, gopher_out(cons(X, Y), cons(X1, Y1))) → U4(U, V, X, Y, samefringe_in(V1, Y1))
samefringe_in(nil, nil) → samefringe_out(nil, nil)
U4(U, V, X, Y, samefringe_out(V1, Y1)) → samefringe_out(cons(U, V), cons(X, Y))

The argument filtering Pi contains the following mapping:
samefringe_in(x1, x2)  =  samefringe_in(x1, x2)
cons(x1, x2)  =  cons(x1, x2)
U2(x1, x2, x3, x4, x5)  =  U2(x3, x4, x5)
gopher_in(x1, x2)  =  gopher_in(x1)
U1(x1, x2, x3, x4, x5)  =  U1(x5)
nil  =  nil
gopher_out(x1, x2)  =  gopher_out(x2)
U3(x1, x2, x3, x4, x5, x6, x7)  =  U3(x6, x7)
U4(x1, x2, x3, x4, x5)  =  U4(x5)
samefringe_out(x1, x2)  =  samefringe_out
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
            ↳ AND
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
              ↳ PiDP

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
            ↳ AND
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ RuleRemovalProof
              ↳ PiDP

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
            ↳ AND
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ RuleRemovalProof
                          ↳ QDP
                            ↳ PisEmptyProof
              ↳ PiDP

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.

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

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

SAMEFRINGE_IN(cons(U, V), cons(X, Y)) → U21(U, V, X, Y, gopher_in(cons(U, V), cons(U1, V1)))
U21(U, V, X, Y, gopher_out(cons(U, V), cons(U1, V1))) → U31(U, V, X, Y, U1, V1, gopher_in(cons(X, Y), cons(X1, Y1)))
U31(U, V, X, Y, U1, V1, gopher_out(cons(X, Y), cons(X1, Y1))) → SAMEFRINGE_IN(V1, Y1)

The TRS R consists of the following rules:

samefringe_in(cons(U, V), cons(X, Y)) → U2(U, V, X, Y, gopher_in(cons(U, V), cons(U1, V1)))
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)
U2(U, V, X, Y, gopher_out(cons(U, V), cons(U1, V1))) → U3(U, V, X, Y, U1, V1, gopher_in(cons(X, Y), cons(X1, Y1)))
U3(U, V, X, Y, U1, V1, gopher_out(cons(X, Y), cons(X1, Y1))) → U4(U, V, X, Y, samefringe_in(V1, Y1))
samefringe_in(nil, nil) → samefringe_out(nil, nil)
U4(U, V, X, Y, samefringe_out(V1, Y1)) → samefringe_out(cons(U, V), cons(X, Y))

The argument filtering Pi contains the following mapping:
samefringe_in(x1, x2)  =  samefringe_in(x1, x2)
cons(x1, x2)  =  cons(x1, x2)
U2(x1, x2, x3, x4, x5)  =  U2(x3, x4, x5)
gopher_in(x1, x2)  =  gopher_in(x1)
U1(x1, x2, x3, x4, x5)  =  U1(x5)
nil  =  nil
gopher_out(x1, x2)  =  gopher_out(x2)
U3(x1, x2, x3, x4, x5, x6, x7)  =  U3(x6, x7)
U4(x1, x2, x3, x4, x5)  =  U4(x5)
samefringe_out(x1, x2)  =  samefringe_out
U31(x1, x2, x3, x4, x5, x6, x7)  =  U31(x6, x7)
SAMEFRINGE_IN(x1, x2)  =  SAMEFRINGE_IN(x1, x2)
U21(x1, x2, x3, x4, x5)  =  U21(x3, x4, x5)

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
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof

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

SAMEFRINGE_IN(cons(U, V), cons(X, Y)) → U21(U, V, X, Y, gopher_in(cons(U, V), cons(U1, V1)))
U21(U, V, X, Y, gopher_out(cons(U, V), cons(U1, V1))) → U31(U, V, X, Y, U1, V1, gopher_in(cons(X, Y), cons(X1, Y1)))
U31(U, V, X, Y, U1, V1, gopher_out(cons(X, Y), cons(X1, Y1))) → SAMEFRINGE_IN(V1, Y1)

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))
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:
cons(x1, x2)  =  cons(x1, x2)
gopher_in(x1, x2)  =  gopher_in(x1)
U1(x1, x2, x3, x4, x5)  =  U1(x5)
nil  =  nil
gopher_out(x1, x2)  =  gopher_out(x2)
U31(x1, x2, x3, x4, x5, x6, x7)  =  U31(x6, x7)
SAMEFRINGE_IN(x1, x2)  =  SAMEFRINGE_IN(x1, x2)
U21(x1, x2, x3, x4, x5)  =  U21(x3, x4, x5)

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
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ RuleRemovalProof

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

U31(V1, gopher_out(cons(X1, Y1))) → SAMEFRINGE_IN(V1, Y1)
U21(X, Y, gopher_out(cons(U1, V1))) → U31(V1, gopher_in(cons(X, Y)))
SAMEFRINGE_IN(cons(U, V), cons(X, Y)) → U21(X, Y, gopher_in(cons(U, V)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

gopher_in(x0)
U1(x0)

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:

U31(V1, gopher_out(cons(X1, Y1))) → SAMEFRINGE_IN(V1, Y1)
SAMEFRINGE_IN(cons(U, V), cons(X, Y)) → U21(X, Y, gopher_in(cons(U, V)))


Used ordering: POLO with Polynomial interpretation [25]:

POL(SAMEFRINGE_IN(x1, x2)) = 2 + 2·x1 + 2·x2   
POL(U1(x1)) = x1   
POL(U21(x1, x2, x3)) = 2·x1 + 2·x2 + 2·x3   
POL(U31(x1, x2)) = 2·x1 + 2·x2   
POL(cons(x1, x2)) = 2 + x1 + x2   
POL(gopher_in(x1)) = x1   
POL(gopher_out(x1)) = x1   
POL(nil) = 0   



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

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

U21(X, Y, gopher_out(cons(U1, V1))) → U31(V1, gopher_in(cons(X, Y)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

gopher_in(x0)
U1(x0)

We have to consider all (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.