Left Termination of the query pattern fold_in_3(g, g, a) w.r.t. the given Prolog program could successfully be proven:



↳ Prolog
  ↳ PrologToPiTRSProof

Clauses:

fold(X, .(Y, Ys), Z) :- ','(myop(X, Y, V), fold(V, Ys, Z)).
fold(X, [], X).
myop(a, b, c).

Queries:

fold(g,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:

fold_in(X, [], X) → fold_out(X, [], X)
fold_in(X, .(Y, Ys), Z) → U1(X, Y, Ys, Z, myop_in(X, Y, V))
myop_in(a, b, c) → myop_out(a, b, c)
U1(X, Y, Ys, Z, myop_out(X, Y, V)) → U2(X, Y, Ys, Z, fold_in(V, Ys, Z))
U2(X, Y, Ys, Z, fold_out(V, Ys, Z)) → fold_out(X, .(Y, Ys), Z)

The argument filtering Pi contains the following mapping:
fold_in(x1, x2, x3)  =  fold_in(x1, x2)
[]  =  []
fold_out(x1, x2, x3)  =  fold_out(x3)
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4, x5)  =  U1(x3, x5)
myop_in(x1, x2, x3)  =  myop_in(x1, x2)
a  =  a
b  =  b
c  =  c
myop_out(x1, x2, x3)  =  myop_out(x3)
U2(x1, x2, x3, x4, x5)  =  U2(x5)

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:

fold_in(X, [], X) → fold_out(X, [], X)
fold_in(X, .(Y, Ys), Z) → U1(X, Y, Ys, Z, myop_in(X, Y, V))
myop_in(a, b, c) → myop_out(a, b, c)
U1(X, Y, Ys, Z, myop_out(X, Y, V)) → U2(X, Y, Ys, Z, fold_in(V, Ys, Z))
U2(X, Y, Ys, Z, fold_out(V, Ys, Z)) → fold_out(X, .(Y, Ys), Z)

The argument filtering Pi contains the following mapping:
fold_in(x1, x2, x3)  =  fold_in(x1, x2)
[]  =  []
fold_out(x1, x2, x3)  =  fold_out(x3)
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4, x5)  =  U1(x3, x5)
myop_in(x1, x2, x3)  =  myop_in(x1, x2)
a  =  a
b  =  b
c  =  c
myop_out(x1, x2, x3)  =  myop_out(x3)
U2(x1, x2, x3, x4, x5)  =  U2(x5)


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

FOLD_IN(X, .(Y, Ys), Z) → U11(X, Y, Ys, Z, myop_in(X, Y, V))
FOLD_IN(X, .(Y, Ys), Z) → MYOP_IN(X, Y, V)
U11(X, Y, Ys, Z, myop_out(X, Y, V)) → U21(X, Y, Ys, Z, fold_in(V, Ys, Z))
U11(X, Y, Ys, Z, myop_out(X, Y, V)) → FOLD_IN(V, Ys, Z)

The TRS R consists of the following rules:

fold_in(X, [], X) → fold_out(X, [], X)
fold_in(X, .(Y, Ys), Z) → U1(X, Y, Ys, Z, myop_in(X, Y, V))
myop_in(a, b, c) → myop_out(a, b, c)
U1(X, Y, Ys, Z, myop_out(X, Y, V)) → U2(X, Y, Ys, Z, fold_in(V, Ys, Z))
U2(X, Y, Ys, Z, fold_out(V, Ys, Z)) → fold_out(X, .(Y, Ys), Z)

The argument filtering Pi contains the following mapping:
fold_in(x1, x2, x3)  =  fold_in(x1, x2)
[]  =  []
fold_out(x1, x2, x3)  =  fold_out(x3)
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4, x5)  =  U1(x3, x5)
myop_in(x1, x2, x3)  =  myop_in(x1, x2)
a  =  a
b  =  b
c  =  c
myop_out(x1, x2, x3)  =  myop_out(x3)
U2(x1, x2, x3, x4, x5)  =  U2(x5)
U21(x1, x2, x3, x4, x5)  =  U21(x5)
MYOP_IN(x1, x2, x3)  =  MYOP_IN(x1, x2)
U11(x1, x2, x3, x4, x5)  =  U11(x3, x5)
FOLD_IN(x1, x2, x3)  =  FOLD_IN(x1, x2)

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:

FOLD_IN(X, .(Y, Ys), Z) → U11(X, Y, Ys, Z, myop_in(X, Y, V))
FOLD_IN(X, .(Y, Ys), Z) → MYOP_IN(X, Y, V)
U11(X, Y, Ys, Z, myop_out(X, Y, V)) → U21(X, Y, Ys, Z, fold_in(V, Ys, Z))
U11(X, Y, Ys, Z, myop_out(X, Y, V)) → FOLD_IN(V, Ys, Z)

The TRS R consists of the following rules:

fold_in(X, [], X) → fold_out(X, [], X)
fold_in(X, .(Y, Ys), Z) → U1(X, Y, Ys, Z, myop_in(X, Y, V))
myop_in(a, b, c) → myop_out(a, b, c)
U1(X, Y, Ys, Z, myop_out(X, Y, V)) → U2(X, Y, Ys, Z, fold_in(V, Ys, Z))
U2(X, Y, Ys, Z, fold_out(V, Ys, Z)) → fold_out(X, .(Y, Ys), Z)

The argument filtering Pi contains the following mapping:
fold_in(x1, x2, x3)  =  fold_in(x1, x2)
[]  =  []
fold_out(x1, x2, x3)  =  fold_out(x3)
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4, x5)  =  U1(x3, x5)
myop_in(x1, x2, x3)  =  myop_in(x1, x2)
a  =  a
b  =  b
c  =  c
myop_out(x1, x2, x3)  =  myop_out(x3)
U2(x1, x2, x3, x4, x5)  =  U2(x5)
U21(x1, x2, x3, x4, x5)  =  U21(x5)
MYOP_IN(x1, x2, x3)  =  MYOP_IN(x1, x2)
U11(x1, x2, x3, x4, x5)  =  U11(x3, x5)
FOLD_IN(x1, x2, x3)  =  FOLD_IN(x1, x2)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 1 SCC with 2 less nodes.

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

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

U11(X, Y, Ys, Z, myop_out(X, Y, V)) → FOLD_IN(V, Ys, Z)
FOLD_IN(X, .(Y, Ys), Z) → U11(X, Y, Ys, Z, myop_in(X, Y, V))

The TRS R consists of the following rules:

fold_in(X, [], X) → fold_out(X, [], X)
fold_in(X, .(Y, Ys), Z) → U1(X, Y, Ys, Z, myop_in(X, Y, V))
myop_in(a, b, c) → myop_out(a, b, c)
U1(X, Y, Ys, Z, myop_out(X, Y, V)) → U2(X, Y, Ys, Z, fold_in(V, Ys, Z))
U2(X, Y, Ys, Z, fold_out(V, Ys, Z)) → fold_out(X, .(Y, Ys), Z)

The argument filtering Pi contains the following mapping:
fold_in(x1, x2, x3)  =  fold_in(x1, x2)
[]  =  []
fold_out(x1, x2, x3)  =  fold_out(x3)
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4, x5)  =  U1(x3, x5)
myop_in(x1, x2, x3)  =  myop_in(x1, x2)
a  =  a
b  =  b
c  =  c
myop_out(x1, x2, x3)  =  myop_out(x3)
U2(x1, x2, x3, x4, x5)  =  U2(x5)
U11(x1, x2, x3, x4, x5)  =  U11(x3, x5)
FOLD_IN(x1, x2, x3)  =  FOLD_IN(x1, x2)

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:

U11(X, Y, Ys, Z, myop_out(X, Y, V)) → FOLD_IN(V, Ys, Z)
FOLD_IN(X, .(Y, Ys), Z) → U11(X, Y, Ys, Z, myop_in(X, Y, V))

The TRS R consists of the following rules:

myop_in(a, b, c) → myop_out(a, b, c)

The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x1, x2)
myop_in(x1, x2, x3)  =  myop_in(x1, x2)
a  =  a
b  =  b
c  =  c
myop_out(x1, x2, x3)  =  myop_out(x3)
U11(x1, x2, x3, x4, x5)  =  U11(x3, x5)
FOLD_IN(x1, x2, x3)  =  FOLD_IN(x1, x2)

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
                      ↳ QDPSizeChangeProof

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

U11(Ys, myop_out(V)) → FOLD_IN(V, Ys)
FOLD_IN(X, .(Y, Ys)) → U11(Ys, myop_in(X, Y))

The TRS R consists of the following rules:

myop_in(a, b) → myop_out(c)

The set Q consists of the following terms:

myop_in(x0, x1)

We have to consider all (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs: