Left Termination of the query pattern fl_in_3(a, a, g) w.r.t. the given Prolog program could not be shown:



Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof

Clauses:

fl([], [], 0).
fl(.(E, X), R, s(Z)) :- ','(append(E, Y, R), fl(X, Y, Z)).
append([], X, X).
append(.(X, Xs), Ys, .(X, Zs)) :- append(Xs, Ys, Zs).

Queries:

fl(a,a,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:

fl_in(.(E, X), R, s(Z)) → U1(E, X, R, Z, append_in(E, Y, R))
append_in(.(X, Xs), Ys, .(X, Zs)) → U3(X, Xs, Ys, Zs, append_in(Xs, Ys, Zs))
append_in([], X, X) → append_out([], X, X)
U3(X, Xs, Ys, Zs, append_out(Xs, Ys, Zs)) → append_out(.(X, Xs), Ys, .(X, Zs))
U1(E, X, R, Z, append_out(E, Y, R)) → U2(E, X, R, Z, fl_in(X, Y, Z))
fl_in([], [], 0) → fl_out([], [], 0)
U2(E, X, R, Z, fl_out(X, Y, Z)) → fl_out(.(E, X), R, s(Z))

The argument filtering Pi contains the following mapping:
fl_in(x1, x2, x3)  =  fl_in(x3)
.(x1, x2)  =  .(x1, x2)
s(x1)  =  s(x1)
U1(x1, x2, x3, x4, x5)  =  U1(x4, x5)
append_in(x1, x2, x3)  =  append_in
U3(x1, x2, x3, x4, x5)  =  U3(x5)
[]  =  []
append_out(x1, x2, x3)  =  append_out
U2(x1, x2, x3, x4, x5)  =  U2(x5)
0  =  0
fl_out(x1, x2, x3)  =  fl_out

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog



↳ Prolog
  ↳ PrologToPiTRSProof
PiTRS
      ↳ DependencyPairsProof
  ↳ PrologToPiTRSProof

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

fl_in(.(E, X), R, s(Z)) → U1(E, X, R, Z, append_in(E, Y, R))
append_in(.(X, Xs), Ys, .(X, Zs)) → U3(X, Xs, Ys, Zs, append_in(Xs, Ys, Zs))
append_in([], X, X) → append_out([], X, X)
U3(X, Xs, Ys, Zs, append_out(Xs, Ys, Zs)) → append_out(.(X, Xs), Ys, .(X, Zs))
U1(E, X, R, Z, append_out(E, Y, R)) → U2(E, X, R, Z, fl_in(X, Y, Z))
fl_in([], [], 0) → fl_out([], [], 0)
U2(E, X, R, Z, fl_out(X, Y, Z)) → fl_out(.(E, X), R, s(Z))

The argument filtering Pi contains the following mapping:
fl_in(x1, x2, x3)  =  fl_in(x3)
.(x1, x2)  =  .(x1, x2)
s(x1)  =  s(x1)
U1(x1, x2, x3, x4, x5)  =  U1(x4, x5)
append_in(x1, x2, x3)  =  append_in
U3(x1, x2, x3, x4, x5)  =  U3(x5)
[]  =  []
append_out(x1, x2, x3)  =  append_out
U2(x1, x2, x3, x4, x5)  =  U2(x5)
0  =  0
fl_out(x1, x2, x3)  =  fl_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:

FL_IN(.(E, X), R, s(Z)) → U11(E, X, R, Z, append_in(E, Y, R))
FL_IN(.(E, X), R, s(Z)) → APPEND_IN(E, Y, R)
APPEND_IN(.(X, Xs), Ys, .(X, Zs)) → U31(X, Xs, Ys, Zs, append_in(Xs, Ys, Zs))
APPEND_IN(.(X, Xs), Ys, .(X, Zs)) → APPEND_IN(Xs, Ys, Zs)
U11(E, X, R, Z, append_out(E, Y, R)) → U21(E, X, R, Z, fl_in(X, Y, Z))
U11(E, X, R, Z, append_out(E, Y, R)) → FL_IN(X, Y, Z)

The TRS R consists of the following rules:

fl_in(.(E, X), R, s(Z)) → U1(E, X, R, Z, append_in(E, Y, R))
append_in(.(X, Xs), Ys, .(X, Zs)) → U3(X, Xs, Ys, Zs, append_in(Xs, Ys, Zs))
append_in([], X, X) → append_out([], X, X)
U3(X, Xs, Ys, Zs, append_out(Xs, Ys, Zs)) → append_out(.(X, Xs), Ys, .(X, Zs))
U1(E, X, R, Z, append_out(E, Y, R)) → U2(E, X, R, Z, fl_in(X, Y, Z))
fl_in([], [], 0) → fl_out([], [], 0)
U2(E, X, R, Z, fl_out(X, Y, Z)) → fl_out(.(E, X), R, s(Z))

The argument filtering Pi contains the following mapping:
fl_in(x1, x2, x3)  =  fl_in(x3)
.(x1, x2)  =  .(x1, x2)
s(x1)  =  s(x1)
U1(x1, x2, x3, x4, x5)  =  U1(x4, x5)
append_in(x1, x2, x3)  =  append_in
U3(x1, x2, x3, x4, x5)  =  U3(x5)
[]  =  []
append_out(x1, x2, x3)  =  append_out
U2(x1, x2, x3, x4, x5)  =  U2(x5)
0  =  0
fl_out(x1, x2, x3)  =  fl_out
U31(x1, x2, x3, x4, x5)  =  U31(x5)
APPEND_IN(x1, x2, x3)  =  APPEND_IN
FL_IN(x1, x2, x3)  =  FL_IN(x3)
U21(x1, x2, x3, x4, x5)  =  U21(x5)
U11(x1, x2, x3, x4, x5)  =  U11(x4, x5)

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

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

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

FL_IN(.(E, X), R, s(Z)) → U11(E, X, R, Z, append_in(E, Y, R))
FL_IN(.(E, X), R, s(Z)) → APPEND_IN(E, Y, R)
APPEND_IN(.(X, Xs), Ys, .(X, Zs)) → U31(X, Xs, Ys, Zs, append_in(Xs, Ys, Zs))
APPEND_IN(.(X, Xs), Ys, .(X, Zs)) → APPEND_IN(Xs, Ys, Zs)
U11(E, X, R, Z, append_out(E, Y, R)) → U21(E, X, R, Z, fl_in(X, Y, Z))
U11(E, X, R, Z, append_out(E, Y, R)) → FL_IN(X, Y, Z)

The TRS R consists of the following rules:

fl_in(.(E, X), R, s(Z)) → U1(E, X, R, Z, append_in(E, Y, R))
append_in(.(X, Xs), Ys, .(X, Zs)) → U3(X, Xs, Ys, Zs, append_in(Xs, Ys, Zs))
append_in([], X, X) → append_out([], X, X)
U3(X, Xs, Ys, Zs, append_out(Xs, Ys, Zs)) → append_out(.(X, Xs), Ys, .(X, Zs))
U1(E, X, R, Z, append_out(E, Y, R)) → U2(E, X, R, Z, fl_in(X, Y, Z))
fl_in([], [], 0) → fl_out([], [], 0)
U2(E, X, R, Z, fl_out(X, Y, Z)) → fl_out(.(E, X), R, s(Z))

The argument filtering Pi contains the following mapping:
fl_in(x1, x2, x3)  =  fl_in(x3)
.(x1, x2)  =  .(x1, x2)
s(x1)  =  s(x1)
U1(x1, x2, x3, x4, x5)  =  U1(x4, x5)
append_in(x1, x2, x3)  =  append_in
U3(x1, x2, x3, x4, x5)  =  U3(x5)
[]  =  []
append_out(x1, x2, x3)  =  append_out
U2(x1, x2, x3, x4, x5)  =  U2(x5)
0  =  0
fl_out(x1, x2, x3)  =  fl_out
U31(x1, x2, x3, x4, x5)  =  U31(x5)
APPEND_IN(x1, x2, x3)  =  APPEND_IN
FL_IN(x1, x2, x3)  =  FL_IN(x3)
U21(x1, x2, x3, x4, x5)  =  U21(x5)
U11(x1, x2, x3, x4, x5)  =  U11(x4, x5)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 2 SCCs with 3 less nodes.

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

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

APPEND_IN(.(X, Xs), Ys, .(X, Zs)) → APPEND_IN(Xs, Ys, Zs)

The TRS R consists of the following rules:

fl_in(.(E, X), R, s(Z)) → U1(E, X, R, Z, append_in(E, Y, R))
append_in(.(X, Xs), Ys, .(X, Zs)) → U3(X, Xs, Ys, Zs, append_in(Xs, Ys, Zs))
append_in([], X, X) → append_out([], X, X)
U3(X, Xs, Ys, Zs, append_out(Xs, Ys, Zs)) → append_out(.(X, Xs), Ys, .(X, Zs))
U1(E, X, R, Z, append_out(E, Y, R)) → U2(E, X, R, Z, fl_in(X, Y, Z))
fl_in([], [], 0) → fl_out([], [], 0)
U2(E, X, R, Z, fl_out(X, Y, Z)) → fl_out(.(E, X), R, s(Z))

The argument filtering Pi contains the following mapping:
fl_in(x1, x2, x3)  =  fl_in(x3)
.(x1, x2)  =  .(x1, x2)
s(x1)  =  s(x1)
U1(x1, x2, x3, x4, x5)  =  U1(x4, x5)
append_in(x1, x2, x3)  =  append_in
U3(x1, x2, x3, x4, x5)  =  U3(x5)
[]  =  []
append_out(x1, x2, x3)  =  append_out
U2(x1, x2, x3, x4, x5)  =  U2(x5)
0  =  0
fl_out(x1, x2, x3)  =  fl_out
APPEND_IN(x1, x2, x3)  =  APPEND_IN

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
  ↳ PrologToPiTRSProof

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

APPEND_IN(.(X, Xs), Ys, .(X, Zs)) → APPEND_IN(Xs, Ys, Zs)

R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x1, x2)
APPEND_IN(x1, x2, x3)  =  APPEND_IN

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
                        ↳ NonTerminationProof
              ↳ PiDP
  ↳ PrologToPiTRSProof

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

APPEND_INAPPEND_IN

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.

The TRS P consists of the following rules:

APPEND_INAPPEND_IN

The TRS R consists of the following rules:none


s = APPEND_IN evaluates to t =APPEND_IN

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:




Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from APPEND_IN to APPEND_IN.





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

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

FL_IN(.(E, X), R, s(Z)) → U11(E, X, R, Z, append_in(E, Y, R))
U11(E, X, R, Z, append_out(E, Y, R)) → FL_IN(X, Y, Z)

The TRS R consists of the following rules:

fl_in(.(E, X), R, s(Z)) → U1(E, X, R, Z, append_in(E, Y, R))
append_in(.(X, Xs), Ys, .(X, Zs)) → U3(X, Xs, Ys, Zs, append_in(Xs, Ys, Zs))
append_in([], X, X) → append_out([], X, X)
U3(X, Xs, Ys, Zs, append_out(Xs, Ys, Zs)) → append_out(.(X, Xs), Ys, .(X, Zs))
U1(E, X, R, Z, append_out(E, Y, R)) → U2(E, X, R, Z, fl_in(X, Y, Z))
fl_in([], [], 0) → fl_out([], [], 0)
U2(E, X, R, Z, fl_out(X, Y, Z)) → fl_out(.(E, X), R, s(Z))

The argument filtering Pi contains the following mapping:
fl_in(x1, x2, x3)  =  fl_in(x3)
.(x1, x2)  =  .(x1, x2)
s(x1)  =  s(x1)
U1(x1, x2, x3, x4, x5)  =  U1(x4, x5)
append_in(x1, x2, x3)  =  append_in
U3(x1, x2, x3, x4, x5)  =  U3(x5)
[]  =  []
append_out(x1, x2, x3)  =  append_out
U2(x1, x2, x3, x4, x5)  =  U2(x5)
0  =  0
fl_out(x1, x2, x3)  =  fl_out
FL_IN(x1, x2, x3)  =  FL_IN(x3)
U11(x1, x2, x3, x4, x5)  =  U11(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
  ↳ PrologToPiTRSProof

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

FL_IN(.(E, X), R, s(Z)) → U11(E, X, R, Z, append_in(E, Y, R))
U11(E, X, R, Z, append_out(E, Y, R)) → FL_IN(X, Y, Z)

The TRS R consists of the following rules:

append_in(.(X, Xs), Ys, .(X, Zs)) → U3(X, Xs, Ys, Zs, append_in(Xs, Ys, Zs))
append_in([], X, X) → append_out([], X, X)
U3(X, Xs, Ys, Zs, append_out(Xs, Ys, Zs)) → append_out(.(X, Xs), Ys, .(X, Zs))

The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x1, x2)
s(x1)  =  s(x1)
append_in(x1, x2, x3)  =  append_in
U3(x1, x2, x3, x4, x5)  =  U3(x5)
[]  =  []
append_out(x1, x2, x3)  =  append_out
FL_IN(x1, x2, x3)  =  FL_IN(x3)
U11(x1, x2, x3, x4, x5)  =  U11(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
                        ↳ QDPSizeChangeProof
  ↳ PrologToPiTRSProof

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

U11(Z, append_out) → FL_IN(Z)
FL_IN(s(Z)) → U11(Z, append_in)

The TRS R consists of the following rules:

append_inU3(append_in)
append_inappend_out
U3(append_out) → append_out

The set Q consists of the following terms:

append_in
U3(x0)

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:


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:

fl_in(.(E, X), R, s(Z)) → U1(E, X, R, Z, append_in(E, Y, R))
append_in(.(X, Xs), Ys, .(X, Zs)) → U3(X, Xs, Ys, Zs, append_in(Xs, Ys, Zs))
append_in([], X, X) → append_out([], X, X)
U3(X, Xs, Ys, Zs, append_out(Xs, Ys, Zs)) → append_out(.(X, Xs), Ys, .(X, Zs))
U1(E, X, R, Z, append_out(E, Y, R)) → U2(E, X, R, Z, fl_in(X, Y, Z))
fl_in([], [], 0) → fl_out([], [], 0)
U2(E, X, R, Z, fl_out(X, Y, Z)) → fl_out(.(E, X), R, s(Z))

The argument filtering Pi contains the following mapping:
fl_in(x1, x2, x3)  =  fl_in(x3)
.(x1, x2)  =  .(x1, x2)
s(x1)  =  s(x1)
U1(x1, x2, x3, x4, x5)  =  U1(x4, x5)
append_in(x1, x2, x3)  =  append_in
U3(x1, x2, x3, x4, x5)  =  U3(x5)
[]  =  []
append_out(x1, x2, x3)  =  append_out
U2(x1, x2, x3, x4, x5)  =  U2(x4, x5)
0  =  0
fl_out(x1, x2, x3)  =  fl_out(x3)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog



↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
PiTRS
      ↳ DependencyPairsProof

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

fl_in(.(E, X), R, s(Z)) → U1(E, X, R, Z, append_in(E, Y, R))
append_in(.(X, Xs), Ys, .(X, Zs)) → U3(X, Xs, Ys, Zs, append_in(Xs, Ys, Zs))
append_in([], X, X) → append_out([], X, X)
U3(X, Xs, Ys, Zs, append_out(Xs, Ys, Zs)) → append_out(.(X, Xs), Ys, .(X, Zs))
U1(E, X, R, Z, append_out(E, Y, R)) → U2(E, X, R, Z, fl_in(X, Y, Z))
fl_in([], [], 0) → fl_out([], [], 0)
U2(E, X, R, Z, fl_out(X, Y, Z)) → fl_out(.(E, X), R, s(Z))

The argument filtering Pi contains the following mapping:
fl_in(x1, x2, x3)  =  fl_in(x3)
.(x1, x2)  =  .(x1, x2)
s(x1)  =  s(x1)
U1(x1, x2, x3, x4, x5)  =  U1(x4, x5)
append_in(x1, x2, x3)  =  append_in
U3(x1, x2, x3, x4, x5)  =  U3(x5)
[]  =  []
append_out(x1, x2, x3)  =  append_out
U2(x1, x2, x3, x4, x5)  =  U2(x4, x5)
0  =  0
fl_out(x1, x2, x3)  =  fl_out(x3)


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

FL_IN(.(E, X), R, s(Z)) → U11(E, X, R, Z, append_in(E, Y, R))
FL_IN(.(E, X), R, s(Z)) → APPEND_IN(E, Y, R)
APPEND_IN(.(X, Xs), Ys, .(X, Zs)) → U31(X, Xs, Ys, Zs, append_in(Xs, Ys, Zs))
APPEND_IN(.(X, Xs), Ys, .(X, Zs)) → APPEND_IN(Xs, Ys, Zs)
U11(E, X, R, Z, append_out(E, Y, R)) → U21(E, X, R, Z, fl_in(X, Y, Z))
U11(E, X, R, Z, append_out(E, Y, R)) → FL_IN(X, Y, Z)

The TRS R consists of the following rules:

fl_in(.(E, X), R, s(Z)) → U1(E, X, R, Z, append_in(E, Y, R))
append_in(.(X, Xs), Ys, .(X, Zs)) → U3(X, Xs, Ys, Zs, append_in(Xs, Ys, Zs))
append_in([], X, X) → append_out([], X, X)
U3(X, Xs, Ys, Zs, append_out(Xs, Ys, Zs)) → append_out(.(X, Xs), Ys, .(X, Zs))
U1(E, X, R, Z, append_out(E, Y, R)) → U2(E, X, R, Z, fl_in(X, Y, Z))
fl_in([], [], 0) → fl_out([], [], 0)
U2(E, X, R, Z, fl_out(X, Y, Z)) → fl_out(.(E, X), R, s(Z))

The argument filtering Pi contains the following mapping:
fl_in(x1, x2, x3)  =  fl_in(x3)
.(x1, x2)  =  .(x1, x2)
s(x1)  =  s(x1)
U1(x1, x2, x3, x4, x5)  =  U1(x4, x5)
append_in(x1, x2, x3)  =  append_in
U3(x1, x2, x3, x4, x5)  =  U3(x5)
[]  =  []
append_out(x1, x2, x3)  =  append_out
U2(x1, x2, x3, x4, x5)  =  U2(x4, x5)
0  =  0
fl_out(x1, x2, x3)  =  fl_out(x3)
U31(x1, x2, x3, x4, x5)  =  U31(x5)
APPEND_IN(x1, x2, x3)  =  APPEND_IN
FL_IN(x1, x2, x3)  =  FL_IN(x3)
U21(x1, x2, x3, x4, x5)  =  U21(x4, x5)
U11(x1, x2, x3, x4, x5)  =  U11(x4, x5)

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

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

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

FL_IN(.(E, X), R, s(Z)) → U11(E, X, R, Z, append_in(E, Y, R))
FL_IN(.(E, X), R, s(Z)) → APPEND_IN(E, Y, R)
APPEND_IN(.(X, Xs), Ys, .(X, Zs)) → U31(X, Xs, Ys, Zs, append_in(Xs, Ys, Zs))
APPEND_IN(.(X, Xs), Ys, .(X, Zs)) → APPEND_IN(Xs, Ys, Zs)
U11(E, X, R, Z, append_out(E, Y, R)) → U21(E, X, R, Z, fl_in(X, Y, Z))
U11(E, X, R, Z, append_out(E, Y, R)) → FL_IN(X, Y, Z)

The TRS R consists of the following rules:

fl_in(.(E, X), R, s(Z)) → U1(E, X, R, Z, append_in(E, Y, R))
append_in(.(X, Xs), Ys, .(X, Zs)) → U3(X, Xs, Ys, Zs, append_in(Xs, Ys, Zs))
append_in([], X, X) → append_out([], X, X)
U3(X, Xs, Ys, Zs, append_out(Xs, Ys, Zs)) → append_out(.(X, Xs), Ys, .(X, Zs))
U1(E, X, R, Z, append_out(E, Y, R)) → U2(E, X, R, Z, fl_in(X, Y, Z))
fl_in([], [], 0) → fl_out([], [], 0)
U2(E, X, R, Z, fl_out(X, Y, Z)) → fl_out(.(E, X), R, s(Z))

The argument filtering Pi contains the following mapping:
fl_in(x1, x2, x3)  =  fl_in(x3)
.(x1, x2)  =  .(x1, x2)
s(x1)  =  s(x1)
U1(x1, x2, x3, x4, x5)  =  U1(x4, x5)
append_in(x1, x2, x3)  =  append_in
U3(x1, x2, x3, x4, x5)  =  U3(x5)
[]  =  []
append_out(x1, x2, x3)  =  append_out
U2(x1, x2, x3, x4, x5)  =  U2(x4, x5)
0  =  0
fl_out(x1, x2, x3)  =  fl_out(x3)
U31(x1, x2, x3, x4, x5)  =  U31(x5)
APPEND_IN(x1, x2, x3)  =  APPEND_IN
FL_IN(x1, x2, x3)  =  FL_IN(x3)
U21(x1, x2, x3, x4, x5)  =  U21(x4, x5)
U11(x1, x2, x3, x4, x5)  =  U11(x4, x5)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 2 SCCs with 3 less nodes.

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

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

APPEND_IN(.(X, Xs), Ys, .(X, Zs)) → APPEND_IN(Xs, Ys, Zs)

The TRS R consists of the following rules:

fl_in(.(E, X), R, s(Z)) → U1(E, X, R, Z, append_in(E, Y, R))
append_in(.(X, Xs), Ys, .(X, Zs)) → U3(X, Xs, Ys, Zs, append_in(Xs, Ys, Zs))
append_in([], X, X) → append_out([], X, X)
U3(X, Xs, Ys, Zs, append_out(Xs, Ys, Zs)) → append_out(.(X, Xs), Ys, .(X, Zs))
U1(E, X, R, Z, append_out(E, Y, R)) → U2(E, X, R, Z, fl_in(X, Y, Z))
fl_in([], [], 0) → fl_out([], [], 0)
U2(E, X, R, Z, fl_out(X, Y, Z)) → fl_out(.(E, X), R, s(Z))

The argument filtering Pi contains the following mapping:
fl_in(x1, x2, x3)  =  fl_in(x3)
.(x1, x2)  =  .(x1, x2)
s(x1)  =  s(x1)
U1(x1, x2, x3, x4, x5)  =  U1(x4, x5)
append_in(x1, x2, x3)  =  append_in
U3(x1, x2, x3, x4, x5)  =  U3(x5)
[]  =  []
append_out(x1, x2, x3)  =  append_out
U2(x1, x2, x3, x4, x5)  =  U2(x4, x5)
0  =  0
fl_out(x1, x2, x3)  =  fl_out(x3)
APPEND_IN(x1, x2, x3)  =  APPEND_IN

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

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

APPEND_IN(.(X, Xs), Ys, .(X, Zs)) → APPEND_IN(Xs, Ys, Zs)

R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x1, x2)
APPEND_IN(x1, x2, x3)  =  APPEND_IN

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
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
QDP
                        ↳ NonTerminationProof
              ↳ PiDP

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

APPEND_INAPPEND_IN

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.

The TRS P consists of the following rules:

APPEND_INAPPEND_IN

The TRS R consists of the following rules:none


s = APPEND_IN evaluates to t =APPEND_IN

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:




Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from APPEND_IN to APPEND_IN.





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

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

FL_IN(.(E, X), R, s(Z)) → U11(E, X, R, Z, append_in(E, Y, R))
U11(E, X, R, Z, append_out(E, Y, R)) → FL_IN(X, Y, Z)

The TRS R consists of the following rules:

fl_in(.(E, X), R, s(Z)) → U1(E, X, R, Z, append_in(E, Y, R))
append_in(.(X, Xs), Ys, .(X, Zs)) → U3(X, Xs, Ys, Zs, append_in(Xs, Ys, Zs))
append_in([], X, X) → append_out([], X, X)
U3(X, Xs, Ys, Zs, append_out(Xs, Ys, Zs)) → append_out(.(X, Xs), Ys, .(X, Zs))
U1(E, X, R, Z, append_out(E, Y, R)) → U2(E, X, R, Z, fl_in(X, Y, Z))
fl_in([], [], 0) → fl_out([], [], 0)
U2(E, X, R, Z, fl_out(X, Y, Z)) → fl_out(.(E, X), R, s(Z))

The argument filtering Pi contains the following mapping:
fl_in(x1, x2, x3)  =  fl_in(x3)
.(x1, x2)  =  .(x1, x2)
s(x1)  =  s(x1)
U1(x1, x2, x3, x4, x5)  =  U1(x4, x5)
append_in(x1, x2, x3)  =  append_in
U3(x1, x2, x3, x4, x5)  =  U3(x5)
[]  =  []
append_out(x1, x2, x3)  =  append_out
U2(x1, x2, x3, x4, x5)  =  U2(x4, x5)
0  =  0
fl_out(x1, x2, x3)  =  fl_out(x3)
FL_IN(x1, x2, x3)  =  FL_IN(x3)
U11(x1, x2, x3, x4, x5)  =  U11(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
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
PiDP
                    ↳ PiDPToQDPProof

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

FL_IN(.(E, X), R, s(Z)) → U11(E, X, R, Z, append_in(E, Y, R))
U11(E, X, R, Z, append_out(E, Y, R)) → FL_IN(X, Y, Z)

The TRS R consists of the following rules:

append_in(.(X, Xs), Ys, .(X, Zs)) → U3(X, Xs, Ys, Zs, append_in(Xs, Ys, Zs))
append_in([], X, X) → append_out([], X, X)
U3(X, Xs, Ys, Zs, append_out(Xs, Ys, Zs)) → append_out(.(X, Xs), Ys, .(X, Zs))

The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x1, x2)
s(x1)  =  s(x1)
append_in(x1, x2, x3)  =  append_in
U3(x1, x2, x3, x4, x5)  =  U3(x5)
[]  =  []
append_out(x1, x2, x3)  =  append_out
FL_IN(x1, x2, x3)  =  FL_IN(x3)
U11(x1, x2, x3, x4, x5)  =  U11(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
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
QDP
                        ↳ QDPSizeChangeProof

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

U11(Z, append_out) → FL_IN(Z)
FL_IN(s(Z)) → U11(Z, append_in)

The TRS R consists of the following rules:

append_inU3(append_in)
append_inappend_out
U3(append_out) → append_out

The set Q consists of the following terms:

append_in
U3(x0)

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: