Left Termination proof of /tmp/tmpU6KKEu/pl4.0.1-oooi.pl

Left Termination of the query pattern append3_in_4(a, a, a, g) w.r.t. the given Prolog program could not be shown:

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

Clauses:

append([], L, L).
append(.(H, L1), L2, .(H, L3)) :- append(L1, L2, L3).
append3(A, B, C, D) :- ','(append(A, B, E), append(E, C, D)).

Queries:

append3(a,a,a,g).

We use the technique of [30]. With regard to the inferred argument filtering the predicates were used in the following modes:
append3_in: (f,f,f,b)
append_in: (f,f,f) (f,f,b)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

append3_in_aaag(A, B, C, D) → U2_aaag(A, B, C, D, append_in_aaa(A, B, E))
append_in_aaa([], L, L) → append_out_aaa([], L, L)
append_in_aaa(.(H, L1), L2, .(H, L3)) → U1_aaa(H, L1, L2, L3, append_in_aaa(L1, L2, L3))
U1_aaa(H, L1, L2, L3, append_out_aaa(L1, L2, L3)) → append_out_aaa(.(H, L1), L2, .(H, L3))
U2_aaag(A, B, C, D, append_out_aaa(A, B, E)) → U3_aaag(A, B, C, D, append_in_aag(E, C, D))
append_in_aag([], L, L) → append_out_aag([], L, L)
append_in_aag(.(H, L1), L2, .(H, L3)) → U1_aag(H, L1, L2, L3, append_in_aag(L1, L2, L3))
U1_aag(H, L1, L2, L3, append_out_aag(L1, L2, L3)) → append_out_aag(.(H, L1), L2, .(H, L3))
U3_aaag(A, B, C, D, append_out_aag(E, C, D)) → append3_out_aaag(A, B, C, D)

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:

append3_in_aaag(A, B, C, D) → U2_aaag(A, B, C, D, append_in_aaa(A, B, E))
append_in_aaa([], L, L) → append_out_aaa([], L, L)
append_in_aaa(.(H, L1), L2, .(H, L3)) → U1_aaa(H, L1, L2, L3, append_in_aaa(L1, L2, L3))
U1_aaa(H, L1, L2, L3, append_out_aaa(L1, L2, L3)) → append_out_aaa(.(H, L1), L2, .(H, L3))
U2_aaag(A, B, C, D, append_out_aaa(A, B, E)) → U3_aaag(A, B, C, D, append_in_aag(E, C, D))
append_in_aag([], L, L) → append_out_aag([], L, L)
append_in_aag(.(H, L1), L2, .(H, L3)) → U1_aag(H, L1, L2, L3, append_in_aag(L1, L2, L3))
U1_aag(H, L1, L2, L3, append_out_aag(L1, L2, L3)) → append_out_aag(.(H, L1), L2, .(H, L3))
U3_aaag(A, B, C, D, append_out_aag(E, C, D)) → append3_out_aaag(A, B, C, D)

APPEND3_IN_AAAG(A, B, C, D) → U2_AAAG(A, B, C, D, append_in_aaa(A, B, E))
APPEND3_IN_AAAG(A, B, C, D) → APPEND_IN_AAA(A, B, E)
APPEND_IN_AAA(.(H, L1), L2, .(H, L3)) → U1_AAA(H, L1, L2, L3, append_in_aaa(L1, L2, L3))
APPEND_IN_AAA(.(H, L1), L2, .(H, L3)) → APPEND_IN_AAA(L1, L2, L3)
U2_AAAG(A, B, C, D, append_out_aaa(A, B, E)) → U3_AAAG(A, B, C, D, append_in_aag(E, C, D))
U2_AAAG(A, B, C, D, append_out_aaa(A, B, E)) → APPEND_IN_AAG(E, C, D)
APPEND_IN_AAG(.(H, L1), L2, .(H, L3)) → U1_AAG(H, L1, L2, L3, append_in_aag(L1, L2, L3))
APPEND_IN_AAG(.(H, L1), L2, .(H, L3)) → APPEND_IN_AAG(L1, L2, L3)

The TRS R consists of the following rules:

append3_in_aaag(A, B, C, D) → U2_aaag(A, B, C, D, append_in_aaa(A, B, E))
append_in_aaa([], L, L) → append_out_aaa([], L, L)
append_in_aaa(.(H, L1), L2, .(H, L3)) → U1_aaa(H, L1, L2, L3, append_in_aaa(L1, L2, L3))
U1_aaa(H, L1, L2, L3, append_out_aaa(L1, L2, L3)) → append_out_aaa(.(H, L1), L2, .(H, L3))
U2_aaag(A, B, C, D, append_out_aaa(A, B, E)) → U3_aaag(A, B, C, D, append_in_aag(E, C, D))
append_in_aag([], L, L) → append_out_aag([], L, L)
append_in_aag(.(H, L1), L2, .(H, L3)) → U1_aag(H, L1, L2, L3, append_in_aag(L1, L2, L3))
U1_aag(H, L1, L2, L3, append_out_aag(L1, L2, L3)) → append_out_aag(.(H, L1), L2, .(H, L3))
U3_aaag(A, B, C, D, append_out_aag(E, C, D)) → append3_out_aaag(A, B, C, D)

The argument filtering Pi contains the following mapping:
append3_in_aaag(x1, x2, x3, x4) = append3_in_aaag(x4)
U2_aaag(x1, x2, x3, x4, x5) = U2_aaag(x4, x5)
append_in_aaa(x1, x2, x3) = append_in_aaa
append_out_aaa(x1, x2, x3) = append_out_aaa(x1)
U1_aaa(x1, x2, x3, x4, x5) = U1_aaa(x5)
.(x1, x2) = .(x2)
U3_aaag(x1, x2, x3, x4, x5) = U3_aaag(x1, x5)
append_in_aag(x1, x2, x3) = append_in_aag(x3)
append_out_aag(x1, x2, x3) = append_out_aag(x1, x2)
U1_aag(x1, x2, x3, x4, x5) = U1_aag(x5)
append3_out_aaag(x1, x2, x3, x4) = append3_out_aaag(x1, x3)
APPEND_IN_AAG(x1, x2, x3) = APPEND_IN_AAG(x3)
U2_AAAG(x1, x2, x3, x4, x5) = U2_AAAG(x4, x5)
U1_AAG(x1, x2, x3, x4, x5) = U1_AAG(x5)
U3_AAAG(x1, x2, x3, x4, x5) = U3_AAAG(x1, x5)
U1_AAA(x1, x2, x3, x4, x5) = U1_AAA(x5)
APPEND_IN_AAA(x1, x2, x3) = APPEND_IN_AAA
APPEND3_IN_AAAG(x1, x2, x3, x4) = APPEND3_IN_AAAG(x4)

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:

APPEND3_IN_AAAG(A, B, C, D) → U2_AAAG(A, B, C, D, append_in_aaa(A, B, E))
APPEND3_IN_AAAG(A, B, C, D) → APPEND_IN_AAA(A, B, E)
APPEND_IN_AAA(.(H, L1), L2, .(H, L3)) → U1_AAA(H, L1, L2, L3, append_in_aaa(L1, L2, L3))
APPEND_IN_AAA(.(H, L1), L2, .(H, L3)) → APPEND_IN_AAA(L1, L2, L3)
U2_AAAG(A, B, C, D, append_out_aaa(A, B, E)) → U3_AAAG(A, B, C, D, append_in_aag(E, C, D))
U2_AAAG(A, B, C, D, append_out_aaa(A, B, E)) → APPEND_IN_AAG(E, C, D)
APPEND_IN_AAG(.(H, L1), L2, .(H, L3)) → U1_AAG(H, L1, L2, L3, append_in_aag(L1, L2, L3))
APPEND_IN_AAG(.(H, L1), L2, .(H, L3)) → APPEND_IN_AAG(L1, L2, L3)

The TRS R consists of the following rules:

append3_in_aaag(A, B, C, D) → U2_aaag(A, B, C, D, append_in_aaa(A, B, E))
append_in_aaa([], L, L) → append_out_aaa([], L, L)
append_in_aaa(.(H, L1), L2, .(H, L3)) → U1_aaa(H, L1, L2, L3, append_in_aaa(L1, L2, L3))
U1_aaa(H, L1, L2, L3, append_out_aaa(L1, L2, L3)) → append_out_aaa(.(H, L1), L2, .(H, L3))
U2_aaag(A, B, C, D, append_out_aaa(A, B, E)) → U3_aaag(A, B, C, D, append_in_aag(E, C, D))
append_in_aag([], L, L) → append_out_aag([], L, L)
append_in_aag(.(H, L1), L2, .(H, L3)) → U1_aag(H, L1, L2, L3, append_in_aag(L1, L2, L3))
U1_aag(H, L1, L2, L3, append_out_aag(L1, L2, L3)) → append_out_aag(.(H, L1), L2, .(H, L3))
U3_aaag(A, B, C, D, append_out_aag(E, C, D)) → append3_out_aaag(A, B, C, D)

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

              ↳ PiDP

  ↳ PrologToPiTRSProof

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

APPEND_IN_AAG(.(H, L1), L2, .(H, L3)) → APPEND_IN_AAG(L1, L2, L3)

The TRS R consists of the following rules:

append3_in_aaag(A, B, C, D) → U2_aaag(A, B, C, D, append_in_aaa(A, B, E))
append_in_aaa([], L, L) → append_out_aaa([], L, L)
append_in_aaa(.(H, L1), L2, .(H, L3)) → U1_aaa(H, L1, L2, L3, append_in_aaa(L1, L2, L3))
U1_aaa(H, L1, L2, L3, append_out_aaa(L1, L2, L3)) → append_out_aaa(.(H, L1), L2, .(H, L3))
U2_aaag(A, B, C, D, append_out_aaa(A, B, E)) → U3_aaag(A, B, C, D, append_in_aag(E, C, D))
append_in_aag([], L, L) → append_out_aag([], L, L)
append_in_aag(.(H, L1), L2, .(H, L3)) → U1_aag(H, L1, L2, L3, append_in_aag(L1, L2, L3))
U1_aag(H, L1, L2, L3, append_out_aag(L1, L2, L3)) → append_out_aag(.(H, L1), L2, .(H, L3))
U3_aaag(A, B, C, D, append_out_aag(E, C, D)) → append3_out_aaag(A, B, C, D)

The argument filtering Pi contains the following mapping:
append3_in_aaag(x1, x2, x3, x4) = append3_in_aaag(x4)
U2_aaag(x1, x2, x3, x4, x5) = U2_aaag(x4, x5)
append_in_aaa(x1, x2, x3) = append_in_aaa
append_out_aaa(x1, x2, x3) = append_out_aaa(x1)
U1_aaa(x1, x2, x3, x4, x5) = U1_aaa(x5)
.(x1, x2) = .(x2)
U3_aaag(x1, x2, x3, x4, x5) = U3_aaag(x1, x5)
append_in_aag(x1, x2, x3) = append_in_aag(x3)
append_out_aag(x1, x2, x3) = append_out_aag(x1, x2)
U1_aag(x1, x2, x3, x4, x5) = U1_aag(x5)
append3_out_aaag(x1, x2, x3, x4) = append3_out_aaag(x1, x3)
APPEND_IN_AAG(x1, x2, x3) = APPEND_IN_AAG(x3)

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_AAG(.(H, L1), L2, .(H, L3)) → APPEND_IN_AAG(L1, L2, L3)

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

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

                        ↳ QDPSizeChangeProof

              ↳ PiDP

  ↳ PrologToPiTRSProof

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

APPEND_IN_AAG(.(L3)) → APPEND_IN_AAG(L3)

R is empty.
Q is empty.
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:

APPEND_IN_AAG(.(L3)) → APPEND_IN_AAG(L3)
The graph contains the following edges 1 > 1

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

  ↳ PrologToPiTRSProof

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

APPEND_IN_AAA(.(H, L1), L2, .(H, L3)) → APPEND_IN_AAA(L1, L2, L3)

The TRS R consists of the following rules:

append3_in_aaag(A, B, C, D) → U2_aaag(A, B, C, D, append_in_aaa(A, B, E))
append_in_aaa([], L, L) → append_out_aaa([], L, L)
append_in_aaa(.(H, L1), L2, .(H, L3)) → U1_aaa(H, L1, L2, L3, append_in_aaa(L1, L2, L3))
U1_aaa(H, L1, L2, L3, append_out_aaa(L1, L2, L3)) → append_out_aaa(.(H, L1), L2, .(H, L3))
U2_aaag(A, B, C, D, append_out_aaa(A, B, E)) → U3_aaag(A, B, C, D, append_in_aag(E, C, D))
append_in_aag([], L, L) → append_out_aag([], L, L)
append_in_aag(.(H, L1), L2, .(H, L3)) → U1_aag(H, L1, L2, L3, append_in_aag(L1, L2, L3))
U1_aag(H, L1, L2, L3, append_out_aag(L1, L2, L3)) → append_out_aag(.(H, L1), L2, .(H, L3))
U3_aaag(A, B, C, D, append_out_aag(E, C, D)) → append3_out_aaag(A, B, C, D)

The argument filtering Pi contains the following mapping:
append3_in_aaag(x1, x2, x3, x4) = append3_in_aaag(x4)
U2_aaag(x1, x2, x3, x4, x5) = U2_aaag(x4, x5)
append_in_aaa(x1, x2, x3) = append_in_aaa
append_out_aaa(x1, x2, x3) = append_out_aaa(x1)
U1_aaa(x1, x2, x3, x4, x5) = U1_aaa(x5)
.(x1, x2) = .(x2)
U3_aaag(x1, x2, x3, x4, x5) = U3_aaag(x1, x5)
append_in_aag(x1, x2, x3) = append_in_aag(x3)
append_out_aag(x1, x2, x3) = append_out_aag(x1, x2)
U1_aag(x1, x2, x3, x4, x5) = U1_aag(x5)
append3_out_aaag(x1, x2, x3, x4) = append3_out_aaag(x1, x3)
APPEND_IN_AAA(x1, x2, x3) = APPEND_IN_AAA

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:

APPEND_IN_AAA(.(H, L1), L2, .(H, L3)) → APPEND_IN_AAA(L1, L2, L3)

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

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

                        ↳ NonTerminationProof

  ↳ PrologToPiTRSProof

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

APPEND_IN_AAA → APPEND_IN_AAA

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_IN_AAA → APPEND_IN_AAA

The TRS R consists of the following rules:none

s = APPEND_IN_AAA evaluates to t =APPEND_IN_AAA

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

Semiunifier: [ ]
Matcher: [ ]

Rewriting sequence

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

We use the technique of [30]. With regard to the inferred argument filtering the predicates were used in the following modes:
append3_in: (f,f,f,b)
append_in: (f,f,f) (f,f,b)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

append3_in_aaag(A, B, C, D) → U2_aaag(A, B, C, D, append_in_aaa(A, B, E))
append_in_aaa([], L, L) → append_out_aaa([], L, L)
append_in_aaa(.(H, L1), L2, .(H, L3)) → U1_aaa(H, L1, L2, L3, append_in_aaa(L1, L2, L3))
U1_aaa(H, L1, L2, L3, append_out_aaa(L1, L2, L3)) → append_out_aaa(.(H, L1), L2, .(H, L3))
U2_aaag(A, B, C, D, append_out_aaa(A, B, E)) → U3_aaag(A, B, C, D, append_in_aag(E, C, D))
append_in_aag([], L, L) → append_out_aag([], L, L)
append_in_aag(.(H, L1), L2, .(H, L3)) → U1_aag(H, L1, L2, L3, append_in_aag(L1, L2, L3))
U1_aag(H, L1, L2, L3, append_out_aag(L1, L2, L3)) → append_out_aag(.(H, L1), L2, .(H, L3))
U3_aaag(A, B, C, D, append_out_aag(E, C, D)) → append3_out_aaag(A, B, C, D)

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:

append3_in_aaag(A, B, C, D) → U2_aaag(A, B, C, D, append_in_aaa(A, B, E))
append_in_aaa([], L, L) → append_out_aaa([], L, L)
append_in_aaa(.(H, L1), L2, .(H, L3)) → U1_aaa(H, L1, L2, L3, append_in_aaa(L1, L2, L3))
U1_aaa(H, L1, L2, L3, append_out_aaa(L1, L2, L3)) → append_out_aaa(.(H, L1), L2, .(H, L3))
U2_aaag(A, B, C, D, append_out_aaa(A, B, E)) → U3_aaag(A, B, C, D, append_in_aag(E, C, D))
append_in_aag([], L, L) → append_out_aag([], L, L)
append_in_aag(.(H, L1), L2, .(H, L3)) → U1_aag(H, L1, L2, L3, append_in_aag(L1, L2, L3))
U1_aag(H, L1, L2, L3, append_out_aag(L1, L2, L3)) → append_out_aag(.(H, L1), L2, .(H, L3))
U3_aaag(A, B, C, D, append_out_aag(E, C, D)) → append3_out_aaag(A, B, C, D)

APPEND3_IN_AAAG(A, B, C, D) → U2_AAAG(A, B, C, D, append_in_aaa(A, B, E))
APPEND3_IN_AAAG(A, B, C, D) → APPEND_IN_AAA(A, B, E)
APPEND_IN_AAA(.(H, L1), L2, .(H, L3)) → U1_AAA(H, L1, L2, L3, append_in_aaa(L1, L2, L3))
APPEND_IN_AAA(.(H, L1), L2, .(H, L3)) → APPEND_IN_AAA(L1, L2, L3)
U2_AAAG(A, B, C, D, append_out_aaa(A, B, E)) → U3_AAAG(A, B, C, D, append_in_aag(E, C, D))
U2_AAAG(A, B, C, D, append_out_aaa(A, B, E)) → APPEND_IN_AAG(E, C, D)
APPEND_IN_AAG(.(H, L1), L2, .(H, L3)) → U1_AAG(H, L1, L2, L3, append_in_aag(L1, L2, L3))
APPEND_IN_AAG(.(H, L1), L2, .(H, L3)) → APPEND_IN_AAG(L1, L2, L3)

The TRS R consists of the following rules:

append3_in_aaag(A, B, C, D) → U2_aaag(A, B, C, D, append_in_aaa(A, B, E))
append_in_aaa([], L, L) → append_out_aaa([], L, L)
append_in_aaa(.(H, L1), L2, .(H, L3)) → U1_aaa(H, L1, L2, L3, append_in_aaa(L1, L2, L3))
U1_aaa(H, L1, L2, L3, append_out_aaa(L1, L2, L3)) → append_out_aaa(.(H, L1), L2, .(H, L3))
U2_aaag(A, B, C, D, append_out_aaa(A, B, E)) → U3_aaag(A, B, C, D, append_in_aag(E, C, D))
append_in_aag([], L, L) → append_out_aag([], L, L)
append_in_aag(.(H, L1), L2, .(H, L3)) → U1_aag(H, L1, L2, L3, append_in_aag(L1, L2, L3))
U1_aag(H, L1, L2, L3, append_out_aag(L1, L2, L3)) → append_out_aag(.(H, L1), L2, .(H, L3))
U3_aaag(A, B, C, D, append_out_aag(E, C, D)) → append3_out_aaag(A, B, C, D)

The argument filtering Pi contains the following mapping:
append3_in_aaag(x1, x2, x3, x4) = append3_in_aaag(x4)
U2_aaag(x1, x2, x3, x4, x5) = U2_aaag(x4, x5)
append_in_aaa(x1, x2, x3) = append_in_aaa
append_out_aaa(x1, x2, x3) = append_out_aaa(x1)
U1_aaa(x1, x2, x3, x4, x5) = U1_aaa(x5)
.(x1, x2) = .(x2)
U3_aaag(x1, x2, x3, x4, x5) = U3_aaag(x1, x4, x5)
append_in_aag(x1, x2, x3) = append_in_aag(x3)
append_out_aag(x1, x2, x3) = append_out_aag(x1, x2, x3)
U1_aag(x1, x2, x3, x4, x5) = U1_aag(x4, x5)
append3_out_aaag(x1, x2, x3, x4) = append3_out_aaag(x1, x3, x4)
APPEND_IN_AAG(x1, x2, x3) = APPEND_IN_AAG(x3)
U2_AAAG(x1, x2, x3, x4, x5) = U2_AAAG(x4, x5)
U1_AAG(x1, x2, x3, x4, x5) = U1_AAG(x4, x5)
U3_AAAG(x1, x2, x3, x4, x5) = U3_AAAG(x1, x4, x5)
U1_AAA(x1, x2, x3, x4, x5) = U1_AAA(x5)
APPEND_IN_AAA(x1, x2, x3) = APPEND_IN_AAA
APPEND3_IN_AAAG(x1, x2, x3, x4) = APPEND3_IN_AAAG(x4)

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:

APPEND3_IN_AAAG(A, B, C, D) → U2_AAAG(A, B, C, D, append_in_aaa(A, B, E))
APPEND3_IN_AAAG(A, B, C, D) → APPEND_IN_AAA(A, B, E)
APPEND_IN_AAA(.(H, L1), L2, .(H, L3)) → U1_AAA(H, L1, L2, L3, append_in_aaa(L1, L2, L3))
APPEND_IN_AAA(.(H, L1), L2, .(H, L3)) → APPEND_IN_AAA(L1, L2, L3)
U2_AAAG(A, B, C, D, append_out_aaa(A, B, E)) → U3_AAAG(A, B, C, D, append_in_aag(E, C, D))
U2_AAAG(A, B, C, D, append_out_aaa(A, B, E)) → APPEND_IN_AAG(E, C, D)
APPEND_IN_AAG(.(H, L1), L2, .(H, L3)) → U1_AAG(H, L1, L2, L3, append_in_aag(L1, L2, L3))
APPEND_IN_AAG(.(H, L1), L2, .(H, L3)) → APPEND_IN_AAG(L1, L2, L3)

The TRS R consists of the following rules:

append3_in_aaag(A, B, C, D) → U2_aaag(A, B, C, D, append_in_aaa(A, B, E))
append_in_aaa([], L, L) → append_out_aaa([], L, L)
append_in_aaa(.(H, L1), L2, .(H, L3)) → U1_aaa(H, L1, L2, L3, append_in_aaa(L1, L2, L3))
U1_aaa(H, L1, L2, L3, append_out_aaa(L1, L2, L3)) → append_out_aaa(.(H, L1), L2, .(H, L3))
U2_aaag(A, B, C, D, append_out_aaa(A, B, E)) → U3_aaag(A, B, C, D, append_in_aag(E, C, D))
append_in_aag([], L, L) → append_out_aag([], L, L)
append_in_aag(.(H, L1), L2, .(H, L3)) → U1_aag(H, L1, L2, L3, append_in_aag(L1, L2, L3))
U1_aag(H, L1, L2, L3, append_out_aag(L1, L2, L3)) → append_out_aag(.(H, L1), L2, .(H, L3))
U3_aaag(A, B, C, D, append_out_aag(E, C, D)) → append3_out_aaag(A, B, C, D)

The argument filtering Pi contains the following mapping:
append3_in_aaag(x1, x2, x3, x4) = append3_in_aaag(x4)
U2_aaag(x1, x2, x3, x4, x5) = U2_aaag(x4, x5)
append_in_aaa(x1, x2, x3) = append_in_aaa
append_out_aaa(x1, x2, x3) = append_out_aaa(x1)
U1_aaa(x1, x2, x3, x4, x5) = U1_aaa(x5)
.(x1, x2) = .(x2)
U3_aaag(x1, x2, x3, x4, x5) = U3_aaag(x1, x4, x5)
append_in_aag(x1, x2, x3) = append_in_aag(x3)
append_out_aag(x1, x2, x3) = append_out_aag(x1, x2, x3)
U1_aag(x1, x2, x3, x4, x5) = U1_aag(x4, x5)
append3_out_aaag(x1, x2, x3, x4) = append3_out_aaag(x1, x3, x4)
APPEND_IN_AAG(x1, x2, x3) = APPEND_IN_AAG(x3)
U2_AAAG(x1, x2, x3, x4, x5) = U2_AAAG(x4, x5)
U1_AAG(x1, x2, x3, x4, x5) = U1_AAG(x4, x5)
U3_AAAG(x1, x2, x3, x4, x5) = U3_AAAG(x1, x4, x5)
U1_AAA(x1, x2, x3, x4, x5) = U1_AAA(x5)
APPEND_IN_AAA(x1, x2, x3) = APPEND_IN_AAA
APPEND3_IN_AAAG(x1, x2, x3, x4) = APPEND3_IN_AAAG(x4)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 2 SCCs with 6 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_AAG(.(H, L1), L2, .(H, L3)) → APPEND_IN_AAG(L1, L2, L3)

The TRS R consists of the following rules:

append3_in_aaag(A, B, C, D) → U2_aaag(A, B, C, D, append_in_aaa(A, B, E))
append_in_aaa([], L, L) → append_out_aaa([], L, L)
append_in_aaa(.(H, L1), L2, .(H, L3)) → U1_aaa(H, L1, L2, L3, append_in_aaa(L1, L2, L3))
U1_aaa(H, L1, L2, L3, append_out_aaa(L1, L2, L3)) → append_out_aaa(.(H, L1), L2, .(H, L3))
U2_aaag(A, B, C, D, append_out_aaa(A, B, E)) → U3_aaag(A, B, C, D, append_in_aag(E, C, D))
append_in_aag([], L, L) → append_out_aag([], L, L)
append_in_aag(.(H, L1), L2, .(H, L3)) → U1_aag(H, L1, L2, L3, append_in_aag(L1, L2, L3))
U1_aag(H, L1, L2, L3, append_out_aag(L1, L2, L3)) → append_out_aag(.(H, L1), L2, .(H, L3))
U3_aaag(A, B, C, D, append_out_aag(E, C, D)) → append3_out_aaag(A, B, C, D)

The argument filtering Pi contains the following mapping:
append3_in_aaag(x1, x2, x3, x4) = append3_in_aaag(x4)
U2_aaag(x1, x2, x3, x4, x5) = U2_aaag(x4, x5)
append_in_aaa(x1, x2, x3) = append_in_aaa
append_out_aaa(x1, x2, x3) = append_out_aaa(x1)
U1_aaa(x1, x2, x3, x4, x5) = U1_aaa(x5)
.(x1, x2) = .(x2)
U3_aaag(x1, x2, x3, x4, x5) = U3_aaag(x1, x4, x5)
append_in_aag(x1, x2, x3) = append_in_aag(x3)
append_out_aag(x1, x2, x3) = append_out_aag(x1, x2, x3)
U1_aag(x1, x2, x3, x4, x5) = U1_aag(x4, x5)
append3_out_aaag(x1, x2, x3, x4) = append3_out_aaag(x1, x3, x4)
APPEND_IN_AAG(x1, x2, x3) = APPEND_IN_AAG(x3)

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_AAG(.(H, L1), L2, .(H, L3)) → APPEND_IN_AAG(L1, L2, L3)

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ PiDP

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

APPEND_IN_AAG(.(L3)) → APPEND_IN_AAG(L3)

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

APPEND_IN_AAG(.(L3)) → APPEND_IN_AAG(L3)
The graph contains the following edges 1 > 1

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

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

APPEND_IN_AAA(.(H, L1), L2, .(H, L3)) → APPEND_IN_AAA(L1, L2, L3)

The TRS R consists of the following rules:

append3_in_aaag(A, B, C, D) → U2_aaag(A, B, C, D, append_in_aaa(A, B, E))
append_in_aaa([], L, L) → append_out_aaa([], L, L)
append_in_aaa(.(H, L1), L2, .(H, L3)) → U1_aaa(H, L1, L2, L3, append_in_aaa(L1, L2, L3))
U1_aaa(H, L1, L2, L3, append_out_aaa(L1, L2, L3)) → append_out_aaa(.(H, L1), L2, .(H, L3))
U2_aaag(A, B, C, D, append_out_aaa(A, B, E)) → U3_aaag(A, B, C, D, append_in_aag(E, C, D))
append_in_aag([], L, L) → append_out_aag([], L, L)
append_in_aag(.(H, L1), L2, .(H, L3)) → U1_aag(H, L1, L2, L3, append_in_aag(L1, L2, L3))
U1_aag(H, L1, L2, L3, append_out_aag(L1, L2, L3)) → append_out_aag(.(H, L1), L2, .(H, L3))
U3_aaag(A, B, C, D, append_out_aag(E, C, D)) → append3_out_aaag(A, B, C, D)

The argument filtering Pi contains the following mapping:
append3_in_aaag(x1, x2, x3, x4) = append3_in_aaag(x4)
U2_aaag(x1, x2, x3, x4, x5) = U2_aaag(x4, x5)
append_in_aaa(x1, x2, x3) = append_in_aaa
append_out_aaa(x1, x2, x3) = append_out_aaa(x1)
U1_aaa(x1, x2, x3, x4, x5) = U1_aaa(x5)
.(x1, x2) = .(x2)
U3_aaag(x1, x2, x3, x4, x5) = U3_aaag(x1, x4, x5)
append_in_aag(x1, x2, x3) = append_in_aag(x3)
append_out_aag(x1, x2, x3) = append_out_aag(x1, x2, x3)
U1_aag(x1, x2, x3, x4, x5) = U1_aag(x4, x5)
append3_out_aaag(x1, x2, x3, x4) = append3_out_aaag(x1, x3, x4)
APPEND_IN_AAA(x1, x2, x3) = APPEND_IN_AAA

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:

APPEND_IN_AAA(.(H, L1), L2, .(H, L3)) → APPEND_IN_AAA(L1, L2, L3)

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ NonTerminationProof

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

APPEND_IN_AAA → APPEND_IN_AAA

The TRS R consists of the following rules:none

s = APPEND_IN_AAA evaluates to t =APPEND_IN_AAA

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

Semiunifier: [ ]
Matcher: [ ]

Rewriting sequence

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