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

append3_in(A, B, C, D) → U2(A, B, C, D, append_in(A, B, E))
append_in(.(H, L1), L2, .(H, L3)) → U1(H, L1, L2, L3, append_in(L1, L2, L3))
append_in([], L, L) → append_out([], L, L)
U1(H, L1, L2, L3, append_out(L1, L2, L3)) → append_out(.(H, L1), L2, .(H, L3))
U2(A, B, C, D, append_out(A, B, E)) → U3(A, B, C, D, append_in(E, C, D))
U3(A, B, C, D, append_out(E, C, D)) → append3_out(A, B, C, D)

The argument filtering Pi contains the following mapping:
append3_in(x1, x2, x3, x4)  =  append3_in(x4)
U2(x1, x2, x3, x4, x5)  =  U2(x5)
append_in(x1, x2, x3)  =  append_in
.(x1, x2)  =  .(x2)
U1(x1, x2, x3, x4, x5)  =  U1(x5)
[]  =  []
append_out(x1, x2, x3)  =  append_out(x1)
U3(x1, x2, x3, x4, x5)  =  U3(x1, x5)
append3_out(x1, x2, x3, x4)  =  append3_out(x1)

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(A, B, C, D) → U2(A, B, C, D, append_in(A, B, E))
append_in(.(H, L1), L2, .(H, L3)) → U1(H, L1, L2, L3, append_in(L1, L2, L3))
append_in([], L, L) → append_out([], L, L)
U1(H, L1, L2, L3, append_out(L1, L2, L3)) → append_out(.(H, L1), L2, .(H, L3))
U2(A, B, C, D, append_out(A, B, E)) → U3(A, B, C, D, append_in(E, C, D))
U3(A, B, C, D, append_out(E, C, D)) → append3_out(A, B, C, D)

The argument filtering Pi contains the following mapping:
append3_in(x1, x2, x3, x4)  =  append3_in(x4)
U2(x1, x2, x3, x4, x5)  =  U2(x5)
append_in(x1, x2, x3)  =  append_in
.(x1, x2)  =  .(x2)
U1(x1, x2, x3, x4, x5)  =  U1(x5)
[]  =  []
append_out(x1, x2, x3)  =  append_out(x1)
U3(x1, x2, x3, x4, x5)  =  U3(x1, x5)
append3_out(x1, x2, x3, x4)  =  append3_out(x1)


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

APPEND3_IN(A, B, C, D) → U21(A, B, C, D, append_in(A, B, E))
APPEND3_IN(A, B, C, D) → APPEND_IN(A, B, E)
APPEND_IN(.(H, L1), L2, .(H, L3)) → U11(H, L1, L2, L3, append_in(L1, L2, L3))
APPEND_IN(.(H, L1), L2, .(H, L3)) → APPEND_IN(L1, L2, L3)
U21(A, B, C, D, append_out(A, B, E)) → U31(A, B, C, D, append_in(E, C, D))
U21(A, B, C, D, append_out(A, B, E)) → APPEND_IN(E, C, D)

The TRS R consists of the following rules:

append3_in(A, B, C, D) → U2(A, B, C, D, append_in(A, B, E))
append_in(.(H, L1), L2, .(H, L3)) → U1(H, L1, L2, L3, append_in(L1, L2, L3))
append_in([], L, L) → append_out([], L, L)
U1(H, L1, L2, L3, append_out(L1, L2, L3)) → append_out(.(H, L1), L2, .(H, L3))
U2(A, B, C, D, append_out(A, B, E)) → U3(A, B, C, D, append_in(E, C, D))
U3(A, B, C, D, append_out(E, C, D)) → append3_out(A, B, C, D)

The argument filtering Pi contains the following mapping:
append3_in(x1, x2, x3, x4)  =  append3_in(x4)
U2(x1, x2, x3, x4, x5)  =  U2(x5)
append_in(x1, x2, x3)  =  append_in
.(x1, x2)  =  .(x2)
U1(x1, x2, x3, x4, x5)  =  U1(x5)
[]  =  []
append_out(x1, x2, x3)  =  append_out(x1)
U3(x1, x2, x3, x4, x5)  =  U3(x1, x5)
append3_out(x1, x2, x3, x4)  =  append3_out(x1)
U31(x1, x2, x3, x4, x5)  =  U31(x1, x5)
APPEND_IN(x1, x2, x3)  =  APPEND_IN
APPEND3_IN(x1, x2, x3, x4)  =  APPEND3_IN(x4)
U21(x1, x2, x3, x4, x5)  =  U21(x5)
U11(x1, x2, x3, x4, x5)  =  U11(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:

APPEND3_IN(A, B, C, D) → U21(A, B, C, D, append_in(A, B, E))
APPEND3_IN(A, B, C, D) → APPEND_IN(A, B, E)
APPEND_IN(.(H, L1), L2, .(H, L3)) → U11(H, L1, L2, L3, append_in(L1, L2, L3))
APPEND_IN(.(H, L1), L2, .(H, L3)) → APPEND_IN(L1, L2, L3)
U21(A, B, C, D, append_out(A, B, E)) → U31(A, B, C, D, append_in(E, C, D))
U21(A, B, C, D, append_out(A, B, E)) → APPEND_IN(E, C, D)

The TRS R consists of the following rules:

append3_in(A, B, C, D) → U2(A, B, C, D, append_in(A, B, E))
append_in(.(H, L1), L2, .(H, L3)) → U1(H, L1, L2, L3, append_in(L1, L2, L3))
append_in([], L, L) → append_out([], L, L)
U1(H, L1, L2, L3, append_out(L1, L2, L3)) → append_out(.(H, L1), L2, .(H, L3))
U2(A, B, C, D, append_out(A, B, E)) → U3(A, B, C, D, append_in(E, C, D))
U3(A, B, C, D, append_out(E, C, D)) → append3_out(A, B, C, D)

The argument filtering Pi contains the following mapping:
append3_in(x1, x2, x3, x4)  =  append3_in(x4)
U2(x1, x2, x3, x4, x5)  =  U2(x5)
append_in(x1, x2, x3)  =  append_in
.(x1, x2)  =  .(x2)
U1(x1, x2, x3, x4, x5)  =  U1(x5)
[]  =  []
append_out(x1, x2, x3)  =  append_out(x1)
U3(x1, x2, x3, x4, x5)  =  U3(x1, x5)
append3_out(x1, x2, x3, x4)  =  append3_out(x1)
U31(x1, x2, x3, x4, x5)  =  U31(x1, x5)
APPEND_IN(x1, x2, x3)  =  APPEND_IN
APPEND3_IN(x1, x2, x3, x4)  =  APPEND3_IN(x4)
U21(x1, x2, x3, x4, x5)  =  U21(x5)
U11(x1, x2, x3, x4, x5)  =  U11(x5)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 1 SCC with 5 less nodes.

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

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

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

The TRS R consists of the following rules:

append3_in(A, B, C, D) → U2(A, B, C, D, append_in(A, B, E))
append_in(.(H, L1), L2, .(H, L3)) → U1(H, L1, L2, L3, append_in(L1, L2, L3))
append_in([], L, L) → append_out([], L, L)
U1(H, L1, L2, L3, append_out(L1, L2, L3)) → append_out(.(H, L1), L2, .(H, L3))
U2(A, B, C, D, append_out(A, B, E)) → U3(A, B, C, D, append_in(E, C, D))
U3(A, B, C, D, append_out(E, C, D)) → append3_out(A, B, C, D)

The argument filtering Pi contains the following mapping:
append3_in(x1, x2, x3, x4)  =  append3_in(x4)
U2(x1, x2, x3, x4, x5)  =  U2(x5)
append_in(x1, x2, x3)  =  append_in
.(x1, x2)  =  .(x2)
U1(x1, x2, x3, x4, x5)  =  U1(x5)
[]  =  []
append_out(x1, x2, x3)  =  append_out(x1)
U3(x1, x2, x3, x4, x5)  =  U3(x1, x5)
append3_out(x1, x2, x3, x4)  =  append3_out(x1)
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
            ↳ PiDP
              ↳ UsableRulesProof
                ↳ PiDP
                  ↳ PiDPToQDPProof
  ↳ PrologToPiTRSProof

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

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

R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(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
            ↳ PiDP
              ↳ UsableRulesProof
                ↳ PiDP
                  ↳ PiDPToQDPProof
                    ↳ QDP
                      ↳ NonTerminationProof
  ↳ PrologToPiTRSProof

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

APPEND_IN → APPEND_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_IN → APPEND_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.




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:

append3_in(A, B, C, D) → U2(A, B, C, D, append_in(A, B, E))
append_in(.(H, L1), L2, .(H, L3)) → U1(H, L1, L2, L3, append_in(L1, L2, L3))
append_in([], L, L) → append_out([], L, L)
U1(H, L1, L2, L3, append_out(L1, L2, L3)) → append_out(.(H, L1), L2, .(H, L3))
U2(A, B, C, D, append_out(A, B, E)) → U3(A, B, C, D, append_in(E, C, D))
U3(A, B, C, D, append_out(E, C, D)) → append3_out(A, B, C, D)

The argument filtering Pi contains the following mapping:
append3_in(x1, x2, x3, x4)  =  append3_in(x4)
U2(x1, x2, x3, x4, x5)  =  U2(x4, x5)
append_in(x1, x2, x3)  =  append_in
.(x1, x2)  =  .(x2)
U1(x1, x2, x3, x4, x5)  =  U1(x5)
[]  =  []
append_out(x1, x2, x3)  =  append_out(x1)
U3(x1, x2, x3, x4, x5)  =  U3(x1, x4, x5)
append3_out(x1, x2, x3, x4)  =  append3_out(x1, x4)

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(A, B, C, D) → U2(A, B, C, D, append_in(A, B, E))
append_in(.(H, L1), L2, .(H, L3)) → U1(H, L1, L2, L3, append_in(L1, L2, L3))
append_in([], L, L) → append_out([], L, L)
U1(H, L1, L2, L3, append_out(L1, L2, L3)) → append_out(.(H, L1), L2, .(H, L3))
U2(A, B, C, D, append_out(A, B, E)) → U3(A, B, C, D, append_in(E, C, D))
U3(A, B, C, D, append_out(E, C, D)) → append3_out(A, B, C, D)

The argument filtering Pi contains the following mapping:
append3_in(x1, x2, x3, x4)  =  append3_in(x4)
U2(x1, x2, x3, x4, x5)  =  U2(x4, x5)
append_in(x1, x2, x3)  =  append_in
.(x1, x2)  =  .(x2)
U1(x1, x2, x3, x4, x5)  =  U1(x5)
[]  =  []
append_out(x1, x2, x3)  =  append_out(x1)
U3(x1, x2, x3, x4, x5)  =  U3(x1, x4, x5)
append3_out(x1, x2, x3, x4)  =  append3_out(x1, x4)


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

APPEND3_IN(A, B, C, D) → U21(A, B, C, D, append_in(A, B, E))
APPEND3_IN(A, B, C, D) → APPEND_IN(A, B, E)
APPEND_IN(.(H, L1), L2, .(H, L3)) → U11(H, L1, L2, L3, append_in(L1, L2, L3))
APPEND_IN(.(H, L1), L2, .(H, L3)) → APPEND_IN(L1, L2, L3)
U21(A, B, C, D, append_out(A, B, E)) → U31(A, B, C, D, append_in(E, C, D))
U21(A, B, C, D, append_out(A, B, E)) → APPEND_IN(E, C, D)

The TRS R consists of the following rules:

append3_in(A, B, C, D) → U2(A, B, C, D, append_in(A, B, E))
append_in(.(H, L1), L2, .(H, L3)) → U1(H, L1, L2, L3, append_in(L1, L2, L3))
append_in([], L, L) → append_out([], L, L)
U1(H, L1, L2, L3, append_out(L1, L2, L3)) → append_out(.(H, L1), L2, .(H, L3))
U2(A, B, C, D, append_out(A, B, E)) → U3(A, B, C, D, append_in(E, C, D))
U3(A, B, C, D, append_out(E, C, D)) → append3_out(A, B, C, D)

The argument filtering Pi contains the following mapping:
append3_in(x1, x2, x3, x4)  =  append3_in(x4)
U2(x1, x2, x3, x4, x5)  =  U2(x4, x5)
append_in(x1, x2, x3)  =  append_in
.(x1, x2)  =  .(x2)
U1(x1, x2, x3, x4, x5)  =  U1(x5)
[]  =  []
append_out(x1, x2, x3)  =  append_out(x1)
U3(x1, x2, x3, x4, x5)  =  U3(x1, x4, x5)
append3_out(x1, x2, x3, x4)  =  append3_out(x1, x4)
U31(x1, x2, x3, x4, x5)  =  U31(x1, x4, x5)
APPEND_IN(x1, x2, x3)  =  APPEND_IN
APPEND3_IN(x1, x2, x3, x4)  =  APPEND3_IN(x4)
U21(x1, x2, x3, x4, x5)  =  U21(x4, x5)
U11(x1, x2, x3, x4, x5)  =  U11(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:

APPEND3_IN(A, B, C, D) → U21(A, B, C, D, append_in(A, B, E))
APPEND3_IN(A, B, C, D) → APPEND_IN(A, B, E)
APPEND_IN(.(H, L1), L2, .(H, L3)) → U11(H, L1, L2, L3, append_in(L1, L2, L3))
APPEND_IN(.(H, L1), L2, .(H, L3)) → APPEND_IN(L1, L2, L3)
U21(A, B, C, D, append_out(A, B, E)) → U31(A, B, C, D, append_in(E, C, D))
U21(A, B, C, D, append_out(A, B, E)) → APPEND_IN(E, C, D)

The TRS R consists of the following rules:

append3_in(A, B, C, D) → U2(A, B, C, D, append_in(A, B, E))
append_in(.(H, L1), L2, .(H, L3)) → U1(H, L1, L2, L3, append_in(L1, L2, L3))
append_in([], L, L) → append_out([], L, L)
U1(H, L1, L2, L3, append_out(L1, L2, L3)) → append_out(.(H, L1), L2, .(H, L3))
U2(A, B, C, D, append_out(A, B, E)) → U3(A, B, C, D, append_in(E, C, D))
U3(A, B, C, D, append_out(E, C, D)) → append3_out(A, B, C, D)

The argument filtering Pi contains the following mapping:
append3_in(x1, x2, x3, x4)  =  append3_in(x4)
U2(x1, x2, x3, x4, x5)  =  U2(x4, x5)
append_in(x1, x2, x3)  =  append_in
.(x1, x2)  =  .(x2)
U1(x1, x2, x3, x4, x5)  =  U1(x5)
[]  =  []
append_out(x1, x2, x3)  =  append_out(x1)
U3(x1, x2, x3, x4, x5)  =  U3(x1, x4, x5)
append3_out(x1, x2, x3, x4)  =  append3_out(x1, x4)
U31(x1, x2, x3, x4, x5)  =  U31(x1, x4, x5)
APPEND_IN(x1, x2, x3)  =  APPEND_IN
APPEND3_IN(x1, x2, x3, x4)  =  APPEND3_IN(x4)
U21(x1, x2, x3, x4, x5)  =  U21(x4, x5)
U11(x1, x2, x3, x4, x5)  =  U11(x5)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 1 SCC with 5 less nodes.

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

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

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

The TRS R consists of the following rules:

append3_in(A, B, C, D) → U2(A, B, C, D, append_in(A, B, E))
append_in(.(H, L1), L2, .(H, L3)) → U1(H, L1, L2, L3, append_in(L1, L2, L3))
append_in([], L, L) → append_out([], L, L)
U1(H, L1, L2, L3, append_out(L1, L2, L3)) → append_out(.(H, L1), L2, .(H, L3))
U2(A, B, C, D, append_out(A, B, E)) → U3(A, B, C, D, append_in(E, C, D))
U3(A, B, C, D, append_out(E, C, D)) → append3_out(A, B, C, D)

The argument filtering Pi contains the following mapping:
append3_in(x1, x2, x3, x4)  =  append3_in(x4)
U2(x1, x2, x3, x4, x5)  =  U2(x4, x5)
append_in(x1, x2, x3)  =  append_in
.(x1, x2)  =  .(x2)
U1(x1, x2, x3, x4, x5)  =  U1(x5)
[]  =  []
append_out(x1, x2, x3)  =  append_out(x1)
U3(x1, x2, x3, x4, x5)  =  U3(x1, x4, x5)
append3_out(x1, x2, x3, x4)  =  append3_out(x1, x4)
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
            ↳ PiDP
              ↳ UsableRulesProof
                ↳ PiDP
                  ↳ PiDPToQDPProof

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

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

R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(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
            ↳ PiDP
              ↳ UsableRulesProof
                ↳ PiDP
                  ↳ PiDPToQDPProof
                    ↳ QDP
                      ↳ NonTerminationProof

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

APPEND_IN → APPEND_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_IN → APPEND_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.