Left Termination of the query pattern sublist_in_2(g, a) w.r.t. the given Prolog program could not be shown:



↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof

Clauses:

sublist(X, Y) :- ','(append(U, X, V), append(V, W, Y)).
append([], Ys, Ys).
append(.(X, Xs), Ys, .(X, Zs)) :- append(Xs, Ys, Zs).

Queries:

sublist(g,a).

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

sublist_in_ga(X, Y) → U1_ga(X, Y, append_in_aga(U, X, V))
append_in_aga([], Ys, Ys) → append_out_aga([], Ys, Ys)
append_in_aga(.(X, Xs), Ys, .(X, Zs)) → U3_aga(X, Xs, Ys, Zs, append_in_aga(Xs, Ys, Zs))
U3_aga(X, Xs, Ys, Zs, append_out_aga(Xs, Ys, Zs)) → append_out_aga(.(X, Xs), Ys, .(X, Zs))
U1_ga(X, Y, append_out_aga(U, X, V)) → U2_ga(X, Y, append_in_aaa(V, W, Y))
append_in_aaa([], Ys, Ys) → append_out_aaa([], Ys, Ys)
append_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U3_aaa(X, Xs, Ys, Zs, append_in_aaa(Xs, Ys, Zs))
U3_aaa(X, Xs, Ys, Zs, append_out_aaa(Xs, Ys, Zs)) → append_out_aaa(.(X, Xs), Ys, .(X, Zs))
U2_ga(X, Y, append_out_aaa(V, W, Y)) → sublist_out_ga(X, Y)

The argument filtering Pi contains the following mapping:
sublist_in_ga(x1, x2)  =  sublist_in_ga(x1)
U1_ga(x1, x2, x3)  =  U1_ga(x3)
append_in_aga(x1, x2, x3)  =  append_in_aga(x2)
append_out_aga(x1, x2, x3)  =  append_out_aga
U3_aga(x1, x2, x3, x4, x5)  =  U3_aga(x5)
U2_ga(x1, x2, x3)  =  U2_ga(x3)
append_in_aaa(x1, x2, x3)  =  append_in_aaa
append_out_aaa(x1, x2, x3)  =  append_out_aaa
U3_aaa(x1, x2, x3, x4, x5)  =  U3_aaa(x5)
sublist_out_ga(x1, x2)  =  sublist_out_ga

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:

sublist_in_ga(X, Y) → U1_ga(X, Y, append_in_aga(U, X, V))
append_in_aga([], Ys, Ys) → append_out_aga([], Ys, Ys)
append_in_aga(.(X, Xs), Ys, .(X, Zs)) → U3_aga(X, Xs, Ys, Zs, append_in_aga(Xs, Ys, Zs))
U3_aga(X, Xs, Ys, Zs, append_out_aga(Xs, Ys, Zs)) → append_out_aga(.(X, Xs), Ys, .(X, Zs))
U1_ga(X, Y, append_out_aga(U, X, V)) → U2_ga(X, Y, append_in_aaa(V, W, Y))
append_in_aaa([], Ys, Ys) → append_out_aaa([], Ys, Ys)
append_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U3_aaa(X, Xs, Ys, Zs, append_in_aaa(Xs, Ys, Zs))
U3_aaa(X, Xs, Ys, Zs, append_out_aaa(Xs, Ys, Zs)) → append_out_aaa(.(X, Xs), Ys, .(X, Zs))
U2_ga(X, Y, append_out_aaa(V, W, Y)) → sublist_out_ga(X, Y)

The argument filtering Pi contains the following mapping:
sublist_in_ga(x1, x2)  =  sublist_in_ga(x1)
U1_ga(x1, x2, x3)  =  U1_ga(x3)
append_in_aga(x1, x2, x3)  =  append_in_aga(x2)
append_out_aga(x1, x2, x3)  =  append_out_aga
U3_aga(x1, x2, x3, x4, x5)  =  U3_aga(x5)
U2_ga(x1, x2, x3)  =  U2_ga(x3)
append_in_aaa(x1, x2, x3)  =  append_in_aaa
append_out_aaa(x1, x2, x3)  =  append_out_aaa
U3_aaa(x1, x2, x3, x4, x5)  =  U3_aaa(x5)
sublist_out_ga(x1, x2)  =  sublist_out_ga


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

SUBLIST_IN_GA(X, Y) → U1_GA(X, Y, append_in_aga(U, X, V))
SUBLIST_IN_GA(X, Y) → APPEND_IN_AGA(U, X, V)
APPEND_IN_AGA(.(X, Xs), Ys, .(X, Zs)) → U3_AGA(X, Xs, Ys, Zs, append_in_aga(Xs, Ys, Zs))
APPEND_IN_AGA(.(X, Xs), Ys, .(X, Zs)) → APPEND_IN_AGA(Xs, Ys, Zs)
U1_GA(X, Y, append_out_aga(U, X, V)) → U2_GA(X, Y, append_in_aaa(V, W, Y))
U1_GA(X, Y, append_out_aga(U, X, V)) → APPEND_IN_AAA(V, W, Y)
APPEND_IN_AAA(.(X, Xs), Ys, .(X, Zs)) → U3_AAA(X, Xs, Ys, Zs, append_in_aaa(Xs, Ys, Zs))
APPEND_IN_AAA(.(X, Xs), Ys, .(X, Zs)) → APPEND_IN_AAA(Xs, Ys, Zs)

The TRS R consists of the following rules:

sublist_in_ga(X, Y) → U1_ga(X, Y, append_in_aga(U, X, V))
append_in_aga([], Ys, Ys) → append_out_aga([], Ys, Ys)
append_in_aga(.(X, Xs), Ys, .(X, Zs)) → U3_aga(X, Xs, Ys, Zs, append_in_aga(Xs, Ys, Zs))
U3_aga(X, Xs, Ys, Zs, append_out_aga(Xs, Ys, Zs)) → append_out_aga(.(X, Xs), Ys, .(X, Zs))
U1_ga(X, Y, append_out_aga(U, X, V)) → U2_ga(X, Y, append_in_aaa(V, W, Y))
append_in_aaa([], Ys, Ys) → append_out_aaa([], Ys, Ys)
append_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U3_aaa(X, Xs, Ys, Zs, append_in_aaa(Xs, Ys, Zs))
U3_aaa(X, Xs, Ys, Zs, append_out_aaa(Xs, Ys, Zs)) → append_out_aaa(.(X, Xs), Ys, .(X, Zs))
U2_ga(X, Y, append_out_aaa(V, W, Y)) → sublist_out_ga(X, Y)

The argument filtering Pi contains the following mapping:
sublist_in_ga(x1, x2)  =  sublist_in_ga(x1)
U1_ga(x1, x2, x3)  =  U1_ga(x3)
append_in_aga(x1, x2, x3)  =  append_in_aga(x2)
append_out_aga(x1, x2, x3)  =  append_out_aga
U3_aga(x1, x2, x3, x4, x5)  =  U3_aga(x5)
U2_ga(x1, x2, x3)  =  U2_ga(x3)
append_in_aaa(x1, x2, x3)  =  append_in_aaa
append_out_aaa(x1, x2, x3)  =  append_out_aaa
U3_aaa(x1, x2, x3, x4, x5)  =  U3_aaa(x5)
sublist_out_ga(x1, x2)  =  sublist_out_ga
SUBLIST_IN_GA(x1, x2)  =  SUBLIST_IN_GA(x1)
APPEND_IN_AGA(x1, x2, x3)  =  APPEND_IN_AGA(x2)
U2_GA(x1, x2, x3)  =  U2_GA(x3)
U1_GA(x1, x2, x3)  =  U1_GA(x3)
APPEND_IN_AAA(x1, x2, x3)  =  APPEND_IN_AAA
U3_AAA(x1, x2, x3, x4, x5)  =  U3_AAA(x5)
U3_AGA(x1, x2, x3, x4, x5)  =  U3_AGA(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:

SUBLIST_IN_GA(X, Y) → U1_GA(X, Y, append_in_aga(U, X, V))
SUBLIST_IN_GA(X, Y) → APPEND_IN_AGA(U, X, V)
APPEND_IN_AGA(.(X, Xs), Ys, .(X, Zs)) → U3_AGA(X, Xs, Ys, Zs, append_in_aga(Xs, Ys, Zs))
APPEND_IN_AGA(.(X, Xs), Ys, .(X, Zs)) → APPEND_IN_AGA(Xs, Ys, Zs)
U1_GA(X, Y, append_out_aga(U, X, V)) → U2_GA(X, Y, append_in_aaa(V, W, Y))
U1_GA(X, Y, append_out_aga(U, X, V)) → APPEND_IN_AAA(V, W, Y)
APPEND_IN_AAA(.(X, Xs), Ys, .(X, Zs)) → U3_AAA(X, Xs, Ys, Zs, append_in_aaa(Xs, Ys, Zs))
APPEND_IN_AAA(.(X, Xs), Ys, .(X, Zs)) → APPEND_IN_AAA(Xs, Ys, Zs)

The TRS R consists of the following rules:

sublist_in_ga(X, Y) → U1_ga(X, Y, append_in_aga(U, X, V))
append_in_aga([], Ys, Ys) → append_out_aga([], Ys, Ys)
append_in_aga(.(X, Xs), Ys, .(X, Zs)) → U3_aga(X, Xs, Ys, Zs, append_in_aga(Xs, Ys, Zs))
U3_aga(X, Xs, Ys, Zs, append_out_aga(Xs, Ys, Zs)) → append_out_aga(.(X, Xs), Ys, .(X, Zs))
U1_ga(X, Y, append_out_aga(U, X, V)) → U2_ga(X, Y, append_in_aaa(V, W, Y))
append_in_aaa([], Ys, Ys) → append_out_aaa([], Ys, Ys)
append_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U3_aaa(X, Xs, Ys, Zs, append_in_aaa(Xs, Ys, Zs))
U3_aaa(X, Xs, Ys, Zs, append_out_aaa(Xs, Ys, Zs)) → append_out_aaa(.(X, Xs), Ys, .(X, Zs))
U2_ga(X, Y, append_out_aaa(V, W, Y)) → sublist_out_ga(X, Y)

The argument filtering Pi contains the following mapping:
sublist_in_ga(x1, x2)  =  sublist_in_ga(x1)
U1_ga(x1, x2, x3)  =  U1_ga(x3)
append_in_aga(x1, x2, x3)  =  append_in_aga(x2)
append_out_aga(x1, x2, x3)  =  append_out_aga
U3_aga(x1, x2, x3, x4, x5)  =  U3_aga(x5)
U2_ga(x1, x2, x3)  =  U2_ga(x3)
append_in_aaa(x1, x2, x3)  =  append_in_aaa
append_out_aaa(x1, x2, x3)  =  append_out_aaa
U3_aaa(x1, x2, x3, x4, x5)  =  U3_aaa(x5)
sublist_out_ga(x1, x2)  =  sublist_out_ga
SUBLIST_IN_GA(x1, x2)  =  SUBLIST_IN_GA(x1)
APPEND_IN_AGA(x1, x2, x3)  =  APPEND_IN_AGA(x2)
U2_GA(x1, x2, x3)  =  U2_GA(x3)
U1_GA(x1, x2, x3)  =  U1_GA(x3)
APPEND_IN_AAA(x1, x2, x3)  =  APPEND_IN_AAA
U3_AAA(x1, x2, x3, x4, x5)  =  U3_AAA(x5)
U3_AGA(x1, x2, x3, x4, x5)  =  U3_AGA(x5)

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

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

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

The TRS R consists of the following rules:

sublist_in_ga(X, Y) → U1_ga(X, Y, append_in_aga(U, X, V))
append_in_aga([], Ys, Ys) → append_out_aga([], Ys, Ys)
append_in_aga(.(X, Xs), Ys, .(X, Zs)) → U3_aga(X, Xs, Ys, Zs, append_in_aga(Xs, Ys, Zs))
U3_aga(X, Xs, Ys, Zs, append_out_aga(Xs, Ys, Zs)) → append_out_aga(.(X, Xs), Ys, .(X, Zs))
U1_ga(X, Y, append_out_aga(U, X, V)) → U2_ga(X, Y, append_in_aaa(V, W, Y))
append_in_aaa([], Ys, Ys) → append_out_aaa([], Ys, Ys)
append_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U3_aaa(X, Xs, Ys, Zs, append_in_aaa(Xs, Ys, Zs))
U3_aaa(X, Xs, Ys, Zs, append_out_aaa(Xs, Ys, Zs)) → append_out_aaa(.(X, Xs), Ys, .(X, Zs))
U2_ga(X, Y, append_out_aaa(V, W, Y)) → sublist_out_ga(X, Y)

The argument filtering Pi contains the following mapping:
sublist_in_ga(x1, x2)  =  sublist_in_ga(x1)
U1_ga(x1, x2, x3)  =  U1_ga(x3)
append_in_aga(x1, x2, x3)  =  append_in_aga(x2)
append_out_aga(x1, x2, x3)  =  append_out_aga
U3_aga(x1, x2, x3, x4, x5)  =  U3_aga(x5)
U2_ga(x1, x2, x3)  =  U2_ga(x3)
append_in_aaa(x1, x2, x3)  =  append_in_aaa
append_out_aaa(x1, x2, x3)  =  append_out_aaa
U3_aaa(x1, x2, x3, x4, x5)  =  U3_aaa(x5)
sublist_out_ga(x1, x2)  =  sublist_out_ga
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
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
              ↳ PiDP
  ↳ PrologToPiTRSProof

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

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

R is empty.
The argument filtering Pi contains the following mapping:
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
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ NonTerminationProof
              ↳ PiDP
  ↳ 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:



Rewriting sequence

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





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

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

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

The TRS R consists of the following rules:

sublist_in_ga(X, Y) → U1_ga(X, Y, append_in_aga(U, X, V))
append_in_aga([], Ys, Ys) → append_out_aga([], Ys, Ys)
append_in_aga(.(X, Xs), Ys, .(X, Zs)) → U3_aga(X, Xs, Ys, Zs, append_in_aga(Xs, Ys, Zs))
U3_aga(X, Xs, Ys, Zs, append_out_aga(Xs, Ys, Zs)) → append_out_aga(.(X, Xs), Ys, .(X, Zs))
U1_ga(X, Y, append_out_aga(U, X, V)) → U2_ga(X, Y, append_in_aaa(V, W, Y))
append_in_aaa([], Ys, Ys) → append_out_aaa([], Ys, Ys)
append_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U3_aaa(X, Xs, Ys, Zs, append_in_aaa(Xs, Ys, Zs))
U3_aaa(X, Xs, Ys, Zs, append_out_aaa(Xs, Ys, Zs)) → append_out_aaa(.(X, Xs), Ys, .(X, Zs))
U2_ga(X, Y, append_out_aaa(V, W, Y)) → sublist_out_ga(X, Y)

The argument filtering Pi contains the following mapping:
sublist_in_ga(x1, x2)  =  sublist_in_ga(x1)
U1_ga(x1, x2, x3)  =  U1_ga(x3)
append_in_aga(x1, x2, x3)  =  append_in_aga(x2)
append_out_aga(x1, x2, x3)  =  append_out_aga
U3_aga(x1, x2, x3, x4, x5)  =  U3_aga(x5)
U2_ga(x1, x2, x3)  =  U2_ga(x3)
append_in_aaa(x1, x2, x3)  =  append_in_aaa
append_out_aaa(x1, x2, x3)  =  append_out_aaa
U3_aaa(x1, x2, x3, x4, x5)  =  U3_aaa(x5)
sublist_out_ga(x1, x2)  =  sublist_out_ga
APPEND_IN_AGA(x1, x2, x3)  =  APPEND_IN_AGA(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
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
  ↳ PrologToPiTRSProof

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

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

R is empty.
The argument filtering Pi contains the following mapping:
APPEND_IN_AGA(x1, x2, x3)  =  APPEND_IN_AGA(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
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ NonTerminationProof
  ↳ PrologToPiTRSProof

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

APPEND_IN_AGA(Ys) → APPEND_IN_AGA(Ys)

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_AGA(Ys) → APPEND_IN_AGA(Ys)

The TRS R consists of the following rules:none


s = APPEND_IN_AGA(Ys) evaluates to t =APPEND_IN_AGA(Ys)

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_AGA(Ys) to APPEND_IN_AGA(Ys).




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

sublist_in_ga(X, Y) → U1_ga(X, Y, append_in_aga(U, X, V))
append_in_aga([], Ys, Ys) → append_out_aga([], Ys, Ys)
append_in_aga(.(X, Xs), Ys, .(X, Zs)) → U3_aga(X, Xs, Ys, Zs, append_in_aga(Xs, Ys, Zs))
U3_aga(X, Xs, Ys, Zs, append_out_aga(Xs, Ys, Zs)) → append_out_aga(.(X, Xs), Ys, .(X, Zs))
U1_ga(X, Y, append_out_aga(U, X, V)) → U2_ga(X, Y, append_in_aaa(V, W, Y))
append_in_aaa([], Ys, Ys) → append_out_aaa([], Ys, Ys)
append_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U3_aaa(X, Xs, Ys, Zs, append_in_aaa(Xs, Ys, Zs))
U3_aaa(X, Xs, Ys, Zs, append_out_aaa(Xs, Ys, Zs)) → append_out_aaa(.(X, Xs), Ys, .(X, Zs))
U2_ga(X, Y, append_out_aaa(V, W, Y)) → sublist_out_ga(X, Y)

The argument filtering Pi contains the following mapping:
sublist_in_ga(x1, x2)  =  sublist_in_ga(x1)
U1_ga(x1, x2, x3)  =  U1_ga(x1, x3)
append_in_aga(x1, x2, x3)  =  append_in_aga(x2)
append_out_aga(x1, x2, x3)  =  append_out_aga(x2)
U3_aga(x1, x2, x3, x4, x5)  =  U3_aga(x3, x5)
U2_ga(x1, x2, x3)  =  U2_ga(x1, x3)
append_in_aaa(x1, x2, x3)  =  append_in_aaa
append_out_aaa(x1, x2, x3)  =  append_out_aaa
U3_aaa(x1, x2, x3, x4, x5)  =  U3_aaa(x5)
sublist_out_ga(x1, x2)  =  sublist_out_ga(x1)

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:

sublist_in_ga(X, Y) → U1_ga(X, Y, append_in_aga(U, X, V))
append_in_aga([], Ys, Ys) → append_out_aga([], Ys, Ys)
append_in_aga(.(X, Xs), Ys, .(X, Zs)) → U3_aga(X, Xs, Ys, Zs, append_in_aga(Xs, Ys, Zs))
U3_aga(X, Xs, Ys, Zs, append_out_aga(Xs, Ys, Zs)) → append_out_aga(.(X, Xs), Ys, .(X, Zs))
U1_ga(X, Y, append_out_aga(U, X, V)) → U2_ga(X, Y, append_in_aaa(V, W, Y))
append_in_aaa([], Ys, Ys) → append_out_aaa([], Ys, Ys)
append_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U3_aaa(X, Xs, Ys, Zs, append_in_aaa(Xs, Ys, Zs))
U3_aaa(X, Xs, Ys, Zs, append_out_aaa(Xs, Ys, Zs)) → append_out_aaa(.(X, Xs), Ys, .(X, Zs))
U2_ga(X, Y, append_out_aaa(V, W, Y)) → sublist_out_ga(X, Y)

The argument filtering Pi contains the following mapping:
sublist_in_ga(x1, x2)  =  sublist_in_ga(x1)
U1_ga(x1, x2, x3)  =  U1_ga(x1, x3)
append_in_aga(x1, x2, x3)  =  append_in_aga(x2)
append_out_aga(x1, x2, x3)  =  append_out_aga(x2)
U3_aga(x1, x2, x3, x4, x5)  =  U3_aga(x3, x5)
U2_ga(x1, x2, x3)  =  U2_ga(x1, x3)
append_in_aaa(x1, x2, x3)  =  append_in_aaa
append_out_aaa(x1, x2, x3)  =  append_out_aaa
U3_aaa(x1, x2, x3, x4, x5)  =  U3_aaa(x5)
sublist_out_ga(x1, x2)  =  sublist_out_ga(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:

SUBLIST_IN_GA(X, Y) → U1_GA(X, Y, append_in_aga(U, X, V))
SUBLIST_IN_GA(X, Y) → APPEND_IN_AGA(U, X, V)
APPEND_IN_AGA(.(X, Xs), Ys, .(X, Zs)) → U3_AGA(X, Xs, Ys, Zs, append_in_aga(Xs, Ys, Zs))
APPEND_IN_AGA(.(X, Xs), Ys, .(X, Zs)) → APPEND_IN_AGA(Xs, Ys, Zs)
U1_GA(X, Y, append_out_aga(U, X, V)) → U2_GA(X, Y, append_in_aaa(V, W, Y))
U1_GA(X, Y, append_out_aga(U, X, V)) → APPEND_IN_AAA(V, W, Y)
APPEND_IN_AAA(.(X, Xs), Ys, .(X, Zs)) → U3_AAA(X, Xs, Ys, Zs, append_in_aaa(Xs, Ys, Zs))
APPEND_IN_AAA(.(X, Xs), Ys, .(X, Zs)) → APPEND_IN_AAA(Xs, Ys, Zs)

The TRS R consists of the following rules:

sublist_in_ga(X, Y) → U1_ga(X, Y, append_in_aga(U, X, V))
append_in_aga([], Ys, Ys) → append_out_aga([], Ys, Ys)
append_in_aga(.(X, Xs), Ys, .(X, Zs)) → U3_aga(X, Xs, Ys, Zs, append_in_aga(Xs, Ys, Zs))
U3_aga(X, Xs, Ys, Zs, append_out_aga(Xs, Ys, Zs)) → append_out_aga(.(X, Xs), Ys, .(X, Zs))
U1_ga(X, Y, append_out_aga(U, X, V)) → U2_ga(X, Y, append_in_aaa(V, W, Y))
append_in_aaa([], Ys, Ys) → append_out_aaa([], Ys, Ys)
append_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U3_aaa(X, Xs, Ys, Zs, append_in_aaa(Xs, Ys, Zs))
U3_aaa(X, Xs, Ys, Zs, append_out_aaa(Xs, Ys, Zs)) → append_out_aaa(.(X, Xs), Ys, .(X, Zs))
U2_ga(X, Y, append_out_aaa(V, W, Y)) → sublist_out_ga(X, Y)

The argument filtering Pi contains the following mapping:
sublist_in_ga(x1, x2)  =  sublist_in_ga(x1)
U1_ga(x1, x2, x3)  =  U1_ga(x1, x3)
append_in_aga(x1, x2, x3)  =  append_in_aga(x2)
append_out_aga(x1, x2, x3)  =  append_out_aga(x2)
U3_aga(x1, x2, x3, x4, x5)  =  U3_aga(x3, x5)
U2_ga(x1, x2, x3)  =  U2_ga(x1, x3)
append_in_aaa(x1, x2, x3)  =  append_in_aaa
append_out_aaa(x1, x2, x3)  =  append_out_aaa
U3_aaa(x1, x2, x3, x4, x5)  =  U3_aaa(x5)
sublist_out_ga(x1, x2)  =  sublist_out_ga(x1)
SUBLIST_IN_GA(x1, x2)  =  SUBLIST_IN_GA(x1)
APPEND_IN_AGA(x1, x2, x3)  =  APPEND_IN_AGA(x2)
U2_GA(x1, x2, x3)  =  U2_GA(x1, x3)
U1_GA(x1, x2, x3)  =  U1_GA(x1, x3)
APPEND_IN_AAA(x1, x2, x3)  =  APPEND_IN_AAA
U3_AAA(x1, x2, x3, x4, x5)  =  U3_AAA(x5)
U3_AGA(x1, x2, x3, x4, x5)  =  U3_AGA(x3, 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:

SUBLIST_IN_GA(X, Y) → U1_GA(X, Y, append_in_aga(U, X, V))
SUBLIST_IN_GA(X, Y) → APPEND_IN_AGA(U, X, V)
APPEND_IN_AGA(.(X, Xs), Ys, .(X, Zs)) → U3_AGA(X, Xs, Ys, Zs, append_in_aga(Xs, Ys, Zs))
APPEND_IN_AGA(.(X, Xs), Ys, .(X, Zs)) → APPEND_IN_AGA(Xs, Ys, Zs)
U1_GA(X, Y, append_out_aga(U, X, V)) → U2_GA(X, Y, append_in_aaa(V, W, Y))
U1_GA(X, Y, append_out_aga(U, X, V)) → APPEND_IN_AAA(V, W, Y)
APPEND_IN_AAA(.(X, Xs), Ys, .(X, Zs)) → U3_AAA(X, Xs, Ys, Zs, append_in_aaa(Xs, Ys, Zs))
APPEND_IN_AAA(.(X, Xs), Ys, .(X, Zs)) → APPEND_IN_AAA(Xs, Ys, Zs)

The TRS R consists of the following rules:

sublist_in_ga(X, Y) → U1_ga(X, Y, append_in_aga(U, X, V))
append_in_aga([], Ys, Ys) → append_out_aga([], Ys, Ys)
append_in_aga(.(X, Xs), Ys, .(X, Zs)) → U3_aga(X, Xs, Ys, Zs, append_in_aga(Xs, Ys, Zs))
U3_aga(X, Xs, Ys, Zs, append_out_aga(Xs, Ys, Zs)) → append_out_aga(.(X, Xs), Ys, .(X, Zs))
U1_ga(X, Y, append_out_aga(U, X, V)) → U2_ga(X, Y, append_in_aaa(V, W, Y))
append_in_aaa([], Ys, Ys) → append_out_aaa([], Ys, Ys)
append_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U3_aaa(X, Xs, Ys, Zs, append_in_aaa(Xs, Ys, Zs))
U3_aaa(X, Xs, Ys, Zs, append_out_aaa(Xs, Ys, Zs)) → append_out_aaa(.(X, Xs), Ys, .(X, Zs))
U2_ga(X, Y, append_out_aaa(V, W, Y)) → sublist_out_ga(X, Y)

The argument filtering Pi contains the following mapping:
sublist_in_ga(x1, x2)  =  sublist_in_ga(x1)
U1_ga(x1, x2, x3)  =  U1_ga(x1, x3)
append_in_aga(x1, x2, x3)  =  append_in_aga(x2)
append_out_aga(x1, x2, x3)  =  append_out_aga(x2)
U3_aga(x1, x2, x3, x4, x5)  =  U3_aga(x3, x5)
U2_ga(x1, x2, x3)  =  U2_ga(x1, x3)
append_in_aaa(x1, x2, x3)  =  append_in_aaa
append_out_aaa(x1, x2, x3)  =  append_out_aaa
U3_aaa(x1, x2, x3, x4, x5)  =  U3_aaa(x5)
sublist_out_ga(x1, x2)  =  sublist_out_ga(x1)
SUBLIST_IN_GA(x1, x2)  =  SUBLIST_IN_GA(x1)
APPEND_IN_AGA(x1, x2, x3)  =  APPEND_IN_AGA(x2)
U2_GA(x1, x2, x3)  =  U2_GA(x1, x3)
U1_GA(x1, x2, x3)  =  U1_GA(x1, x3)
APPEND_IN_AAA(x1, x2, x3)  =  APPEND_IN_AAA
U3_AAA(x1, x2, x3, x4, x5)  =  U3_AAA(x5)
U3_AGA(x1, x2, x3, x4, x5)  =  U3_AGA(x3, x5)

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_AAA(.(X, Xs), Ys, .(X, Zs)) → APPEND_IN_AAA(Xs, Ys, Zs)

The TRS R consists of the following rules:

sublist_in_ga(X, Y) → U1_ga(X, Y, append_in_aga(U, X, V))
append_in_aga([], Ys, Ys) → append_out_aga([], Ys, Ys)
append_in_aga(.(X, Xs), Ys, .(X, Zs)) → U3_aga(X, Xs, Ys, Zs, append_in_aga(Xs, Ys, Zs))
U3_aga(X, Xs, Ys, Zs, append_out_aga(Xs, Ys, Zs)) → append_out_aga(.(X, Xs), Ys, .(X, Zs))
U1_ga(X, Y, append_out_aga(U, X, V)) → U2_ga(X, Y, append_in_aaa(V, W, Y))
append_in_aaa([], Ys, Ys) → append_out_aaa([], Ys, Ys)
append_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U3_aaa(X, Xs, Ys, Zs, append_in_aaa(Xs, Ys, Zs))
U3_aaa(X, Xs, Ys, Zs, append_out_aaa(Xs, Ys, Zs)) → append_out_aaa(.(X, Xs), Ys, .(X, Zs))
U2_ga(X, Y, append_out_aaa(V, W, Y)) → sublist_out_ga(X, Y)

The argument filtering Pi contains the following mapping:
sublist_in_ga(x1, x2)  =  sublist_in_ga(x1)
U1_ga(x1, x2, x3)  =  U1_ga(x1, x3)
append_in_aga(x1, x2, x3)  =  append_in_aga(x2)
append_out_aga(x1, x2, x3)  =  append_out_aga(x2)
U3_aga(x1, x2, x3, x4, x5)  =  U3_aga(x3, x5)
U2_ga(x1, x2, x3)  =  U2_ga(x1, x3)
append_in_aaa(x1, x2, x3)  =  append_in_aaa
append_out_aaa(x1, x2, x3)  =  append_out_aaa
U3_aaa(x1, x2, x3, x4, x5)  =  U3_aaa(x5)
sublist_out_ga(x1, x2)  =  sublist_out_ga(x1)
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
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
              ↳ PiDP

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

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

R is empty.
The argument filtering Pi contains the following mapping:
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
  ↳ 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_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:



Rewriting sequence

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





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

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

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

The TRS R consists of the following rules:

sublist_in_ga(X, Y) → U1_ga(X, Y, append_in_aga(U, X, V))
append_in_aga([], Ys, Ys) → append_out_aga([], Ys, Ys)
append_in_aga(.(X, Xs), Ys, .(X, Zs)) → U3_aga(X, Xs, Ys, Zs, append_in_aga(Xs, Ys, Zs))
U3_aga(X, Xs, Ys, Zs, append_out_aga(Xs, Ys, Zs)) → append_out_aga(.(X, Xs), Ys, .(X, Zs))
U1_ga(X, Y, append_out_aga(U, X, V)) → U2_ga(X, Y, append_in_aaa(V, W, Y))
append_in_aaa([], Ys, Ys) → append_out_aaa([], Ys, Ys)
append_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U3_aaa(X, Xs, Ys, Zs, append_in_aaa(Xs, Ys, Zs))
U3_aaa(X, Xs, Ys, Zs, append_out_aaa(Xs, Ys, Zs)) → append_out_aaa(.(X, Xs), Ys, .(X, Zs))
U2_ga(X, Y, append_out_aaa(V, W, Y)) → sublist_out_ga(X, Y)

The argument filtering Pi contains the following mapping:
sublist_in_ga(x1, x2)  =  sublist_in_ga(x1)
U1_ga(x1, x2, x3)  =  U1_ga(x1, x3)
append_in_aga(x1, x2, x3)  =  append_in_aga(x2)
append_out_aga(x1, x2, x3)  =  append_out_aga(x2)
U3_aga(x1, x2, x3, x4, x5)  =  U3_aga(x3, x5)
U2_ga(x1, x2, x3)  =  U2_ga(x1, x3)
append_in_aaa(x1, x2, x3)  =  append_in_aaa
append_out_aaa(x1, x2, x3)  =  append_out_aaa
U3_aaa(x1, x2, x3, x4, x5)  =  U3_aaa(x5)
sublist_out_ga(x1, x2)  =  sublist_out_ga(x1)
APPEND_IN_AGA(x1, x2, x3)  =  APPEND_IN_AGA(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
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof

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

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

R is empty.
The argument filtering Pi contains the following mapping:
APPEND_IN_AGA(x1, x2, x3)  =  APPEND_IN_AGA(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
  ↳ 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_AGA(Ys) → APPEND_IN_AGA(Ys)

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_AGA(Ys) → APPEND_IN_AGA(Ys)

The TRS R consists of the following rules:none


s = APPEND_IN_AGA(Ys) evaluates to t =APPEND_IN_AGA(Ys)

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_AGA(Ys) to APPEND_IN_AGA(Ys).