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(Xs, Ys) :- ','(app(X, Zs, Ys), app(Xs, X1, Zs)).
app([], X, X).
app(.(X, Xs), Ys, .(X, Zs)) :- app(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)
app_in: (f,f,f) (b,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(Xs, Ys) → U1_ga(Xs, Ys, app_in_aaa(X, Zs, Ys))
app_in_aaa([], X, X) → app_out_aaa([], X, X)
app_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U3_aaa(X, Xs, Ys, Zs, app_in_aaa(Xs, Ys, Zs))
U3_aaa(X, Xs, Ys, Zs, app_out_aaa(Xs, Ys, Zs)) → app_out_aaa(.(X, Xs), Ys, .(X, Zs))
U1_ga(Xs, Ys, app_out_aaa(X, Zs, Ys)) → U2_ga(Xs, Ys, app_in_gaa(Xs, X1, Zs))
app_in_gaa([], X, X) → app_out_gaa([], X, X)
app_in_gaa(.(X, Xs), Ys, .(X, Zs)) → U3_gaa(X, Xs, Ys, Zs, app_in_gaa(Xs, Ys, Zs))
U3_gaa(X, Xs, Ys, Zs, app_out_gaa(Xs, Ys, Zs)) → app_out_gaa(.(X, Xs), Ys, .(X, Zs))
U2_ga(Xs, Ys, app_out_gaa(Xs, X1, Zs)) → sublist_out_ga(Xs, Ys)

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)
app_in_aaa(x1, x2, x3)  =  app_in_aaa
app_out_aaa(x1, x2, x3)  =  app_out_aaa
U3_aaa(x1, x2, x3, x4, x5)  =  U3_aaa(x5)
U2_ga(x1, x2, x3)  =  U2_ga(x3)
app_in_gaa(x1, x2, x3)  =  app_in_gaa(x1)
[]  =  []
app_out_gaa(x1, x2, x3)  =  app_out_gaa
.(x1, x2)  =  .(x1, x2)
U3_gaa(x1, x2, x3, x4, x5)  =  U3_gaa(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(Xs, Ys) → U1_ga(Xs, Ys, app_in_aaa(X, Zs, Ys))
app_in_aaa([], X, X) → app_out_aaa([], X, X)
app_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U3_aaa(X, Xs, Ys, Zs, app_in_aaa(Xs, Ys, Zs))
U3_aaa(X, Xs, Ys, Zs, app_out_aaa(Xs, Ys, Zs)) → app_out_aaa(.(X, Xs), Ys, .(X, Zs))
U1_ga(Xs, Ys, app_out_aaa(X, Zs, Ys)) → U2_ga(Xs, Ys, app_in_gaa(Xs, X1, Zs))
app_in_gaa([], X, X) → app_out_gaa([], X, X)
app_in_gaa(.(X, Xs), Ys, .(X, Zs)) → U3_gaa(X, Xs, Ys, Zs, app_in_gaa(Xs, Ys, Zs))
U3_gaa(X, Xs, Ys, Zs, app_out_gaa(Xs, Ys, Zs)) → app_out_gaa(.(X, Xs), Ys, .(X, Zs))
U2_ga(Xs, Ys, app_out_gaa(Xs, X1, Zs)) → sublist_out_ga(Xs, Ys)

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)
app_in_aaa(x1, x2, x3)  =  app_in_aaa
app_out_aaa(x1, x2, x3)  =  app_out_aaa
U3_aaa(x1, x2, x3, x4, x5)  =  U3_aaa(x5)
U2_ga(x1, x2, x3)  =  U2_ga(x3)
app_in_gaa(x1, x2, x3)  =  app_in_gaa(x1)
[]  =  []
app_out_gaa(x1, x2, x3)  =  app_out_gaa
.(x1, x2)  =  .(x1, x2)
U3_gaa(x1, x2, x3, x4, x5)  =  U3_gaa(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(Xs, Ys) → U1_GA(Xs, Ys, app_in_aaa(X, Zs, Ys))
SUBLIST_IN_GA(Xs, Ys) → APP_IN_AAA(X, Zs, Ys)
APP_IN_AAA(.(X, Xs), Ys, .(X, Zs)) → U3_AAA(X, Xs, Ys, Zs, app_in_aaa(Xs, Ys, Zs))
APP_IN_AAA(.(X, Xs), Ys, .(X, Zs)) → APP_IN_AAA(Xs, Ys, Zs)
U1_GA(Xs, Ys, app_out_aaa(X, Zs, Ys)) → U2_GA(Xs, Ys, app_in_gaa(Xs, X1, Zs))
U1_GA(Xs, Ys, app_out_aaa(X, Zs, Ys)) → APP_IN_GAA(Xs, X1, Zs)
APP_IN_GAA(.(X, Xs), Ys, .(X, Zs)) → U3_GAA(X, Xs, Ys, Zs, app_in_gaa(Xs, Ys, Zs))
APP_IN_GAA(.(X, Xs), Ys, .(X, Zs)) → APP_IN_GAA(Xs, Ys, Zs)

The TRS R consists of the following rules:

sublist_in_ga(Xs, Ys) → U1_ga(Xs, Ys, app_in_aaa(X, Zs, Ys))
app_in_aaa([], X, X) → app_out_aaa([], X, X)
app_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U3_aaa(X, Xs, Ys, Zs, app_in_aaa(Xs, Ys, Zs))
U3_aaa(X, Xs, Ys, Zs, app_out_aaa(Xs, Ys, Zs)) → app_out_aaa(.(X, Xs), Ys, .(X, Zs))
U1_ga(Xs, Ys, app_out_aaa(X, Zs, Ys)) → U2_ga(Xs, Ys, app_in_gaa(Xs, X1, Zs))
app_in_gaa([], X, X) → app_out_gaa([], X, X)
app_in_gaa(.(X, Xs), Ys, .(X, Zs)) → U3_gaa(X, Xs, Ys, Zs, app_in_gaa(Xs, Ys, Zs))
U3_gaa(X, Xs, Ys, Zs, app_out_gaa(Xs, Ys, Zs)) → app_out_gaa(.(X, Xs), Ys, .(X, Zs))
U2_ga(Xs, Ys, app_out_gaa(Xs, X1, Zs)) → sublist_out_ga(Xs, Ys)

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)
app_in_aaa(x1, x2, x3)  =  app_in_aaa
app_out_aaa(x1, x2, x3)  =  app_out_aaa
U3_aaa(x1, x2, x3, x4, x5)  =  U3_aaa(x5)
U2_ga(x1, x2, x3)  =  U2_ga(x3)
app_in_gaa(x1, x2, x3)  =  app_in_gaa(x1)
[]  =  []
app_out_gaa(x1, x2, x3)  =  app_out_gaa
.(x1, x2)  =  .(x1, x2)
U3_gaa(x1, x2, x3, x4, x5)  =  U3_gaa(x5)
sublist_out_ga(x1, x2)  =  sublist_out_ga
APP_IN_AAA(x1, x2, x3)  =  APP_IN_AAA
SUBLIST_IN_GA(x1, x2)  =  SUBLIST_IN_GA(x1)
U3_GAA(x1, x2, x3, x4, x5)  =  U3_GAA(x5)
U2_GA(x1, x2, x3)  =  U2_GA(x3)
U1_GA(x1, x2, x3)  =  U1_GA(x1, x3)
APP_IN_GAA(x1, x2, x3)  =  APP_IN_GAA(x1)
U3_AAA(x1, x2, x3, x4, x5)  =  U3_AAA(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(Xs, Ys) → U1_GA(Xs, Ys, app_in_aaa(X, Zs, Ys))
SUBLIST_IN_GA(Xs, Ys) → APP_IN_AAA(X, Zs, Ys)
APP_IN_AAA(.(X, Xs), Ys, .(X, Zs)) → U3_AAA(X, Xs, Ys, Zs, app_in_aaa(Xs, Ys, Zs))
APP_IN_AAA(.(X, Xs), Ys, .(X, Zs)) → APP_IN_AAA(Xs, Ys, Zs)
U1_GA(Xs, Ys, app_out_aaa(X, Zs, Ys)) → U2_GA(Xs, Ys, app_in_gaa(Xs, X1, Zs))
U1_GA(Xs, Ys, app_out_aaa(X, Zs, Ys)) → APP_IN_GAA(Xs, X1, Zs)
APP_IN_GAA(.(X, Xs), Ys, .(X, Zs)) → U3_GAA(X, Xs, Ys, Zs, app_in_gaa(Xs, Ys, Zs))
APP_IN_GAA(.(X, Xs), Ys, .(X, Zs)) → APP_IN_GAA(Xs, Ys, Zs)

The TRS R consists of the following rules:

sublist_in_ga(Xs, Ys) → U1_ga(Xs, Ys, app_in_aaa(X, Zs, Ys))
app_in_aaa([], X, X) → app_out_aaa([], X, X)
app_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U3_aaa(X, Xs, Ys, Zs, app_in_aaa(Xs, Ys, Zs))
U3_aaa(X, Xs, Ys, Zs, app_out_aaa(Xs, Ys, Zs)) → app_out_aaa(.(X, Xs), Ys, .(X, Zs))
U1_ga(Xs, Ys, app_out_aaa(X, Zs, Ys)) → U2_ga(Xs, Ys, app_in_gaa(Xs, X1, Zs))
app_in_gaa([], X, X) → app_out_gaa([], X, X)
app_in_gaa(.(X, Xs), Ys, .(X, Zs)) → U3_gaa(X, Xs, Ys, Zs, app_in_gaa(Xs, Ys, Zs))
U3_gaa(X, Xs, Ys, Zs, app_out_gaa(Xs, Ys, Zs)) → app_out_gaa(.(X, Xs), Ys, .(X, Zs))
U2_ga(Xs, Ys, app_out_gaa(Xs, X1, Zs)) → sublist_out_ga(Xs, Ys)

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)
app_in_aaa(x1, x2, x3)  =  app_in_aaa
app_out_aaa(x1, x2, x3)  =  app_out_aaa
U3_aaa(x1, x2, x3, x4, x5)  =  U3_aaa(x5)
U2_ga(x1, x2, x3)  =  U2_ga(x3)
app_in_gaa(x1, x2, x3)  =  app_in_gaa(x1)
[]  =  []
app_out_gaa(x1, x2, x3)  =  app_out_gaa
.(x1, x2)  =  .(x1, x2)
U3_gaa(x1, x2, x3, x4, x5)  =  U3_gaa(x5)
sublist_out_ga(x1, x2)  =  sublist_out_ga
APP_IN_AAA(x1, x2, x3)  =  APP_IN_AAA
SUBLIST_IN_GA(x1, x2)  =  SUBLIST_IN_GA(x1)
U3_GAA(x1, x2, x3, x4, x5)  =  U3_GAA(x5)
U2_GA(x1, x2, x3)  =  U2_GA(x3)
U1_GA(x1, x2, x3)  =  U1_GA(x1, x3)
APP_IN_GAA(x1, x2, x3)  =  APP_IN_GAA(x1)
U3_AAA(x1, x2, x3, x4, x5)  =  U3_AAA(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:

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

The TRS R consists of the following rules:

sublist_in_ga(Xs, Ys) → U1_ga(Xs, Ys, app_in_aaa(X, Zs, Ys))
app_in_aaa([], X, X) → app_out_aaa([], X, X)
app_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U3_aaa(X, Xs, Ys, Zs, app_in_aaa(Xs, Ys, Zs))
U3_aaa(X, Xs, Ys, Zs, app_out_aaa(Xs, Ys, Zs)) → app_out_aaa(.(X, Xs), Ys, .(X, Zs))
U1_ga(Xs, Ys, app_out_aaa(X, Zs, Ys)) → U2_ga(Xs, Ys, app_in_gaa(Xs, X1, Zs))
app_in_gaa([], X, X) → app_out_gaa([], X, X)
app_in_gaa(.(X, Xs), Ys, .(X, Zs)) → U3_gaa(X, Xs, Ys, Zs, app_in_gaa(Xs, Ys, Zs))
U3_gaa(X, Xs, Ys, Zs, app_out_gaa(Xs, Ys, Zs)) → app_out_gaa(.(X, Xs), Ys, .(X, Zs))
U2_ga(Xs, Ys, app_out_gaa(Xs, X1, Zs)) → sublist_out_ga(Xs, Ys)

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)
app_in_aaa(x1, x2, x3)  =  app_in_aaa
app_out_aaa(x1, x2, x3)  =  app_out_aaa
U3_aaa(x1, x2, x3, x4, x5)  =  U3_aaa(x5)
U2_ga(x1, x2, x3)  =  U2_ga(x3)
app_in_gaa(x1, x2, x3)  =  app_in_gaa(x1)
[]  =  []
app_out_gaa(x1, x2, x3)  =  app_out_gaa
.(x1, x2)  =  .(x1, x2)
U3_gaa(x1, x2, x3, x4, x5)  =  U3_gaa(x5)
sublist_out_ga(x1, x2)  =  sublist_out_ga
APP_IN_GAA(x1, x2, x3)  =  APP_IN_GAA(x1)

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:

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

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

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:

APP_IN_GAA(.(X, Xs)) → APP_IN_GAA(Xs)

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: 



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

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

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

The TRS R consists of the following rules:

sublist_in_ga(Xs, Ys) → U1_ga(Xs, Ys, app_in_aaa(X, Zs, Ys))
app_in_aaa([], X, X) → app_out_aaa([], X, X)
app_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U3_aaa(X, Xs, Ys, Zs, app_in_aaa(Xs, Ys, Zs))
U3_aaa(X, Xs, Ys, Zs, app_out_aaa(Xs, Ys, Zs)) → app_out_aaa(.(X, Xs), Ys, .(X, Zs))
U1_ga(Xs, Ys, app_out_aaa(X, Zs, Ys)) → U2_ga(Xs, Ys, app_in_gaa(Xs, X1, Zs))
app_in_gaa([], X, X) → app_out_gaa([], X, X)
app_in_gaa(.(X, Xs), Ys, .(X, Zs)) → U3_gaa(X, Xs, Ys, Zs, app_in_gaa(Xs, Ys, Zs))
U3_gaa(X, Xs, Ys, Zs, app_out_gaa(Xs, Ys, Zs)) → app_out_gaa(.(X, Xs), Ys, .(X, Zs))
U2_ga(Xs, Ys, app_out_gaa(Xs, X1, Zs)) → sublist_out_ga(Xs, Ys)

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)
app_in_aaa(x1, x2, x3)  =  app_in_aaa
app_out_aaa(x1, x2, x3)  =  app_out_aaa
U3_aaa(x1, x2, x3, x4, x5)  =  U3_aaa(x5)
U2_ga(x1, x2, x3)  =  U2_ga(x3)
app_in_gaa(x1, x2, x3)  =  app_in_gaa(x1)
[]  =  []
app_out_gaa(x1, x2, x3)  =  app_out_gaa
.(x1, x2)  =  .(x1, x2)
U3_gaa(x1, x2, x3, x4, x5)  =  U3_gaa(x5)
sublist_out_ga(x1, x2)  =  sublist_out_ga
APP_IN_AAA(x1, x2, x3)  =  APP_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:

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

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

APP_IN_AAA → APP_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:

APP_IN_AAA → APP_IN_AAA

The TRS R consists of the following rules:none


s = APP_IN_AAA evaluates to t =APP_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 APP_IN_AAA to APP_IN_AAA.




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)
app_in: (f,f,f) (b,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(Xs, Ys) → U1_ga(Xs, Ys, app_in_aaa(X, Zs, Ys))
app_in_aaa([], X, X) → app_out_aaa([], X, X)
app_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U3_aaa(X, Xs, Ys, Zs, app_in_aaa(Xs, Ys, Zs))
U3_aaa(X, Xs, Ys, Zs, app_out_aaa(Xs, Ys, Zs)) → app_out_aaa(.(X, Xs), Ys, .(X, Zs))
U1_ga(Xs, Ys, app_out_aaa(X, Zs, Ys)) → U2_ga(Xs, Ys, app_in_gaa(Xs, X1, Zs))
app_in_gaa([], X, X) → app_out_gaa([], X, X)
app_in_gaa(.(X, Xs), Ys, .(X, Zs)) → U3_gaa(X, Xs, Ys, Zs, app_in_gaa(Xs, Ys, Zs))
U3_gaa(X, Xs, Ys, Zs, app_out_gaa(Xs, Ys, Zs)) → app_out_gaa(.(X, Xs), Ys, .(X, Zs))
U2_ga(Xs, Ys, app_out_gaa(Xs, X1, Zs)) → sublist_out_ga(Xs, Ys)

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)
app_in_aaa(x1, x2, x3)  =  app_in_aaa
app_out_aaa(x1, x2, x3)  =  app_out_aaa
U3_aaa(x1, x2, x3, x4, x5)  =  U3_aaa(x5)
U2_ga(x1, x2, x3)  =  U2_ga(x1, x3)
app_in_gaa(x1, x2, x3)  =  app_in_gaa(x1)
[]  =  []
app_out_gaa(x1, x2, x3)  =  app_out_gaa(x1)
.(x1, x2)  =  .(x1, x2)
U3_gaa(x1, x2, x3, x4, x5)  =  U3_gaa(x1, x2, 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(Xs, Ys) → U1_ga(Xs, Ys, app_in_aaa(X, Zs, Ys))
app_in_aaa([], X, X) → app_out_aaa([], X, X)
app_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U3_aaa(X, Xs, Ys, Zs, app_in_aaa(Xs, Ys, Zs))
U3_aaa(X, Xs, Ys, Zs, app_out_aaa(Xs, Ys, Zs)) → app_out_aaa(.(X, Xs), Ys, .(X, Zs))
U1_ga(Xs, Ys, app_out_aaa(X, Zs, Ys)) → U2_ga(Xs, Ys, app_in_gaa(Xs, X1, Zs))
app_in_gaa([], X, X) → app_out_gaa([], X, X)
app_in_gaa(.(X, Xs), Ys, .(X, Zs)) → U3_gaa(X, Xs, Ys, Zs, app_in_gaa(Xs, Ys, Zs))
U3_gaa(X, Xs, Ys, Zs, app_out_gaa(Xs, Ys, Zs)) → app_out_gaa(.(X, Xs), Ys, .(X, Zs))
U2_ga(Xs, Ys, app_out_gaa(Xs, X1, Zs)) → sublist_out_ga(Xs, Ys)

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)
app_in_aaa(x1, x2, x3)  =  app_in_aaa
app_out_aaa(x1, x2, x3)  =  app_out_aaa
U3_aaa(x1, x2, x3, x4, x5)  =  U3_aaa(x5)
U2_ga(x1, x2, x3)  =  U2_ga(x1, x3)
app_in_gaa(x1, x2, x3)  =  app_in_gaa(x1)
[]  =  []
app_out_gaa(x1, x2, x3)  =  app_out_gaa(x1)
.(x1, x2)  =  .(x1, x2)
U3_gaa(x1, x2, x3, x4, x5)  =  U3_gaa(x1, x2, 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(Xs, Ys) → U1_GA(Xs, Ys, app_in_aaa(X, Zs, Ys))
SUBLIST_IN_GA(Xs, Ys) → APP_IN_AAA(X, Zs, Ys)
APP_IN_AAA(.(X, Xs), Ys, .(X, Zs)) → U3_AAA(X, Xs, Ys, Zs, app_in_aaa(Xs, Ys, Zs))
APP_IN_AAA(.(X, Xs), Ys, .(X, Zs)) → APP_IN_AAA(Xs, Ys, Zs)
U1_GA(Xs, Ys, app_out_aaa(X, Zs, Ys)) → U2_GA(Xs, Ys, app_in_gaa(Xs, X1, Zs))
U1_GA(Xs, Ys, app_out_aaa(X, Zs, Ys)) → APP_IN_GAA(Xs, X1, Zs)
APP_IN_GAA(.(X, Xs), Ys, .(X, Zs)) → U3_GAA(X, Xs, Ys, Zs, app_in_gaa(Xs, Ys, Zs))
APP_IN_GAA(.(X, Xs), Ys, .(X, Zs)) → APP_IN_GAA(Xs, Ys, Zs)

The TRS R consists of the following rules:

sublist_in_ga(Xs, Ys) → U1_ga(Xs, Ys, app_in_aaa(X, Zs, Ys))
app_in_aaa([], X, X) → app_out_aaa([], X, X)
app_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U3_aaa(X, Xs, Ys, Zs, app_in_aaa(Xs, Ys, Zs))
U3_aaa(X, Xs, Ys, Zs, app_out_aaa(Xs, Ys, Zs)) → app_out_aaa(.(X, Xs), Ys, .(X, Zs))
U1_ga(Xs, Ys, app_out_aaa(X, Zs, Ys)) → U2_ga(Xs, Ys, app_in_gaa(Xs, X1, Zs))
app_in_gaa([], X, X) → app_out_gaa([], X, X)
app_in_gaa(.(X, Xs), Ys, .(X, Zs)) → U3_gaa(X, Xs, Ys, Zs, app_in_gaa(Xs, Ys, Zs))
U3_gaa(X, Xs, Ys, Zs, app_out_gaa(Xs, Ys, Zs)) → app_out_gaa(.(X, Xs), Ys, .(X, Zs))
U2_ga(Xs, Ys, app_out_gaa(Xs, X1, Zs)) → sublist_out_ga(Xs, Ys)

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)
app_in_aaa(x1, x2, x3)  =  app_in_aaa
app_out_aaa(x1, x2, x3)  =  app_out_aaa
U3_aaa(x1, x2, x3, x4, x5)  =  U3_aaa(x5)
U2_ga(x1, x2, x3)  =  U2_ga(x1, x3)
app_in_gaa(x1, x2, x3)  =  app_in_gaa(x1)
[]  =  []
app_out_gaa(x1, x2, x3)  =  app_out_gaa(x1)
.(x1, x2)  =  .(x1, x2)
U3_gaa(x1, x2, x3, x4, x5)  =  U3_gaa(x1, x2, x5)
sublist_out_ga(x1, x2)  =  sublist_out_ga(x1)
APP_IN_AAA(x1, x2, x3)  =  APP_IN_AAA
SUBLIST_IN_GA(x1, x2)  =  SUBLIST_IN_GA(x1)
U3_GAA(x1, x2, x3, x4, x5)  =  U3_GAA(x1, x2, x5)
U2_GA(x1, x2, x3)  =  U2_GA(x1, x3)
U1_GA(x1, x2, x3)  =  U1_GA(x1, x3)
APP_IN_GAA(x1, x2, x3)  =  APP_IN_GAA(x1)
U3_AAA(x1, x2, x3, x4, x5)  =  U3_AAA(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(Xs, Ys) → U1_GA(Xs, Ys, app_in_aaa(X, Zs, Ys))
SUBLIST_IN_GA(Xs, Ys) → APP_IN_AAA(X, Zs, Ys)
APP_IN_AAA(.(X, Xs), Ys, .(X, Zs)) → U3_AAA(X, Xs, Ys, Zs, app_in_aaa(Xs, Ys, Zs))
APP_IN_AAA(.(X, Xs), Ys, .(X, Zs)) → APP_IN_AAA(Xs, Ys, Zs)
U1_GA(Xs, Ys, app_out_aaa(X, Zs, Ys)) → U2_GA(Xs, Ys, app_in_gaa(Xs, X1, Zs))
U1_GA(Xs, Ys, app_out_aaa(X, Zs, Ys)) → APP_IN_GAA(Xs, X1, Zs)
APP_IN_GAA(.(X, Xs), Ys, .(X, Zs)) → U3_GAA(X, Xs, Ys, Zs, app_in_gaa(Xs, Ys, Zs))
APP_IN_GAA(.(X, Xs), Ys, .(X, Zs)) → APP_IN_GAA(Xs, Ys, Zs)

The TRS R consists of the following rules:

sublist_in_ga(Xs, Ys) → U1_ga(Xs, Ys, app_in_aaa(X, Zs, Ys))
app_in_aaa([], X, X) → app_out_aaa([], X, X)
app_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U3_aaa(X, Xs, Ys, Zs, app_in_aaa(Xs, Ys, Zs))
U3_aaa(X, Xs, Ys, Zs, app_out_aaa(Xs, Ys, Zs)) → app_out_aaa(.(X, Xs), Ys, .(X, Zs))
U1_ga(Xs, Ys, app_out_aaa(X, Zs, Ys)) → U2_ga(Xs, Ys, app_in_gaa(Xs, X1, Zs))
app_in_gaa([], X, X) → app_out_gaa([], X, X)
app_in_gaa(.(X, Xs), Ys, .(X, Zs)) → U3_gaa(X, Xs, Ys, Zs, app_in_gaa(Xs, Ys, Zs))
U3_gaa(X, Xs, Ys, Zs, app_out_gaa(Xs, Ys, Zs)) → app_out_gaa(.(X, Xs), Ys, .(X, Zs))
U2_ga(Xs, Ys, app_out_gaa(Xs, X1, Zs)) → sublist_out_ga(Xs, Ys)

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)
app_in_aaa(x1, x2, x3)  =  app_in_aaa
app_out_aaa(x1, x2, x3)  =  app_out_aaa
U3_aaa(x1, x2, x3, x4, x5)  =  U3_aaa(x5)
U2_ga(x1, x2, x3)  =  U2_ga(x1, x3)
app_in_gaa(x1, x2, x3)  =  app_in_gaa(x1)
[]  =  []
app_out_gaa(x1, x2, x3)  =  app_out_gaa(x1)
.(x1, x2)  =  .(x1, x2)
U3_gaa(x1, x2, x3, x4, x5)  =  U3_gaa(x1, x2, x5)
sublist_out_ga(x1, x2)  =  sublist_out_ga(x1)
APP_IN_AAA(x1, x2, x3)  =  APP_IN_AAA
SUBLIST_IN_GA(x1, x2)  =  SUBLIST_IN_GA(x1)
U3_GAA(x1, x2, x3, x4, x5)  =  U3_GAA(x1, x2, x5)
U2_GA(x1, x2, x3)  =  U2_GA(x1, x3)
U1_GA(x1, x2, x3)  =  U1_GA(x1, x3)
APP_IN_GAA(x1, x2, x3)  =  APP_IN_GAA(x1)
U3_AAA(x1, x2, x3, x4, x5)  =  U3_AAA(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:

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

The TRS R consists of the following rules:

sublist_in_ga(Xs, Ys) → U1_ga(Xs, Ys, app_in_aaa(X, Zs, Ys))
app_in_aaa([], X, X) → app_out_aaa([], X, X)
app_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U3_aaa(X, Xs, Ys, Zs, app_in_aaa(Xs, Ys, Zs))
U3_aaa(X, Xs, Ys, Zs, app_out_aaa(Xs, Ys, Zs)) → app_out_aaa(.(X, Xs), Ys, .(X, Zs))
U1_ga(Xs, Ys, app_out_aaa(X, Zs, Ys)) → U2_ga(Xs, Ys, app_in_gaa(Xs, X1, Zs))
app_in_gaa([], X, X) → app_out_gaa([], X, X)
app_in_gaa(.(X, Xs), Ys, .(X, Zs)) → U3_gaa(X, Xs, Ys, Zs, app_in_gaa(Xs, Ys, Zs))
U3_gaa(X, Xs, Ys, Zs, app_out_gaa(Xs, Ys, Zs)) → app_out_gaa(.(X, Xs), Ys, .(X, Zs))
U2_ga(Xs, Ys, app_out_gaa(Xs, X1, Zs)) → sublist_out_ga(Xs, Ys)

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)
app_in_aaa(x1, x2, x3)  =  app_in_aaa
app_out_aaa(x1, x2, x3)  =  app_out_aaa
U3_aaa(x1, x2, x3, x4, x5)  =  U3_aaa(x5)
U2_ga(x1, x2, x3)  =  U2_ga(x1, x3)
app_in_gaa(x1, x2, x3)  =  app_in_gaa(x1)
[]  =  []
app_out_gaa(x1, x2, x3)  =  app_out_gaa(x1)
.(x1, x2)  =  .(x1, x2)
U3_gaa(x1, x2, x3, x4, x5)  =  U3_gaa(x1, x2, x5)
sublist_out_ga(x1, x2)  =  sublist_out_ga(x1)
APP_IN_GAA(x1, x2, x3)  =  APP_IN_GAA(x1)

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:

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

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

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
                        ↳ QDPSizeChangeProof
              ↳ PiDP

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

APP_IN_GAA(.(X, Xs)) → APP_IN_GAA(Xs)

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: 



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

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

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

The TRS R consists of the following rules:

sublist_in_ga(Xs, Ys) → U1_ga(Xs, Ys, app_in_aaa(X, Zs, Ys))
app_in_aaa([], X, X) → app_out_aaa([], X, X)
app_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U3_aaa(X, Xs, Ys, Zs, app_in_aaa(Xs, Ys, Zs))
U3_aaa(X, Xs, Ys, Zs, app_out_aaa(Xs, Ys, Zs)) → app_out_aaa(.(X, Xs), Ys, .(X, Zs))
U1_ga(Xs, Ys, app_out_aaa(X, Zs, Ys)) → U2_ga(Xs, Ys, app_in_gaa(Xs, X1, Zs))
app_in_gaa([], X, X) → app_out_gaa([], X, X)
app_in_gaa(.(X, Xs), Ys, .(X, Zs)) → U3_gaa(X, Xs, Ys, Zs, app_in_gaa(Xs, Ys, Zs))
U3_gaa(X, Xs, Ys, Zs, app_out_gaa(Xs, Ys, Zs)) → app_out_gaa(.(X, Xs), Ys, .(X, Zs))
U2_ga(Xs, Ys, app_out_gaa(Xs, X1, Zs)) → sublist_out_ga(Xs, Ys)

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)
app_in_aaa(x1, x2, x3)  =  app_in_aaa
app_out_aaa(x1, x2, x3)  =  app_out_aaa
U3_aaa(x1, x2, x3, x4, x5)  =  U3_aaa(x5)
U2_ga(x1, x2, x3)  =  U2_ga(x1, x3)
app_in_gaa(x1, x2, x3)  =  app_in_gaa(x1)
[]  =  []
app_out_gaa(x1, x2, x3)  =  app_out_gaa(x1)
.(x1, x2)  =  .(x1, x2)
U3_gaa(x1, x2, x3, x4, x5)  =  U3_gaa(x1, x2, x5)
sublist_out_ga(x1, x2)  =  sublist_out_ga(x1)
APP_IN_AAA(x1, x2, x3)  =  APP_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:

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

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

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

APP_IN_AAA → APP_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:

APP_IN_AAA → APP_IN_AAA

The TRS R consists of the following rules:none


s = APP_IN_AAA evaluates to t =APP_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 APP_IN_AAA to APP_IN_AAA.