Left Termination of the query pattern sublist_in_2(g, g) w.r.t. the given Prolog program could successfully be proven:



↳ Prolog
  ↳ PrologToPiTRSProof

Clauses:

append1([], Ys, Ys).
append1(.(X, Xs), Ys, .(X, Zs)) :- append1(Xs, Ys, Zs).
append2([], Ys, Ys).
append2(.(X, Xs), Ys, .(X, Zs)) :- append2(Xs, Ys, Zs).
sublist(X, Y) :- ','(append1(P, X1, Y), append2(X2, X, P)).

Queries:

sublist(g,g).

We use the technique of [30]. With regard to the inferred argument filtering the predicates were used in the following modes:
sublist_in: (b,b)
append1_in: (f,f,b)
append2_in: (f,b,b)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

sublist_in_gg(X, Y) → U3_gg(X, Y, append1_in_aag(P, X1, Y))
append1_in_aag([], Ys, Ys) → append1_out_aag([], Ys, Ys)
append1_in_aag(.(X, Xs), Ys, .(X, Zs)) → U1_aag(X, Xs, Ys, Zs, append1_in_aag(Xs, Ys, Zs))
U1_aag(X, Xs, Ys, Zs, append1_out_aag(Xs, Ys, Zs)) → append1_out_aag(.(X, Xs), Ys, .(X, Zs))
U3_gg(X, Y, append1_out_aag(P, X1, Y)) → U4_gg(X, Y, append2_in_agg(X2, X, P))
append2_in_agg([], Ys, Ys) → append2_out_agg([], Ys, Ys)
append2_in_agg(.(X, Xs), Ys, .(X, Zs)) → U2_agg(X, Xs, Ys, Zs, append2_in_agg(Xs, Ys, Zs))
U2_agg(X, Xs, Ys, Zs, append2_out_agg(Xs, Ys, Zs)) → append2_out_agg(.(X, Xs), Ys, .(X, Zs))
U4_gg(X, Y, append2_out_agg(X2, X, P)) → sublist_out_gg(X, Y)

The argument filtering Pi contains the following mapping:
sublist_in_gg(x1, x2)  =  sublist_in_gg(x1, x2)
U3_gg(x1, x2, x3)  =  U3_gg(x1, x3)
append1_in_aag(x1, x2, x3)  =  append1_in_aag(x3)
append1_out_aag(x1, x2, x3)  =  append1_out_aag(x1, x2)
.(x1, x2)  =  .(x1, x2)
U1_aag(x1, x2, x3, x4, x5)  =  U1_aag(x1, x5)
U4_gg(x1, x2, x3)  =  U4_gg(x3)
append2_in_agg(x1, x2, x3)  =  append2_in_agg(x2, x3)
append2_out_agg(x1, x2, x3)  =  append2_out_agg(x1)
U2_agg(x1, x2, x3, x4, x5)  =  U2_agg(x1, x5)
sublist_out_gg(x1, x2)  =  sublist_out_gg

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog



↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof

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

sublist_in_gg(X, Y) → U3_gg(X, Y, append1_in_aag(P, X1, Y))
append1_in_aag([], Ys, Ys) → append1_out_aag([], Ys, Ys)
append1_in_aag(.(X, Xs), Ys, .(X, Zs)) → U1_aag(X, Xs, Ys, Zs, append1_in_aag(Xs, Ys, Zs))
U1_aag(X, Xs, Ys, Zs, append1_out_aag(Xs, Ys, Zs)) → append1_out_aag(.(X, Xs), Ys, .(X, Zs))
U3_gg(X, Y, append1_out_aag(P, X1, Y)) → U4_gg(X, Y, append2_in_agg(X2, X, P))
append2_in_agg([], Ys, Ys) → append2_out_agg([], Ys, Ys)
append2_in_agg(.(X, Xs), Ys, .(X, Zs)) → U2_agg(X, Xs, Ys, Zs, append2_in_agg(Xs, Ys, Zs))
U2_agg(X, Xs, Ys, Zs, append2_out_agg(Xs, Ys, Zs)) → append2_out_agg(.(X, Xs), Ys, .(X, Zs))
U4_gg(X, Y, append2_out_agg(X2, X, P)) → sublist_out_gg(X, Y)

The argument filtering Pi contains the following mapping:
sublist_in_gg(x1, x2)  =  sublist_in_gg(x1, x2)
U3_gg(x1, x2, x3)  =  U3_gg(x1, x3)
append1_in_aag(x1, x2, x3)  =  append1_in_aag(x3)
append1_out_aag(x1, x2, x3)  =  append1_out_aag(x1, x2)
.(x1, x2)  =  .(x1, x2)
U1_aag(x1, x2, x3, x4, x5)  =  U1_aag(x1, x5)
U4_gg(x1, x2, x3)  =  U4_gg(x3)
append2_in_agg(x1, x2, x3)  =  append2_in_agg(x2, x3)
append2_out_agg(x1, x2, x3)  =  append2_out_agg(x1)
U2_agg(x1, x2, x3, x4, x5)  =  U2_agg(x1, x5)
sublist_out_gg(x1, x2)  =  sublist_out_gg


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_GG(X, Y) → U3_GG(X, Y, append1_in_aag(P, X1, Y))
SUBLIST_IN_GG(X, Y) → APPEND1_IN_AAG(P, X1, Y)
APPEND1_IN_AAG(.(X, Xs), Ys, .(X, Zs)) → U1_AAG(X, Xs, Ys, Zs, append1_in_aag(Xs, Ys, Zs))
APPEND1_IN_AAG(.(X, Xs), Ys, .(X, Zs)) → APPEND1_IN_AAG(Xs, Ys, Zs)
U3_GG(X, Y, append1_out_aag(P, X1, Y)) → U4_GG(X, Y, append2_in_agg(X2, X, P))
U3_GG(X, Y, append1_out_aag(P, X1, Y)) → APPEND2_IN_AGG(X2, X, P)
APPEND2_IN_AGG(.(X, Xs), Ys, .(X, Zs)) → U2_AGG(X, Xs, Ys, Zs, append2_in_agg(Xs, Ys, Zs))
APPEND2_IN_AGG(.(X, Xs), Ys, .(X, Zs)) → APPEND2_IN_AGG(Xs, Ys, Zs)

The TRS R consists of the following rules:

sublist_in_gg(X, Y) → U3_gg(X, Y, append1_in_aag(P, X1, Y))
append1_in_aag([], Ys, Ys) → append1_out_aag([], Ys, Ys)
append1_in_aag(.(X, Xs), Ys, .(X, Zs)) → U1_aag(X, Xs, Ys, Zs, append1_in_aag(Xs, Ys, Zs))
U1_aag(X, Xs, Ys, Zs, append1_out_aag(Xs, Ys, Zs)) → append1_out_aag(.(X, Xs), Ys, .(X, Zs))
U3_gg(X, Y, append1_out_aag(P, X1, Y)) → U4_gg(X, Y, append2_in_agg(X2, X, P))
append2_in_agg([], Ys, Ys) → append2_out_agg([], Ys, Ys)
append2_in_agg(.(X, Xs), Ys, .(X, Zs)) → U2_agg(X, Xs, Ys, Zs, append2_in_agg(Xs, Ys, Zs))
U2_agg(X, Xs, Ys, Zs, append2_out_agg(Xs, Ys, Zs)) → append2_out_agg(.(X, Xs), Ys, .(X, Zs))
U4_gg(X, Y, append2_out_agg(X2, X, P)) → sublist_out_gg(X, Y)

The argument filtering Pi contains the following mapping:
sublist_in_gg(x1, x2)  =  sublist_in_gg(x1, x2)
U3_gg(x1, x2, x3)  =  U3_gg(x1, x3)
append1_in_aag(x1, x2, x3)  =  append1_in_aag(x3)
append1_out_aag(x1, x2, x3)  =  append1_out_aag(x1, x2)
.(x1, x2)  =  .(x1, x2)
U1_aag(x1, x2, x3, x4, x5)  =  U1_aag(x1, x5)
U4_gg(x1, x2, x3)  =  U4_gg(x3)
append2_in_agg(x1, x2, x3)  =  append2_in_agg(x2, x3)
append2_out_agg(x1, x2, x3)  =  append2_out_agg(x1)
U2_agg(x1, x2, x3, x4, x5)  =  U2_agg(x1, x5)
sublist_out_gg(x1, x2)  =  sublist_out_gg
U4_GG(x1, x2, x3)  =  U4_GG(x3)
SUBLIST_IN_GG(x1, x2)  =  SUBLIST_IN_GG(x1, x2)
U1_AAG(x1, x2, x3, x4, x5)  =  U1_AAG(x1, x5)
U2_AGG(x1, x2, x3, x4, x5)  =  U2_AGG(x1, x5)
APPEND2_IN_AGG(x1, x2, x3)  =  APPEND2_IN_AGG(x2, x3)
APPEND1_IN_AAG(x1, x2, x3)  =  APPEND1_IN_AAG(x3)
U3_GG(x1, x2, x3)  =  U3_GG(x1, x3)

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

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

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

SUBLIST_IN_GG(X, Y) → U3_GG(X, Y, append1_in_aag(P, X1, Y))
SUBLIST_IN_GG(X, Y) → APPEND1_IN_AAG(P, X1, Y)
APPEND1_IN_AAG(.(X, Xs), Ys, .(X, Zs)) → U1_AAG(X, Xs, Ys, Zs, append1_in_aag(Xs, Ys, Zs))
APPEND1_IN_AAG(.(X, Xs), Ys, .(X, Zs)) → APPEND1_IN_AAG(Xs, Ys, Zs)
U3_GG(X, Y, append1_out_aag(P, X1, Y)) → U4_GG(X, Y, append2_in_agg(X2, X, P))
U3_GG(X, Y, append1_out_aag(P, X1, Y)) → APPEND2_IN_AGG(X2, X, P)
APPEND2_IN_AGG(.(X, Xs), Ys, .(X, Zs)) → U2_AGG(X, Xs, Ys, Zs, append2_in_agg(Xs, Ys, Zs))
APPEND2_IN_AGG(.(X, Xs), Ys, .(X, Zs)) → APPEND2_IN_AGG(Xs, Ys, Zs)

The TRS R consists of the following rules:

sublist_in_gg(X, Y) → U3_gg(X, Y, append1_in_aag(P, X1, Y))
append1_in_aag([], Ys, Ys) → append1_out_aag([], Ys, Ys)
append1_in_aag(.(X, Xs), Ys, .(X, Zs)) → U1_aag(X, Xs, Ys, Zs, append1_in_aag(Xs, Ys, Zs))
U1_aag(X, Xs, Ys, Zs, append1_out_aag(Xs, Ys, Zs)) → append1_out_aag(.(X, Xs), Ys, .(X, Zs))
U3_gg(X, Y, append1_out_aag(P, X1, Y)) → U4_gg(X, Y, append2_in_agg(X2, X, P))
append2_in_agg([], Ys, Ys) → append2_out_agg([], Ys, Ys)
append2_in_agg(.(X, Xs), Ys, .(X, Zs)) → U2_agg(X, Xs, Ys, Zs, append2_in_agg(Xs, Ys, Zs))
U2_agg(X, Xs, Ys, Zs, append2_out_agg(Xs, Ys, Zs)) → append2_out_agg(.(X, Xs), Ys, .(X, Zs))
U4_gg(X, Y, append2_out_agg(X2, X, P)) → sublist_out_gg(X, Y)

The argument filtering Pi contains the following mapping:
sublist_in_gg(x1, x2)  =  sublist_in_gg(x1, x2)
U3_gg(x1, x2, x3)  =  U3_gg(x1, x3)
append1_in_aag(x1, x2, x3)  =  append1_in_aag(x3)
append1_out_aag(x1, x2, x3)  =  append1_out_aag(x1, x2)
.(x1, x2)  =  .(x1, x2)
U1_aag(x1, x2, x3, x4, x5)  =  U1_aag(x1, x5)
U4_gg(x1, x2, x3)  =  U4_gg(x3)
append2_in_agg(x1, x2, x3)  =  append2_in_agg(x2, x3)
append2_out_agg(x1, x2, x3)  =  append2_out_agg(x1)
U2_agg(x1, x2, x3, x4, x5)  =  U2_agg(x1, x5)
sublist_out_gg(x1, x2)  =  sublist_out_gg
U4_GG(x1, x2, x3)  =  U4_GG(x3)
SUBLIST_IN_GG(x1, x2)  =  SUBLIST_IN_GG(x1, x2)
U1_AAG(x1, x2, x3, x4, x5)  =  U1_AAG(x1, x5)
U2_AGG(x1, x2, x3, x4, x5)  =  U2_AGG(x1, x5)
APPEND2_IN_AGG(x1, x2, x3)  =  APPEND2_IN_AGG(x2, x3)
APPEND1_IN_AAG(x1, x2, x3)  =  APPEND1_IN_AAG(x3)
U3_GG(x1, x2, x3)  =  U3_GG(x1, x3)

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

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

APPEND2_IN_AGG(.(X, Xs), Ys, .(X, Zs)) → APPEND2_IN_AGG(Xs, Ys, Zs)

The TRS R consists of the following rules:

sublist_in_gg(X, Y) → U3_gg(X, Y, append1_in_aag(P, X1, Y))
append1_in_aag([], Ys, Ys) → append1_out_aag([], Ys, Ys)
append1_in_aag(.(X, Xs), Ys, .(X, Zs)) → U1_aag(X, Xs, Ys, Zs, append1_in_aag(Xs, Ys, Zs))
U1_aag(X, Xs, Ys, Zs, append1_out_aag(Xs, Ys, Zs)) → append1_out_aag(.(X, Xs), Ys, .(X, Zs))
U3_gg(X, Y, append1_out_aag(P, X1, Y)) → U4_gg(X, Y, append2_in_agg(X2, X, P))
append2_in_agg([], Ys, Ys) → append2_out_agg([], Ys, Ys)
append2_in_agg(.(X, Xs), Ys, .(X, Zs)) → U2_agg(X, Xs, Ys, Zs, append2_in_agg(Xs, Ys, Zs))
U2_agg(X, Xs, Ys, Zs, append2_out_agg(Xs, Ys, Zs)) → append2_out_agg(.(X, Xs), Ys, .(X, Zs))
U4_gg(X, Y, append2_out_agg(X2, X, P)) → sublist_out_gg(X, Y)

The argument filtering Pi contains the following mapping:
sublist_in_gg(x1, x2)  =  sublist_in_gg(x1, x2)
U3_gg(x1, x2, x3)  =  U3_gg(x1, x3)
append1_in_aag(x1, x2, x3)  =  append1_in_aag(x3)
append1_out_aag(x1, x2, x3)  =  append1_out_aag(x1, x2)
.(x1, x2)  =  .(x1, x2)
U1_aag(x1, x2, x3, x4, x5)  =  U1_aag(x1, x5)
U4_gg(x1, x2, x3)  =  U4_gg(x3)
append2_in_agg(x1, x2, x3)  =  append2_in_agg(x2, x3)
append2_out_agg(x1, x2, x3)  =  append2_out_agg(x1)
U2_agg(x1, x2, x3, x4, x5)  =  U2_agg(x1, x5)
sublist_out_gg(x1, x2)  =  sublist_out_gg
APPEND2_IN_AGG(x1, x2, x3)  =  APPEND2_IN_AGG(x2, 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

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

APPEND2_IN_AGG(.(X, Xs), Ys, .(X, Zs)) → APPEND2_IN_AGG(Xs, Ys, Zs)

R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x1, x2)
APPEND2_IN_AGG(x1, x2, x3)  =  APPEND2_IN_AGG(x2, 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

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

APPEND2_IN_AGG(Ys, .(X, Zs)) → APPEND2_IN_AGG(Ys, Zs)

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

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

APPEND1_IN_AAG(.(X, Xs), Ys, .(X, Zs)) → APPEND1_IN_AAG(Xs, Ys, Zs)

The TRS R consists of the following rules:

sublist_in_gg(X, Y) → U3_gg(X, Y, append1_in_aag(P, X1, Y))
append1_in_aag([], Ys, Ys) → append1_out_aag([], Ys, Ys)
append1_in_aag(.(X, Xs), Ys, .(X, Zs)) → U1_aag(X, Xs, Ys, Zs, append1_in_aag(Xs, Ys, Zs))
U1_aag(X, Xs, Ys, Zs, append1_out_aag(Xs, Ys, Zs)) → append1_out_aag(.(X, Xs), Ys, .(X, Zs))
U3_gg(X, Y, append1_out_aag(P, X1, Y)) → U4_gg(X, Y, append2_in_agg(X2, X, P))
append2_in_agg([], Ys, Ys) → append2_out_agg([], Ys, Ys)
append2_in_agg(.(X, Xs), Ys, .(X, Zs)) → U2_agg(X, Xs, Ys, Zs, append2_in_agg(Xs, Ys, Zs))
U2_agg(X, Xs, Ys, Zs, append2_out_agg(Xs, Ys, Zs)) → append2_out_agg(.(X, Xs), Ys, .(X, Zs))
U4_gg(X, Y, append2_out_agg(X2, X, P)) → sublist_out_gg(X, Y)

The argument filtering Pi contains the following mapping:
sublist_in_gg(x1, x2)  =  sublist_in_gg(x1, x2)
U3_gg(x1, x2, x3)  =  U3_gg(x1, x3)
append1_in_aag(x1, x2, x3)  =  append1_in_aag(x3)
append1_out_aag(x1, x2, x3)  =  append1_out_aag(x1, x2)
.(x1, x2)  =  .(x1, x2)
U1_aag(x1, x2, x3, x4, x5)  =  U1_aag(x1, x5)
U4_gg(x1, x2, x3)  =  U4_gg(x3)
append2_in_agg(x1, x2, x3)  =  append2_in_agg(x2, x3)
append2_out_agg(x1, x2, x3)  =  append2_out_agg(x1)
U2_agg(x1, x2, x3, x4, x5)  =  U2_agg(x1, x5)
sublist_out_gg(x1, x2)  =  sublist_out_gg
APPEND1_IN_AAG(x1, x2, x3)  =  APPEND1_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
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof

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

APPEND1_IN_AAG(.(X, Xs), Ys, .(X, Zs)) → APPEND1_IN_AAG(Xs, Ys, Zs)

R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x1, x2)
APPEND1_IN_AAG(x1, x2, x3)  =  APPEND1_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
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ QDPSizeChangeProof

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

APPEND1_IN_AAG(.(X, Zs)) → APPEND1_IN_AAG(Zs)

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: