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

sublist_in(X, Y) → U3(X, Y, append1_in(P, X1, Y))
append1_in(.(X, Xs), Ys, .(X, Zs)) → U1(X, Xs, Ys, Zs, append1_in(Xs, Ys, Zs))
append1_in([], Ys, Ys) → append1_out([], Ys, Ys)
U1(X, Xs, Ys, Zs, append1_out(Xs, Ys, Zs)) → append1_out(.(X, Xs), Ys, .(X, Zs))
U3(X, Y, append1_out(P, X1, Y)) → U4(X, Y, append2_in(X2, X, P))
append2_in(.(X, Xs), Ys, .(X, Zs)) → U2(X, Xs, Ys, Zs, append2_in(Xs, Ys, Zs))
append2_in([], Ys, Ys) → append2_out([], Ys, Ys)
U2(X, Xs, Ys, Zs, append2_out(Xs, Ys, Zs)) → append2_out(.(X, Xs), Ys, .(X, Zs))
U4(X, Y, append2_out(X2, X, P)) → sublist_out(X, Y)

The argument filtering Pi contains the following mapping:
sublist_in(x1, x2)  =  sublist_in(x1, x2)
U3(x1, x2, x3)  =  U3(x1, x3)
append1_in(x1, x2, x3)  =  append1_in(x3)
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4, x5)  =  U1(x1, x5)
[]  =  []
append1_out(x1, x2, x3)  =  append1_out(x1, x2)
U4(x1, x2, x3)  =  U4(x3)
append2_in(x1, x2, x3)  =  append2_in(x2, x3)
U2(x1, x2, x3, x4, x5)  =  U2(x1, x5)
append2_out(x1, x2, x3)  =  append2_out(x1)
sublist_out(x1, x2)  =  sublist_out

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(X, Y) → U3(X, Y, append1_in(P, X1, Y))
append1_in(.(X, Xs), Ys, .(X, Zs)) → U1(X, Xs, Ys, Zs, append1_in(Xs, Ys, Zs))
append1_in([], Ys, Ys) → append1_out([], Ys, Ys)
U1(X, Xs, Ys, Zs, append1_out(Xs, Ys, Zs)) → append1_out(.(X, Xs), Ys, .(X, Zs))
U3(X, Y, append1_out(P, X1, Y)) → U4(X, Y, append2_in(X2, X, P))
append2_in(.(X, Xs), Ys, .(X, Zs)) → U2(X, Xs, Ys, Zs, append2_in(Xs, Ys, Zs))
append2_in([], Ys, Ys) → append2_out([], Ys, Ys)
U2(X, Xs, Ys, Zs, append2_out(Xs, Ys, Zs)) → append2_out(.(X, Xs), Ys, .(X, Zs))
U4(X, Y, append2_out(X2, X, P)) → sublist_out(X, Y)

The argument filtering Pi contains the following mapping:
sublist_in(x1, x2)  =  sublist_in(x1, x2)
U3(x1, x2, x3)  =  U3(x1, x3)
append1_in(x1, x2, x3)  =  append1_in(x3)
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4, x5)  =  U1(x1, x5)
[]  =  []
append1_out(x1, x2, x3)  =  append1_out(x1, x2)
U4(x1, x2, x3)  =  U4(x3)
append2_in(x1, x2, x3)  =  append2_in(x2, x3)
U2(x1, x2, x3, x4, x5)  =  U2(x1, x5)
append2_out(x1, x2, x3)  =  append2_out(x1)
sublist_out(x1, x2)  =  sublist_out


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(X, Y) → U31(X, Y, append1_in(P, X1, Y))
SUBLIST_IN(X, Y) → APPEND1_IN(P, X1, Y)
APPEND1_IN(.(X, Xs), Ys, .(X, Zs)) → U11(X, Xs, Ys, Zs, append1_in(Xs, Ys, Zs))
APPEND1_IN(.(X, Xs), Ys, .(X, Zs)) → APPEND1_IN(Xs, Ys, Zs)
U31(X, Y, append1_out(P, X1, Y)) → U41(X, Y, append2_in(X2, X, P))
U31(X, Y, append1_out(P, X1, Y)) → APPEND2_IN(X2, X, P)
APPEND2_IN(.(X, Xs), Ys, .(X, Zs)) → U21(X, Xs, Ys, Zs, append2_in(Xs, Ys, Zs))
APPEND2_IN(.(X, Xs), Ys, .(X, Zs)) → APPEND2_IN(Xs, Ys, Zs)

The TRS R consists of the following rules:

sublist_in(X, Y) → U3(X, Y, append1_in(P, X1, Y))
append1_in(.(X, Xs), Ys, .(X, Zs)) → U1(X, Xs, Ys, Zs, append1_in(Xs, Ys, Zs))
append1_in([], Ys, Ys) → append1_out([], Ys, Ys)
U1(X, Xs, Ys, Zs, append1_out(Xs, Ys, Zs)) → append1_out(.(X, Xs), Ys, .(X, Zs))
U3(X, Y, append1_out(P, X1, Y)) → U4(X, Y, append2_in(X2, X, P))
append2_in(.(X, Xs), Ys, .(X, Zs)) → U2(X, Xs, Ys, Zs, append2_in(Xs, Ys, Zs))
append2_in([], Ys, Ys) → append2_out([], Ys, Ys)
U2(X, Xs, Ys, Zs, append2_out(Xs, Ys, Zs)) → append2_out(.(X, Xs), Ys, .(X, Zs))
U4(X, Y, append2_out(X2, X, P)) → sublist_out(X, Y)

The argument filtering Pi contains the following mapping:
sublist_in(x1, x2)  =  sublist_in(x1, x2)
U3(x1, x2, x3)  =  U3(x1, x3)
append1_in(x1, x2, x3)  =  append1_in(x3)
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4, x5)  =  U1(x1, x5)
[]  =  []
append1_out(x1, x2, x3)  =  append1_out(x1, x2)
U4(x1, x2, x3)  =  U4(x3)
append2_in(x1, x2, x3)  =  append2_in(x2, x3)
U2(x1, x2, x3, x4, x5)  =  U2(x1, x5)
append2_out(x1, x2, x3)  =  append2_out(x1)
sublist_out(x1, x2)  =  sublist_out
APPEND1_IN(x1, x2, x3)  =  APPEND1_IN(x3)
U41(x1, x2, x3)  =  U41(x3)
U31(x1, x2, x3)  =  U31(x1, x3)
APPEND2_IN(x1, x2, x3)  =  APPEND2_IN(x2, x3)
U21(x1, x2, x3, x4, x5)  =  U21(x1, x5)
U11(x1, x2, x3, x4, x5)  =  U11(x1, x5)
SUBLIST_IN(x1, x2)  =  SUBLIST_IN(x1, x2)

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(X, Y) → U31(X, Y, append1_in(P, X1, Y))
SUBLIST_IN(X, Y) → APPEND1_IN(P, X1, Y)
APPEND1_IN(.(X, Xs), Ys, .(X, Zs)) → U11(X, Xs, Ys, Zs, append1_in(Xs, Ys, Zs))
APPEND1_IN(.(X, Xs), Ys, .(X, Zs)) → APPEND1_IN(Xs, Ys, Zs)
U31(X, Y, append1_out(P, X1, Y)) → U41(X, Y, append2_in(X2, X, P))
U31(X, Y, append1_out(P, X1, Y)) → APPEND2_IN(X2, X, P)
APPEND2_IN(.(X, Xs), Ys, .(X, Zs)) → U21(X, Xs, Ys, Zs, append2_in(Xs, Ys, Zs))
APPEND2_IN(.(X, Xs), Ys, .(X, Zs)) → APPEND2_IN(Xs, Ys, Zs)

The TRS R consists of the following rules:

sublist_in(X, Y) → U3(X, Y, append1_in(P, X1, Y))
append1_in(.(X, Xs), Ys, .(X, Zs)) → U1(X, Xs, Ys, Zs, append1_in(Xs, Ys, Zs))
append1_in([], Ys, Ys) → append1_out([], Ys, Ys)
U1(X, Xs, Ys, Zs, append1_out(Xs, Ys, Zs)) → append1_out(.(X, Xs), Ys, .(X, Zs))
U3(X, Y, append1_out(P, X1, Y)) → U4(X, Y, append2_in(X2, X, P))
append2_in(.(X, Xs), Ys, .(X, Zs)) → U2(X, Xs, Ys, Zs, append2_in(Xs, Ys, Zs))
append2_in([], Ys, Ys) → append2_out([], Ys, Ys)
U2(X, Xs, Ys, Zs, append2_out(Xs, Ys, Zs)) → append2_out(.(X, Xs), Ys, .(X, Zs))
U4(X, Y, append2_out(X2, X, P)) → sublist_out(X, Y)

The argument filtering Pi contains the following mapping:
sublist_in(x1, x2)  =  sublist_in(x1, x2)
U3(x1, x2, x3)  =  U3(x1, x3)
append1_in(x1, x2, x3)  =  append1_in(x3)
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4, x5)  =  U1(x1, x5)
[]  =  []
append1_out(x1, x2, x3)  =  append1_out(x1, x2)
U4(x1, x2, x3)  =  U4(x3)
append2_in(x1, x2, x3)  =  append2_in(x2, x3)
U2(x1, x2, x3, x4, x5)  =  U2(x1, x5)
append2_out(x1, x2, x3)  =  append2_out(x1)
sublist_out(x1, x2)  =  sublist_out
APPEND1_IN(x1, x2, x3)  =  APPEND1_IN(x3)
U41(x1, x2, x3)  =  U41(x3)
U31(x1, x2, x3)  =  U31(x1, x3)
APPEND2_IN(x1, x2, x3)  =  APPEND2_IN(x2, x3)
U21(x1, x2, x3, x4, x5)  =  U21(x1, x5)
U11(x1, x2, x3, x4, x5)  =  U11(x1, x5)
SUBLIST_IN(x1, x2)  =  SUBLIST_IN(x1, x2)

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(.(X, Xs), Ys, .(X, Zs)) → APPEND2_IN(Xs, Ys, Zs)

The TRS R consists of the following rules:

sublist_in(X, Y) → U3(X, Y, append1_in(P, X1, Y))
append1_in(.(X, Xs), Ys, .(X, Zs)) → U1(X, Xs, Ys, Zs, append1_in(Xs, Ys, Zs))
append1_in([], Ys, Ys) → append1_out([], Ys, Ys)
U1(X, Xs, Ys, Zs, append1_out(Xs, Ys, Zs)) → append1_out(.(X, Xs), Ys, .(X, Zs))
U3(X, Y, append1_out(P, X1, Y)) → U4(X, Y, append2_in(X2, X, P))
append2_in(.(X, Xs), Ys, .(X, Zs)) → U2(X, Xs, Ys, Zs, append2_in(Xs, Ys, Zs))
append2_in([], Ys, Ys) → append2_out([], Ys, Ys)
U2(X, Xs, Ys, Zs, append2_out(Xs, Ys, Zs)) → append2_out(.(X, Xs), Ys, .(X, Zs))
U4(X, Y, append2_out(X2, X, P)) → sublist_out(X, Y)

The argument filtering Pi contains the following mapping:
sublist_in(x1, x2)  =  sublist_in(x1, x2)
U3(x1, x2, x3)  =  U3(x1, x3)
append1_in(x1, x2, x3)  =  append1_in(x3)
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4, x5)  =  U1(x1, x5)
[]  =  []
append1_out(x1, x2, x3)  =  append1_out(x1, x2)
U4(x1, x2, x3)  =  U4(x3)
append2_in(x1, x2, x3)  =  append2_in(x2, x3)
U2(x1, x2, x3, x4, x5)  =  U2(x1, x5)
append2_out(x1, x2, x3)  =  append2_out(x1)
sublist_out(x1, x2)  =  sublist_out
APPEND2_IN(x1, x2, x3)  =  APPEND2_IN(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(.(X, Xs), Ys, .(X, Zs)) → APPEND2_IN(Xs, Ys, Zs)

R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x1, x2)
APPEND2_IN(x1, x2, x3)  =  APPEND2_IN(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(Ys, .(X, Zs)) → APPEND2_IN(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(.(X, Xs), Ys, .(X, Zs)) → APPEND1_IN(Xs, Ys, Zs)

The TRS R consists of the following rules:

sublist_in(X, Y) → U3(X, Y, append1_in(P, X1, Y))
append1_in(.(X, Xs), Ys, .(X, Zs)) → U1(X, Xs, Ys, Zs, append1_in(Xs, Ys, Zs))
append1_in([], Ys, Ys) → append1_out([], Ys, Ys)
U1(X, Xs, Ys, Zs, append1_out(Xs, Ys, Zs)) → append1_out(.(X, Xs), Ys, .(X, Zs))
U3(X, Y, append1_out(P, X1, Y)) → U4(X, Y, append2_in(X2, X, P))
append2_in(.(X, Xs), Ys, .(X, Zs)) → U2(X, Xs, Ys, Zs, append2_in(Xs, Ys, Zs))
append2_in([], Ys, Ys) → append2_out([], Ys, Ys)
U2(X, Xs, Ys, Zs, append2_out(Xs, Ys, Zs)) → append2_out(.(X, Xs), Ys, .(X, Zs))
U4(X, Y, append2_out(X2, X, P)) → sublist_out(X, Y)

The argument filtering Pi contains the following mapping:
sublist_in(x1, x2)  =  sublist_in(x1, x2)
U3(x1, x2, x3)  =  U3(x1, x3)
append1_in(x1, x2, x3)  =  append1_in(x3)
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4, x5)  =  U1(x1, x5)
[]  =  []
append1_out(x1, x2, x3)  =  append1_out(x1, x2)
U4(x1, x2, x3)  =  U4(x3)
append2_in(x1, x2, x3)  =  append2_in(x2, x3)
U2(x1, x2, x3, x4, x5)  =  U2(x1, x5)
append2_out(x1, x2, x3)  =  append2_out(x1)
sublist_out(x1, x2)  =  sublist_out
APPEND1_IN(x1, x2, x3)  =  APPEND1_IN(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(.(X, Xs), Ys, .(X, Zs)) → APPEND1_IN(Xs, Ys, Zs)

R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x1, x2)
APPEND1_IN(x1, x2, x3)  =  APPEND1_IN(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(.(X, Zs)) → APPEND1_IN(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: