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

0 Prolog
  ↳1 PrologToPiTRSProof (⇐)
    ↳2 PiTRS
    ↳3 DependencyPairsProof (⇔)
    ↳4 PiDP
    ↳5 DependencyGraphProof (⇔)
    ↳6 AND
      ↳7 PiDP
        ↳8 UsableRulesProof (⇔)
        ↳9 PiDP
        ↳10 PiDPToQDPProof (⇐)
        ↳11 QDP
        ↳12 QDPSizeChangeProof (⇔)
        ↳13 TRUE
      ↳14 PiDP
        ↳15 UsableRulesProof (⇔)
        ↳16 PiDP
        ↳17 PiDPToQDPProof (⇐)
        ↳18 QDP
        ↳19 QDPSizeChangeProof (⇔)
        ↳20 TRUE
  ↳21 PrologToPiTRSProof (⇐)
    ↳22 PiTRS
    ↳23 DependencyPairsProof (⇔)
    ↳24 PiDP
    ↳25 DependencyGraphProof (⇔)
    ↳26 AND
      ↳27 PiDP
        ↳28 UsableRulesProof (⇔)
        ↳29 PiDP
        ↳30 PiDPToQDPProof (⇐)
        ↳31 QDP
        ↳32 QDPSizeChangeProof (⇔)
        ↳33 TRUE
      ↳34 PiDP
        ↳35 UsableRulesProof (⇔)
        ↳36 PiDP
        ↳37 PiDPToQDPProof (⇐)
        ↳38 QDP

(0) Obligation:

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).

(1) PrologToPiTRSProof (SOUND transformation)

We use the technique of [LOPSTR]. 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

(2) Obligation:

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

(3) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LOPSTR] 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
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)
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)
U2_AGG(x1, x2, x3, x4, x5)  =  U2_AGG(x1, x5)

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

(4) Obligation:

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
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)
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)
U2_AGG(x1, x2, x3, x4, x5)  =  U2_AGG(x1, x5)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 2 SCCs with 6 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

(8) UsableRulesProof (EQUIVALENT transformation)

For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.

(9) Obligation:

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

(10) PiDPToQDPProof (SOUND transformation)

Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.

(11) Obligation:

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.

(12) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] 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:

  • APPEND2_IN_AGG(Ys, .(X, Zs)) → APPEND2_IN_AGG(Ys, Zs)
    The graph contains the following edges 1 >= 1, 2 > 2

(13) TRUE

(14) Obligation:

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

(15) UsableRulesProof (EQUIVALENT transformation)

For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.

(16) Obligation:

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

(17) PiDPToQDPProof (SOUND transformation)

Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.

(18) Obligation:

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.

(19) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] 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:

  • APPEND1_IN_AAG(.(X, Zs)) → APPEND1_IN_AAG(Zs)
    The graph contains the following edges 1 > 1

(20) TRUE

(21) PrologToPiTRSProof (SOUND transformation)

We use the technique of [LOPSTR]. 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, x2, x3)
append1_in_aag(x1, x2, x3)  =  append1_in_aag(x3)
append1_out_aag(x1, x2, x3)  =  append1_out_aag(x1, x2, x3)
.(x1, x2)  =  .(x1, x2)
U1_aag(x1, x2, x3, x4, x5)  =  U1_aag(x1, x4, x5)
U4_gg(x1, x2, x3)  =  U4_gg(x1, x2, x3)
append2_in_agg(x1, x2, x3)  =  append2_in_agg(x2, x3)
append2_out_agg(x1, x2, x3)  =  append2_out_agg(x1, x2, x3)
U2_agg(x1, x2, x3, x4, x5)  =  U2_agg(x1, x3, x4, x5)
sublist_out_gg(x1, x2)  =  sublist_out_gg(x1, x2)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(22) Obligation:

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, x2, x3)
append1_in_aag(x1, x2, x3)  =  append1_in_aag(x3)
append1_out_aag(x1, x2, x3)  =  append1_out_aag(x1, x2, x3)
.(x1, x2)  =  .(x1, x2)
U1_aag(x1, x2, x3, x4, x5)  =  U1_aag(x1, x4, x5)
U4_gg(x1, x2, x3)  =  U4_gg(x1, x2, x3)
append2_in_agg(x1, x2, x3)  =  append2_in_agg(x2, x3)
append2_out_agg(x1, x2, x3)  =  append2_out_agg(x1, x2, x3)
U2_agg(x1, x2, x3, x4, x5)  =  U2_agg(x1, x3, x4, x5)
sublist_out_gg(x1, x2)  =  sublist_out_gg(x1, x2)

(23) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LOPSTR] 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, x2, x3)
append1_in_aag(x1, x2, x3)  =  append1_in_aag(x3)
append1_out_aag(x1, x2, x3)  =  append1_out_aag(x1, x2, x3)
.(x1, x2)  =  .(x1, x2)
U1_aag(x1, x2, x3, x4, x5)  =  U1_aag(x1, x4, x5)
U4_gg(x1, x2, x3)  =  U4_gg(x1, x2, x3)
append2_in_agg(x1, x2, x3)  =  append2_in_agg(x2, x3)
append2_out_agg(x1, x2, x3)  =  append2_out_agg(x1, x2, x3)
U2_agg(x1, x2, x3, x4, x5)  =  U2_agg(x1, x3, x4, x5)
sublist_out_gg(x1, x2)  =  sublist_out_gg(x1, x2)
SUBLIST_IN_GG(x1, x2)  =  SUBLIST_IN_GG(x1, x2)
U3_GG(x1, x2, x3)  =  U3_GG(x1, x2, x3)
APPEND1_IN_AAG(x1, x2, x3)  =  APPEND1_IN_AAG(x3)
U1_AAG(x1, x2, x3, x4, x5)  =  U1_AAG(x1, x4, x5)
U4_GG(x1, x2, x3)  =  U4_GG(x1, x2, x3)
APPEND2_IN_AGG(x1, x2, x3)  =  APPEND2_IN_AGG(x2, x3)
U2_AGG(x1, x2, x3, x4, x5)  =  U2_AGG(x1, x3, x4, x5)

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

(24) Obligation:

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, x2, x3)
append1_in_aag(x1, x2, x3)  =  append1_in_aag(x3)
append1_out_aag(x1, x2, x3)  =  append1_out_aag(x1, x2, x3)
.(x1, x2)  =  .(x1, x2)
U1_aag(x1, x2, x3, x4, x5)  =  U1_aag(x1, x4, x5)
U4_gg(x1, x2, x3)  =  U4_gg(x1, x2, x3)
append2_in_agg(x1, x2, x3)  =  append2_in_agg(x2, x3)
append2_out_agg(x1, x2, x3)  =  append2_out_agg(x1, x2, x3)
U2_agg(x1, x2, x3, x4, x5)  =  U2_agg(x1, x3, x4, x5)
sublist_out_gg(x1, x2)  =  sublist_out_gg(x1, x2)
SUBLIST_IN_GG(x1, x2)  =  SUBLIST_IN_GG(x1, x2)
U3_GG(x1, x2, x3)  =  U3_GG(x1, x2, x3)
APPEND1_IN_AAG(x1, x2, x3)  =  APPEND1_IN_AAG(x3)
U1_AAG(x1, x2, x3, x4, x5)  =  U1_AAG(x1, x4, x5)
U4_GG(x1, x2, x3)  =  U4_GG(x1, x2, x3)
APPEND2_IN_AGG(x1, x2, x3)  =  APPEND2_IN_AGG(x2, x3)
U2_AGG(x1, x2, x3, x4, x5)  =  U2_AGG(x1, x3, x4, x5)

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

(25) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 2 SCCs with 6 less nodes.

(26) Complex Obligation (AND)

(27) Obligation:

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, x2, x3)
append1_in_aag(x1, x2, x3)  =  append1_in_aag(x3)
append1_out_aag(x1, x2, x3)  =  append1_out_aag(x1, x2, x3)
.(x1, x2)  =  .(x1, x2)
U1_aag(x1, x2, x3, x4, x5)  =  U1_aag(x1, x4, x5)
U4_gg(x1, x2, x3)  =  U4_gg(x1, x2, x3)
append2_in_agg(x1, x2, x3)  =  append2_in_agg(x2, x3)
append2_out_agg(x1, x2, x3)  =  append2_out_agg(x1, x2, x3)
U2_agg(x1, x2, x3, x4, x5)  =  U2_agg(x1, x3, x4, x5)
sublist_out_gg(x1, x2)  =  sublist_out_gg(x1, x2)
APPEND2_IN_AGG(x1, x2, x3)  =  APPEND2_IN_AGG(x2, x3)

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

(28) UsableRulesProof (EQUIVALENT transformation)

For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.

(29) Obligation:

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

(30) PiDPToQDPProof (SOUND transformation)

Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.

(31) Obligation:

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.

(32) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] 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:

  • APPEND2_IN_AGG(Ys, .(X, Zs)) → APPEND2_IN_AGG(Ys, Zs)
    The graph contains the following edges 1 >= 1, 2 > 2

(33) TRUE

(34) Obligation:

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, x2, x3)
append1_in_aag(x1, x2, x3)  =  append1_in_aag(x3)
append1_out_aag(x1, x2, x3)  =  append1_out_aag(x1, x2, x3)
.(x1, x2)  =  .(x1, x2)
U1_aag(x1, x2, x3, x4, x5)  =  U1_aag(x1, x4, x5)
U4_gg(x1, x2, x3)  =  U4_gg(x1, x2, x3)
append2_in_agg(x1, x2, x3)  =  append2_in_agg(x2, x3)
append2_out_agg(x1, x2, x3)  =  append2_out_agg(x1, x2, x3)
U2_agg(x1, x2, x3, x4, x5)  =  U2_agg(x1, x3, x4, x5)
sublist_out_gg(x1, x2)  =  sublist_out_gg(x1, x2)
APPEND1_IN_AAG(x1, x2, x3)  =  APPEND1_IN_AAG(x3)

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

(35) UsableRulesProof (EQUIVALENT transformation)

For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.

(36) Obligation:

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

(37) PiDPToQDPProof (SOUND transformation)

Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.

(38) Obligation:

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.