YES
proof of /hpcwork/ff862203/termcomp26/benchmarks/xtggD.pl
# AProVE Commit ID: 23a904c96b029b0a549cde0d0d17dbccf967db59 jckassing 20260626 unpublished dirty


Left Termination of the query pattern

sublist(g,g)

w.r.t. the given Prolog program could successfully be proven:

(0) Prolog
(1) PrologToPiTRSProof [SOUND, 0 ms]
(2) PiTRS
(3) DependencyPairsProof [EQUIVALENT, 0 ms]
(4) PiDP
(5) DependencyGraphProof [EQUIVALENT, 0 ms]
(6) AND
    (7) PiDP
        (8) UsableRulesProof [EQUIVALENT, 0 ms]
        (9) PiDP
        (10) PiDPToQDPProof [SOUND, 0 ms]
        (11) QDP
        (12) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (13) YES
    (14) PiDP
        (15) UsableRulesProof [EQUIVALENT, 0 ms]
        (16) PiDP
        (17) PiDPToQDPProof [SOUND, 0 ms]
        (18) QDP
        (19) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (20) YES


----------------------------------------

(0)
Obligation:
Clauses:

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


Query: sublist(g,g)
----------------------------------------

(1) PrologToPiTRSProof (SOUND)
We use the technique of [TOCL09]. With regard to the inferred argument filtering the predicates were used in the following modes:

sublist_in_2: (b,b)

append_in_3: (f,f,b) (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) -> U2_gg(X, Y, append_in_aag(P, X1, Y))
   append_in_aag([], Ys, Ys) -> append_out_aag([], Ys, Ys)
   append_in_aag(.(X, Xs), Ys, .(X, Zs)) -> U1_aag(X, Xs, Ys, Zs, append_in_aag(Xs, Ys, Zs))
   U1_aag(X, Xs, Ys, Zs, append_out_aag(Xs, Ys, Zs)) -> append_out_aag(.(X, Xs), Ys, .(X, Zs))
   U2_gg(X, Y, append_out_aag(P, X1, Y)) -> U3_gg(X, Y, append_in_agg(X2, X, P))
   append_in_agg([], Ys, Ys) -> append_out_agg([], Ys, Ys)
   append_in_agg(.(X, Xs), Ys, .(X, Zs)) -> U1_agg(X, Xs, Ys, Zs, append_in_agg(Xs, Ys, Zs))
   U1_agg(X, Xs, Ys, Zs, append_out_agg(Xs, Ys, Zs)) -> append_out_agg(.(X, Xs), Ys, .(X, Zs))
   U3_gg(X, Y, append_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)

U2_gg(x1, x2, x3)  =  U2_gg(x1, x3)

append_in_aag(x1, x2, x3)  =  append_in_aag(x3)

append_out_aag(x1, x2, x3)  =  append_out_aag(x1, x2)

.(x1, x2)  =  .(x1, x2)

U1_aag(x1, x2, x3, x4, x5)  =  U1_aag(x1, x5)

U3_gg(x1, x2, x3)  =  U3_gg(x3)

append_in_agg(x1, x2, x3)  =  append_in_agg(x2, x3)

append_out_agg(x1, x2, x3)  =  append_out_agg(x1)

U1_agg(x1, x2, x3, x4, x5)  =  U1_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) -> U2_gg(X, Y, append_in_aag(P, X1, Y))
   append_in_aag([], Ys, Ys) -> append_out_aag([], Ys, Ys)
   append_in_aag(.(X, Xs), Ys, .(X, Zs)) -> U1_aag(X, Xs, Ys, Zs, append_in_aag(Xs, Ys, Zs))
   U1_aag(X, Xs, Ys, Zs, append_out_aag(Xs, Ys, Zs)) -> append_out_aag(.(X, Xs), Ys, .(X, Zs))
   U2_gg(X, Y, append_out_aag(P, X1, Y)) -> U3_gg(X, Y, append_in_agg(X2, X, P))
   append_in_agg([], Ys, Ys) -> append_out_agg([], Ys, Ys)
   append_in_agg(.(X, Xs), Ys, .(X, Zs)) -> U1_agg(X, Xs, Ys, Zs, append_in_agg(Xs, Ys, Zs))
   U1_agg(X, Xs, Ys, Zs, append_out_agg(Xs, Ys, Zs)) -> append_out_agg(.(X, Xs), Ys, .(X, Zs))
   U3_gg(X, Y, append_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)

U2_gg(x1, x2, x3)  =  U2_gg(x1, x3)

append_in_aag(x1, x2, x3)  =  append_in_aag(x3)

append_out_aag(x1, x2, x3)  =  append_out_aag(x1, x2)

.(x1, x2)  =  .(x1, x2)

U1_aag(x1, x2, x3, x4, x5)  =  U1_aag(x1, x5)

U3_gg(x1, x2, x3)  =  U3_gg(x3)

append_in_agg(x1, x2, x3)  =  append_in_agg(x2, x3)

append_out_agg(x1, x2, x3)  =  append_out_agg(x1)

U1_agg(x1, x2, x3, x4, x5)  =  U1_agg(x1, x5)

sublist_out_gg(x1, x2)  =  sublist_out_gg



----------------------------------------

(3) DependencyPairsProof (EQUIVALENT)
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) -> U2_GG(X, Y, append_in_aag(P, X1, Y))
   SUBLIST_IN_GG(X, Y) -> APPEND_IN_AAG(P, X1, Y)
   APPEND_IN_AAG(.(X, Xs), Ys, .(X, Zs)) -> U1_AAG(X, Xs, Ys, Zs, append_in_aag(Xs, Ys, Zs))
   APPEND_IN_AAG(.(X, Xs), Ys, .(X, Zs)) -> APPEND_IN_AAG(Xs, Ys, Zs)
   U2_GG(X, Y, append_out_aag(P, X1, Y)) -> U3_GG(X, Y, append_in_agg(X2, X, P))
   U2_GG(X, Y, append_out_aag(P, X1, Y)) -> APPEND_IN_AGG(X2, X, P)
   APPEND_IN_AGG(.(X, Xs), Ys, .(X, Zs)) -> U1_AGG(X, Xs, Ys, Zs, append_in_agg(Xs, Ys, Zs))
   APPEND_IN_AGG(.(X, Xs), Ys, .(X, Zs)) -> APPEND_IN_AGG(Xs, Ys, Zs)

The TRS R consists of the following rules:

   sublist_in_gg(X, Y) -> U2_gg(X, Y, append_in_aag(P, X1, Y))
   append_in_aag([], Ys, Ys) -> append_out_aag([], Ys, Ys)
   append_in_aag(.(X, Xs), Ys, .(X, Zs)) -> U1_aag(X, Xs, Ys, Zs, append_in_aag(Xs, Ys, Zs))
   U1_aag(X, Xs, Ys, Zs, append_out_aag(Xs, Ys, Zs)) -> append_out_aag(.(X, Xs), Ys, .(X, Zs))
   U2_gg(X, Y, append_out_aag(P, X1, Y)) -> U3_gg(X, Y, append_in_agg(X2, X, P))
   append_in_agg([], Ys, Ys) -> append_out_agg([], Ys, Ys)
   append_in_agg(.(X, Xs), Ys, .(X, Zs)) -> U1_agg(X, Xs, Ys, Zs, append_in_agg(Xs, Ys, Zs))
   U1_agg(X, Xs, Ys, Zs, append_out_agg(Xs, Ys, Zs)) -> append_out_agg(.(X, Xs), Ys, .(X, Zs))
   U3_gg(X, Y, append_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)

U2_gg(x1, x2, x3)  =  U2_gg(x1, x3)

append_in_aag(x1, x2, x3)  =  append_in_aag(x3)

append_out_aag(x1, x2, x3)  =  append_out_aag(x1, x2)

.(x1, x2)  =  .(x1, x2)

U1_aag(x1, x2, x3, x4, x5)  =  U1_aag(x1, x5)

U3_gg(x1, x2, x3)  =  U3_gg(x3)

append_in_agg(x1, x2, x3)  =  append_in_agg(x2, x3)

append_out_agg(x1, x2, x3)  =  append_out_agg(x1)

U1_agg(x1, x2, x3, x4, x5)  =  U1_agg(x1, x5)

sublist_out_gg(x1, x2)  =  sublist_out_gg

SUBLIST_IN_GG(x1, x2)  =  SUBLIST_IN_GG(x1, x2)

U2_GG(x1, x2, x3)  =  U2_GG(x1, x3)

APPEND_IN_AAG(x1, x2, x3)  =  APPEND_IN_AAG(x3)

U1_AAG(x1, x2, x3, x4, x5)  =  U1_AAG(x1, x5)

U3_GG(x1, x2, x3)  =  U3_GG(x3)

APPEND_IN_AGG(x1, x2, x3)  =  APPEND_IN_AGG(x2, x3)

U1_AGG(x1, x2, x3, x4, x5)  =  U1_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) -> U2_GG(X, Y, append_in_aag(P, X1, Y))
   SUBLIST_IN_GG(X, Y) -> APPEND_IN_AAG(P, X1, Y)
   APPEND_IN_AAG(.(X, Xs), Ys, .(X, Zs)) -> U1_AAG(X, Xs, Ys, Zs, append_in_aag(Xs, Ys, Zs))
   APPEND_IN_AAG(.(X, Xs), Ys, .(X, Zs)) -> APPEND_IN_AAG(Xs, Ys, Zs)
   U2_GG(X, Y, append_out_aag(P, X1, Y)) -> U3_GG(X, Y, append_in_agg(X2, X, P))
   U2_GG(X, Y, append_out_aag(P, X1, Y)) -> APPEND_IN_AGG(X2, X, P)
   APPEND_IN_AGG(.(X, Xs), Ys, .(X, Zs)) -> U1_AGG(X, Xs, Ys, Zs, append_in_agg(Xs, Ys, Zs))
   APPEND_IN_AGG(.(X, Xs), Ys, .(X, Zs)) -> APPEND_IN_AGG(Xs, Ys, Zs)

The TRS R consists of the following rules:

   sublist_in_gg(X, Y) -> U2_gg(X, Y, append_in_aag(P, X1, Y))
   append_in_aag([], Ys, Ys) -> append_out_aag([], Ys, Ys)
   append_in_aag(.(X, Xs), Ys, .(X, Zs)) -> U1_aag(X, Xs, Ys, Zs, append_in_aag(Xs, Ys, Zs))
   U1_aag(X, Xs, Ys, Zs, append_out_aag(Xs, Ys, Zs)) -> append_out_aag(.(X, Xs), Ys, .(X, Zs))
   U2_gg(X, Y, append_out_aag(P, X1, Y)) -> U3_gg(X, Y, append_in_agg(X2, X, P))
   append_in_agg([], Ys, Ys) -> append_out_agg([], Ys, Ys)
   append_in_agg(.(X, Xs), Ys, .(X, Zs)) -> U1_agg(X, Xs, Ys, Zs, append_in_agg(Xs, Ys, Zs))
   U1_agg(X, Xs, Ys, Zs, append_out_agg(Xs, Ys, Zs)) -> append_out_agg(.(X, Xs), Ys, .(X, Zs))
   U3_gg(X, Y, append_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)

U2_gg(x1, x2, x3)  =  U2_gg(x1, x3)

append_in_aag(x1, x2, x3)  =  append_in_aag(x3)

append_out_aag(x1, x2, x3)  =  append_out_aag(x1, x2)

.(x1, x2)  =  .(x1, x2)

U1_aag(x1, x2, x3, x4, x5)  =  U1_aag(x1, x5)

U3_gg(x1, x2, x3)  =  U3_gg(x3)

append_in_agg(x1, x2, x3)  =  append_in_agg(x2, x3)

append_out_agg(x1, x2, x3)  =  append_out_agg(x1)

U1_agg(x1, x2, x3, x4, x5)  =  U1_agg(x1, x5)

sublist_out_gg(x1, x2)  =  sublist_out_gg

SUBLIST_IN_GG(x1, x2)  =  SUBLIST_IN_GG(x1, x2)

U2_GG(x1, x2, x3)  =  U2_GG(x1, x3)

APPEND_IN_AAG(x1, x2, x3)  =  APPEND_IN_AAG(x3)

U1_AAG(x1, x2, x3, x4, x5)  =  U1_AAG(x1, x5)

U3_GG(x1, x2, x3)  =  U3_GG(x3)

APPEND_IN_AGG(x1, x2, x3)  =  APPEND_IN_AGG(x2, x3)

U1_AGG(x1, x2, x3, x4, x5)  =  U1_AGG(x1, x5)


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

(5) DependencyGraphProof (EQUIVALENT)
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:

   APPEND_IN_AGG(.(X, Xs), Ys, .(X, Zs)) -> APPEND_IN_AGG(Xs, Ys, Zs)

The TRS R consists of the following rules:

   sublist_in_gg(X, Y) -> U2_gg(X, Y, append_in_aag(P, X1, Y))
   append_in_aag([], Ys, Ys) -> append_out_aag([], Ys, Ys)
   append_in_aag(.(X, Xs), Ys, .(X, Zs)) -> U1_aag(X, Xs, Ys, Zs, append_in_aag(Xs, Ys, Zs))
   U1_aag(X, Xs, Ys, Zs, append_out_aag(Xs, Ys, Zs)) -> append_out_aag(.(X, Xs), Ys, .(X, Zs))
   U2_gg(X, Y, append_out_aag(P, X1, Y)) -> U3_gg(X, Y, append_in_agg(X2, X, P))
   append_in_agg([], Ys, Ys) -> append_out_agg([], Ys, Ys)
   append_in_agg(.(X, Xs), Ys, .(X, Zs)) -> U1_agg(X, Xs, Ys, Zs, append_in_agg(Xs, Ys, Zs))
   U1_agg(X, Xs, Ys, Zs, append_out_agg(Xs, Ys, Zs)) -> append_out_agg(.(X, Xs), Ys, .(X, Zs))
   U3_gg(X, Y, append_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)

U2_gg(x1, x2, x3)  =  U2_gg(x1, x3)

append_in_aag(x1, x2, x3)  =  append_in_aag(x3)

append_out_aag(x1, x2, x3)  =  append_out_aag(x1, x2)

.(x1, x2)  =  .(x1, x2)

U1_aag(x1, x2, x3, x4, x5)  =  U1_aag(x1, x5)

U3_gg(x1, x2, x3)  =  U3_gg(x3)

append_in_agg(x1, x2, x3)  =  append_in_agg(x2, x3)

append_out_agg(x1, x2, x3)  =  append_out_agg(x1)

U1_agg(x1, x2, x3, x4, x5)  =  U1_agg(x1, x5)

sublist_out_gg(x1, x2)  =  sublist_out_gg

APPEND_IN_AGG(x1, x2, x3)  =  APPEND_IN_AGG(x2, x3)


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

(8) UsableRulesProof (EQUIVALENT)
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:

   APPEND_IN_AGG(.(X, Xs), Ys, .(X, Zs)) -> APPEND_IN_AGG(Xs, Ys, Zs)

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

APPEND_IN_AGG(x1, x2, x3)  =  APPEND_IN_AGG(x2, x3)


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

(10) PiDPToQDPProof (SOUND)
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:

   APPEND_IN_AGG(Ys, .(X, Zs)) -> APPEND_IN_AGG(Ys, Zs)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
----------------------------------------

(12) QDPSizeChangeProof (EQUIVALENT)
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:
*APPEND_IN_AGG(Ys, .(X, Zs)) -> APPEND_IN_AGG(Ys, Zs)
The graph contains the following edges 1 >= 1, 2 > 2


----------------------------------------

(13)
YES

----------------------------------------

(14)
Obligation:
Pi DP problem:
The TRS P consists of the following rules:

   APPEND_IN_AAG(.(X, Xs), Ys, .(X, Zs)) -> APPEND_IN_AAG(Xs, Ys, Zs)

The TRS R consists of the following rules:

   sublist_in_gg(X, Y) -> U2_gg(X, Y, append_in_aag(P, X1, Y))
   append_in_aag([], Ys, Ys) -> append_out_aag([], Ys, Ys)
   append_in_aag(.(X, Xs), Ys, .(X, Zs)) -> U1_aag(X, Xs, Ys, Zs, append_in_aag(Xs, Ys, Zs))
   U1_aag(X, Xs, Ys, Zs, append_out_aag(Xs, Ys, Zs)) -> append_out_aag(.(X, Xs), Ys, .(X, Zs))
   U2_gg(X, Y, append_out_aag(P, X1, Y)) -> U3_gg(X, Y, append_in_agg(X2, X, P))
   append_in_agg([], Ys, Ys) -> append_out_agg([], Ys, Ys)
   append_in_agg(.(X, Xs), Ys, .(X, Zs)) -> U1_agg(X, Xs, Ys, Zs, append_in_agg(Xs, Ys, Zs))
   U1_agg(X, Xs, Ys, Zs, append_out_agg(Xs, Ys, Zs)) -> append_out_agg(.(X, Xs), Ys, .(X, Zs))
   U3_gg(X, Y, append_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)

U2_gg(x1, x2, x3)  =  U2_gg(x1, x3)

append_in_aag(x1, x2, x3)  =  append_in_aag(x3)

append_out_aag(x1, x2, x3)  =  append_out_aag(x1, x2)

.(x1, x2)  =  .(x1, x2)

U1_aag(x1, x2, x3, x4, x5)  =  U1_aag(x1, x5)

U3_gg(x1, x2, x3)  =  U3_gg(x3)

append_in_agg(x1, x2, x3)  =  append_in_agg(x2, x3)

append_out_agg(x1, x2, x3)  =  append_out_agg(x1)

U1_agg(x1, x2, x3, x4, x5)  =  U1_agg(x1, x5)

sublist_out_gg(x1, x2)  =  sublist_out_gg

APPEND_IN_AAG(x1, x2, x3)  =  APPEND_IN_AAG(x3)


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

(15) UsableRulesProof (EQUIVALENT)
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:

   APPEND_IN_AAG(.(X, Xs), Ys, .(X, Zs)) -> APPEND_IN_AAG(Xs, Ys, Zs)

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

APPEND_IN_AAG(x1, x2, x3)  =  APPEND_IN_AAG(x3)


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

(17) PiDPToQDPProof (SOUND)
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:

   APPEND_IN_AAG(.(X, Zs)) -> APPEND_IN_AAG(Zs)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
----------------------------------------

(19) QDPSizeChangeProof (EQUIVALENT)
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:
*APPEND_IN_AAG(.(X, Zs)) -> APPEND_IN_AAG(Zs)
The graph contains the following edges 1 > 1


----------------------------------------

(20)
YES
