YES
proof of /hpcwork/ff862203/termcomp26/benchmarks/x7hlc.pl
# AProVE Commit ID: 23a904c96b029b0a549cde0d0d17dbccf967db59 jckassing 20260626 unpublished dirty


Left Termination of the query pattern

sublist(a,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) PiDP
(7) UsableRulesProof [EQUIVALENT, 0 ms]
(8) PiDP
(9) PiDPToQDPProof [SOUND, 0 ms]
(10) QDP
(11) QDPSizeChangeProof [EQUIVALENT, 0 ms]
(12) 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(a,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: (f,b)

append_in_3: (f,f,b)

Transforming Prolog into the following Term Rewriting System:

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

   sublist_in_ag(X, Y) -> U2_ag(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_ag(X, Y, append_out_aag(P, X1, Y)) -> U3_ag(X, Y, append_in_aag(X2, X, P))
   U3_ag(X, Y, append_out_aag(X2, X, P)) -> sublist_out_ag(X, Y)

The argument filtering Pi contains the following mapping:
sublist_in_ag(x1, x2)  =  sublist_in_ag(x2)

U2_ag(x1, x2, x3)  =  U2_ag(x2, x3)

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

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

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

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

U3_ag(x1, x2, x3)  =  U3_ag(x2, x3)

sublist_out_ag(x1, x2)  =  sublist_out_ag(x1, x2)





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_ag(X, Y) -> U2_ag(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_ag(X, Y, append_out_aag(P, X1, Y)) -> U3_ag(X, Y, append_in_aag(X2, X, P))
   U3_ag(X, Y, append_out_aag(X2, X, P)) -> sublist_out_ag(X, Y)

The argument filtering Pi contains the following mapping:
sublist_in_ag(x1, x2)  =  sublist_in_ag(x2)

U2_ag(x1, x2, x3)  =  U2_ag(x2, x3)

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

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

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

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

U3_ag(x1, x2, x3)  =  U3_ag(x2, x3)

sublist_out_ag(x1, x2)  =  sublist_out_ag(x1, x2)



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

(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_AG(X, Y) -> U2_AG(X, Y, append_in_aag(P, X1, Y))
   SUBLIST_IN_AG(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_AG(X, Y, append_out_aag(P, X1, Y)) -> U3_AG(X, Y, append_in_aag(X2, X, P))
   U2_AG(X, Y, append_out_aag(P, X1, Y)) -> APPEND_IN_AAG(X2, X, P)

The TRS R consists of the following rules:

   sublist_in_ag(X, Y) -> U2_ag(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_ag(X, Y, append_out_aag(P, X1, Y)) -> U3_ag(X, Y, append_in_aag(X2, X, P))
   U3_ag(X, Y, append_out_aag(X2, X, P)) -> sublist_out_ag(X, Y)

The argument filtering Pi contains the following mapping:
sublist_in_ag(x1, x2)  =  sublist_in_ag(x2)

U2_ag(x1, x2, x3)  =  U2_ag(x2, x3)

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

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

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

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

U3_ag(x1, x2, x3)  =  U3_ag(x2, x3)

sublist_out_ag(x1, x2)  =  sublist_out_ag(x1, x2)

SUBLIST_IN_AG(x1, x2)  =  SUBLIST_IN_AG(x2)

U2_AG(x1, x2, x3)  =  U2_AG(x2, x3)

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

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

U3_AG(x1, x2, x3)  =  U3_AG(x2, x3)


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

(4)
Obligation:
Pi DP problem:
The TRS P consists of the following rules:

   SUBLIST_IN_AG(X, Y) -> U2_AG(X, Y, append_in_aag(P, X1, Y))
   SUBLIST_IN_AG(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_AG(X, Y, append_out_aag(P, X1, Y)) -> U3_AG(X, Y, append_in_aag(X2, X, P))
   U2_AG(X, Y, append_out_aag(P, X1, Y)) -> APPEND_IN_AAG(X2, X, P)

The TRS R consists of the following rules:

   sublist_in_ag(X, Y) -> U2_ag(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_ag(X, Y, append_out_aag(P, X1, Y)) -> U3_ag(X, Y, append_in_aag(X2, X, P))
   U3_ag(X, Y, append_out_aag(X2, X, P)) -> sublist_out_ag(X, Y)

The argument filtering Pi contains the following mapping:
sublist_in_ag(x1, x2)  =  sublist_in_ag(x2)

U2_ag(x1, x2, x3)  =  U2_ag(x2, x3)

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

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

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

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

U3_ag(x1, x2, x3)  =  U3_ag(x2, x3)

sublist_out_ag(x1, x2)  =  sublist_out_ag(x1, x2)

SUBLIST_IN_AG(x1, x2)  =  SUBLIST_IN_AG(x2)

U2_AG(x1, x2, x3)  =  U2_AG(x2, x3)

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

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

U3_AG(x1, x2, x3)  =  U3_AG(x2, x3)


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

(5) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 5 less nodes.
----------------------------------------

(6)
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_ag(X, Y) -> U2_ag(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_ag(X, Y, append_out_aag(P, X1, Y)) -> U3_ag(X, Y, append_in_aag(X2, X, P))
   U3_ag(X, Y, append_out_aag(X2, X, P)) -> sublist_out_ag(X, Y)

The argument filtering Pi contains the following mapping:
sublist_in_ag(x1, x2)  =  sublist_in_ag(x2)

U2_ag(x1, x2, x3)  =  U2_ag(x2, x3)

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

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

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

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

U3_ag(x1, x2, x3)  =  U3_ag(x2, x3)

sublist_out_ag(x1, x2)  =  sublist_out_ag(x1, x2)

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


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

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

(8)
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
----------------------------------------

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

(10)
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.
----------------------------------------

(11) 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


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

(12)
YES
