Left Termination of the query pattern sublist_in_2(g, a) w.r.t. the given Prolog program could not be shown:

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 NonTerminationProof (⇔)
        ↳20 FALSE
  ↳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
        ↳39 NonTerminationProof (⇔)
        ↳40 FALSE

(0) Obligation:

Clauses:

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

Queries:

sublist(g,a).

(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,f)
app_in: (f,f,f) (b,f,f)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

sublist_in_ga(Xs, Ys) → U1_ga(Xs, Ys, app_in_aaa(X1, Zs, Ys))
app_in_aaa([], X, X) → app_out_aaa([], X, X)
app_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U3_aaa(X, Xs, Ys, Zs, app_in_aaa(Xs, Ys, Zs))
U3_aaa(X, Xs, Ys, Zs, app_out_aaa(Xs, Ys, Zs)) → app_out_aaa(.(X, Xs), Ys, .(X, Zs))
U1_ga(Xs, Ys, app_out_aaa(X1, Zs, Ys)) → U2_ga(Xs, Ys, app_in_gaa(Xs, X2, Zs))
app_in_gaa([], X, X) → app_out_gaa([], X, X)
app_in_gaa(.(X, Xs), Ys, .(X, Zs)) → U3_gaa(X, Xs, Ys, Zs, app_in_gaa(Xs, Ys, Zs))
U3_gaa(X, Xs, Ys, Zs, app_out_gaa(Xs, Ys, Zs)) → app_out_gaa(.(X, Xs), Ys, .(X, Zs))
U2_ga(Xs, Ys, app_out_gaa(Xs, X2, Zs)) → sublist_out_ga(Xs, Ys)

The argument filtering Pi contains the following mapping:
sublist_in_ga(x1, x2)  =  sublist_in_ga(x1)
U1_ga(x1, x2, x3)  =  U1_ga(x1, x3)
app_in_aaa(x1, x2, x3)  =  app_in_aaa
app_out_aaa(x1, x2, x3)  =  app_out_aaa(x1)
U3_aaa(x1, x2, x3, x4, x5)  =  U3_aaa(x5)
.(x1, x2)  =  .(x2)
U2_ga(x1, x2, x3)  =  U2_ga(x3)
app_in_gaa(x1, x2, x3)  =  app_in_gaa(x1)
[]  =  []
app_out_gaa(x1, x2, x3)  =  app_out_gaa
U3_gaa(x1, x2, x3, x4, x5)  =  U3_gaa(x5)
sublist_out_ga(x1, x2)  =  sublist_out_ga

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_ga(Xs, Ys) → U1_ga(Xs, Ys, app_in_aaa(X1, Zs, Ys))
app_in_aaa([], X, X) → app_out_aaa([], X, X)
app_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U3_aaa(X, Xs, Ys, Zs, app_in_aaa(Xs, Ys, Zs))
U3_aaa(X, Xs, Ys, Zs, app_out_aaa(Xs, Ys, Zs)) → app_out_aaa(.(X, Xs), Ys, .(X, Zs))
U1_ga(Xs, Ys, app_out_aaa(X1, Zs, Ys)) → U2_ga(Xs, Ys, app_in_gaa(Xs, X2, Zs))
app_in_gaa([], X, X) → app_out_gaa([], X, X)
app_in_gaa(.(X, Xs), Ys, .(X, Zs)) → U3_gaa(X, Xs, Ys, Zs, app_in_gaa(Xs, Ys, Zs))
U3_gaa(X, Xs, Ys, Zs, app_out_gaa(Xs, Ys, Zs)) → app_out_gaa(.(X, Xs), Ys, .(X, Zs))
U2_ga(Xs, Ys, app_out_gaa(Xs, X2, Zs)) → sublist_out_ga(Xs, Ys)

The argument filtering Pi contains the following mapping:
sublist_in_ga(x1, x2)  =  sublist_in_ga(x1)
U1_ga(x1, x2, x3)  =  U1_ga(x1, x3)
app_in_aaa(x1, x2, x3)  =  app_in_aaa
app_out_aaa(x1, x2, x3)  =  app_out_aaa(x1)
U3_aaa(x1, x2, x3, x4, x5)  =  U3_aaa(x5)
.(x1, x2)  =  .(x2)
U2_ga(x1, x2, x3)  =  U2_ga(x3)
app_in_gaa(x1, x2, x3)  =  app_in_gaa(x1)
[]  =  []
app_out_gaa(x1, x2, x3)  =  app_out_gaa
U3_gaa(x1, x2, x3, x4, x5)  =  U3_gaa(x5)
sublist_out_ga(x1, x2)  =  sublist_out_ga

(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_GA(Xs, Ys) → U1_GA(Xs, Ys, app_in_aaa(X1, Zs, Ys))
SUBLIST_IN_GA(Xs, Ys) → APP_IN_AAA(X1, Zs, Ys)
APP_IN_AAA(.(X, Xs), Ys, .(X, Zs)) → U3_AAA(X, Xs, Ys, Zs, app_in_aaa(Xs, Ys, Zs))
APP_IN_AAA(.(X, Xs), Ys, .(X, Zs)) → APP_IN_AAA(Xs, Ys, Zs)
U1_GA(Xs, Ys, app_out_aaa(X1, Zs, Ys)) → U2_GA(Xs, Ys, app_in_gaa(Xs, X2, Zs))
U1_GA(Xs, Ys, app_out_aaa(X1, Zs, Ys)) → APP_IN_GAA(Xs, X2, Zs)
APP_IN_GAA(.(X, Xs), Ys, .(X, Zs)) → U3_GAA(X, Xs, Ys, Zs, app_in_gaa(Xs, Ys, Zs))
APP_IN_GAA(.(X, Xs), Ys, .(X, Zs)) → APP_IN_GAA(Xs, Ys, Zs)

The TRS R consists of the following rules:

sublist_in_ga(Xs, Ys) → U1_ga(Xs, Ys, app_in_aaa(X1, Zs, Ys))
app_in_aaa([], X, X) → app_out_aaa([], X, X)
app_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U3_aaa(X, Xs, Ys, Zs, app_in_aaa(Xs, Ys, Zs))
U3_aaa(X, Xs, Ys, Zs, app_out_aaa(Xs, Ys, Zs)) → app_out_aaa(.(X, Xs), Ys, .(X, Zs))
U1_ga(Xs, Ys, app_out_aaa(X1, Zs, Ys)) → U2_ga(Xs, Ys, app_in_gaa(Xs, X2, Zs))
app_in_gaa([], X, X) → app_out_gaa([], X, X)
app_in_gaa(.(X, Xs), Ys, .(X, Zs)) → U3_gaa(X, Xs, Ys, Zs, app_in_gaa(Xs, Ys, Zs))
U3_gaa(X, Xs, Ys, Zs, app_out_gaa(Xs, Ys, Zs)) → app_out_gaa(.(X, Xs), Ys, .(X, Zs))
U2_ga(Xs, Ys, app_out_gaa(Xs, X2, Zs)) → sublist_out_ga(Xs, Ys)

The argument filtering Pi contains the following mapping:
sublist_in_ga(x1, x2)  =  sublist_in_ga(x1)
U1_ga(x1, x2, x3)  =  U1_ga(x1, x3)
app_in_aaa(x1, x2, x3)  =  app_in_aaa
app_out_aaa(x1, x2, x3)  =  app_out_aaa(x1)
U3_aaa(x1, x2, x3, x4, x5)  =  U3_aaa(x5)
.(x1, x2)  =  .(x2)
U2_ga(x1, x2, x3)  =  U2_ga(x3)
app_in_gaa(x1, x2, x3)  =  app_in_gaa(x1)
[]  =  []
app_out_gaa(x1, x2, x3)  =  app_out_gaa
U3_gaa(x1, x2, x3, x4, x5)  =  U3_gaa(x5)
sublist_out_ga(x1, x2)  =  sublist_out_ga
SUBLIST_IN_GA(x1, x2)  =  SUBLIST_IN_GA(x1)
U1_GA(x1, x2, x3)  =  U1_GA(x1, x3)
APP_IN_AAA(x1, x2, x3)  =  APP_IN_AAA
U3_AAA(x1, x2, x3, x4, x5)  =  U3_AAA(x5)
U2_GA(x1, x2, x3)  =  U2_GA(x3)
APP_IN_GAA(x1, x2, x3)  =  APP_IN_GAA(x1)
U3_GAA(x1, x2, x3, x4, x5)  =  U3_GAA(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_GA(Xs, Ys) → U1_GA(Xs, Ys, app_in_aaa(X1, Zs, Ys))
SUBLIST_IN_GA(Xs, Ys) → APP_IN_AAA(X1, Zs, Ys)
APP_IN_AAA(.(X, Xs), Ys, .(X, Zs)) → U3_AAA(X, Xs, Ys, Zs, app_in_aaa(Xs, Ys, Zs))
APP_IN_AAA(.(X, Xs), Ys, .(X, Zs)) → APP_IN_AAA(Xs, Ys, Zs)
U1_GA(Xs, Ys, app_out_aaa(X1, Zs, Ys)) → U2_GA(Xs, Ys, app_in_gaa(Xs, X2, Zs))
U1_GA(Xs, Ys, app_out_aaa(X1, Zs, Ys)) → APP_IN_GAA(Xs, X2, Zs)
APP_IN_GAA(.(X, Xs), Ys, .(X, Zs)) → U3_GAA(X, Xs, Ys, Zs, app_in_gaa(Xs, Ys, Zs))
APP_IN_GAA(.(X, Xs), Ys, .(X, Zs)) → APP_IN_GAA(Xs, Ys, Zs)

The TRS R consists of the following rules:

sublist_in_ga(Xs, Ys) → U1_ga(Xs, Ys, app_in_aaa(X1, Zs, Ys))
app_in_aaa([], X, X) → app_out_aaa([], X, X)
app_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U3_aaa(X, Xs, Ys, Zs, app_in_aaa(Xs, Ys, Zs))
U3_aaa(X, Xs, Ys, Zs, app_out_aaa(Xs, Ys, Zs)) → app_out_aaa(.(X, Xs), Ys, .(X, Zs))
U1_ga(Xs, Ys, app_out_aaa(X1, Zs, Ys)) → U2_ga(Xs, Ys, app_in_gaa(Xs, X2, Zs))
app_in_gaa([], X, X) → app_out_gaa([], X, X)
app_in_gaa(.(X, Xs), Ys, .(X, Zs)) → U3_gaa(X, Xs, Ys, Zs, app_in_gaa(Xs, Ys, Zs))
U3_gaa(X, Xs, Ys, Zs, app_out_gaa(Xs, Ys, Zs)) → app_out_gaa(.(X, Xs), Ys, .(X, Zs))
U2_ga(Xs, Ys, app_out_gaa(Xs, X2, Zs)) → sublist_out_ga(Xs, Ys)

The argument filtering Pi contains the following mapping:
sublist_in_ga(x1, x2)  =  sublist_in_ga(x1)
U1_ga(x1, x2, x3)  =  U1_ga(x1, x3)
app_in_aaa(x1, x2, x3)  =  app_in_aaa
app_out_aaa(x1, x2, x3)  =  app_out_aaa(x1)
U3_aaa(x1, x2, x3, x4, x5)  =  U3_aaa(x5)
.(x1, x2)  =  .(x2)
U2_ga(x1, x2, x3)  =  U2_ga(x3)
app_in_gaa(x1, x2, x3)  =  app_in_gaa(x1)
[]  =  []
app_out_gaa(x1, x2, x3)  =  app_out_gaa
U3_gaa(x1, x2, x3, x4, x5)  =  U3_gaa(x5)
sublist_out_ga(x1, x2)  =  sublist_out_ga
SUBLIST_IN_GA(x1, x2)  =  SUBLIST_IN_GA(x1)
U1_GA(x1, x2, x3)  =  U1_GA(x1, x3)
APP_IN_AAA(x1, x2, x3)  =  APP_IN_AAA
U3_AAA(x1, x2, x3, x4, x5)  =  U3_AAA(x5)
U2_GA(x1, x2, x3)  =  U2_GA(x3)
APP_IN_GAA(x1, x2, x3)  =  APP_IN_GAA(x1)
U3_GAA(x1, x2, x3, x4, x5)  =  U3_GAA(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:

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

The TRS R consists of the following rules:

sublist_in_ga(Xs, Ys) → U1_ga(Xs, Ys, app_in_aaa(X1, Zs, Ys))
app_in_aaa([], X, X) → app_out_aaa([], X, X)
app_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U3_aaa(X, Xs, Ys, Zs, app_in_aaa(Xs, Ys, Zs))
U3_aaa(X, Xs, Ys, Zs, app_out_aaa(Xs, Ys, Zs)) → app_out_aaa(.(X, Xs), Ys, .(X, Zs))
U1_ga(Xs, Ys, app_out_aaa(X1, Zs, Ys)) → U2_ga(Xs, Ys, app_in_gaa(Xs, X2, Zs))
app_in_gaa([], X, X) → app_out_gaa([], X, X)
app_in_gaa(.(X, Xs), Ys, .(X, Zs)) → U3_gaa(X, Xs, Ys, Zs, app_in_gaa(Xs, Ys, Zs))
U3_gaa(X, Xs, Ys, Zs, app_out_gaa(Xs, Ys, Zs)) → app_out_gaa(.(X, Xs), Ys, .(X, Zs))
U2_ga(Xs, Ys, app_out_gaa(Xs, X2, Zs)) → sublist_out_ga(Xs, Ys)

The argument filtering Pi contains the following mapping:
sublist_in_ga(x1, x2)  =  sublist_in_ga(x1)
U1_ga(x1, x2, x3)  =  U1_ga(x1, x3)
app_in_aaa(x1, x2, x3)  =  app_in_aaa
app_out_aaa(x1, x2, x3)  =  app_out_aaa(x1)
U3_aaa(x1, x2, x3, x4, x5)  =  U3_aaa(x5)
.(x1, x2)  =  .(x2)
U2_ga(x1, x2, x3)  =  U2_ga(x3)
app_in_gaa(x1, x2, x3)  =  app_in_gaa(x1)
[]  =  []
app_out_gaa(x1, x2, x3)  =  app_out_gaa
U3_gaa(x1, x2, x3, x4, x5)  =  U3_gaa(x5)
sublist_out_ga(x1, x2)  =  sublist_out_ga
APP_IN_GAA(x1, x2, x3)  =  APP_IN_GAA(x1)

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:

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

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

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:

APP_IN_GAA(.(Xs)) → APP_IN_GAA(Xs)

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:

  • APP_IN_GAA(.(Xs)) → APP_IN_GAA(Xs)
    The graph contains the following edges 1 > 1

(13) TRUE

(14) Obligation:

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

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

The TRS R consists of the following rules:

sublist_in_ga(Xs, Ys) → U1_ga(Xs, Ys, app_in_aaa(X1, Zs, Ys))
app_in_aaa([], X, X) → app_out_aaa([], X, X)
app_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U3_aaa(X, Xs, Ys, Zs, app_in_aaa(Xs, Ys, Zs))
U3_aaa(X, Xs, Ys, Zs, app_out_aaa(Xs, Ys, Zs)) → app_out_aaa(.(X, Xs), Ys, .(X, Zs))
U1_ga(Xs, Ys, app_out_aaa(X1, Zs, Ys)) → U2_ga(Xs, Ys, app_in_gaa(Xs, X2, Zs))
app_in_gaa([], X, X) → app_out_gaa([], X, X)
app_in_gaa(.(X, Xs), Ys, .(X, Zs)) → U3_gaa(X, Xs, Ys, Zs, app_in_gaa(Xs, Ys, Zs))
U3_gaa(X, Xs, Ys, Zs, app_out_gaa(Xs, Ys, Zs)) → app_out_gaa(.(X, Xs), Ys, .(X, Zs))
U2_ga(Xs, Ys, app_out_gaa(Xs, X2, Zs)) → sublist_out_ga(Xs, Ys)

The argument filtering Pi contains the following mapping:
sublist_in_ga(x1, x2)  =  sublist_in_ga(x1)
U1_ga(x1, x2, x3)  =  U1_ga(x1, x3)
app_in_aaa(x1, x2, x3)  =  app_in_aaa
app_out_aaa(x1, x2, x3)  =  app_out_aaa(x1)
U3_aaa(x1, x2, x3, x4, x5)  =  U3_aaa(x5)
.(x1, x2)  =  .(x2)
U2_ga(x1, x2, x3)  =  U2_ga(x3)
app_in_gaa(x1, x2, x3)  =  app_in_gaa(x1)
[]  =  []
app_out_gaa(x1, x2, x3)  =  app_out_gaa
U3_gaa(x1, x2, x3, x4, x5)  =  U3_gaa(x5)
sublist_out_ga(x1, x2)  =  sublist_out_ga
APP_IN_AAA(x1, x2, x3)  =  APP_IN_AAA

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:

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

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

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:

APP_IN_AAA → APP_IN_AAA

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

(19) NonTerminationProof (EQUIVALENT transformation)

We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.

s = APP_IN_AAA evaluates to t =APP_IN_AAA

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
  • Semiunifier: [ ]
  • Matcher: [ ]



Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from APP_IN_AAA to APP_IN_AAA.



(20) FALSE

(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,f)
app_in: (f,f,f) (b,f,f)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

sublist_in_ga(Xs, Ys) → U1_ga(Xs, Ys, app_in_aaa(X1, Zs, Ys))
app_in_aaa([], X, X) → app_out_aaa([], X, X)
app_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U3_aaa(X, Xs, Ys, Zs, app_in_aaa(Xs, Ys, Zs))
U3_aaa(X, Xs, Ys, Zs, app_out_aaa(Xs, Ys, Zs)) → app_out_aaa(.(X, Xs), Ys, .(X, Zs))
U1_ga(Xs, Ys, app_out_aaa(X1, Zs, Ys)) → U2_ga(Xs, Ys, app_in_gaa(Xs, X2, Zs))
app_in_gaa([], X, X) → app_out_gaa([], X, X)
app_in_gaa(.(X, Xs), Ys, .(X, Zs)) → U3_gaa(X, Xs, Ys, Zs, app_in_gaa(Xs, Ys, Zs))
U3_gaa(X, Xs, Ys, Zs, app_out_gaa(Xs, Ys, Zs)) → app_out_gaa(.(X, Xs), Ys, .(X, Zs))
U2_ga(Xs, Ys, app_out_gaa(Xs, X2, Zs)) → sublist_out_ga(Xs, Ys)

The argument filtering Pi contains the following mapping:
sublist_in_ga(x1, x2)  =  sublist_in_ga(x1)
U1_ga(x1, x2, x3)  =  U1_ga(x1, x3)
app_in_aaa(x1, x2, x3)  =  app_in_aaa
app_out_aaa(x1, x2, x3)  =  app_out_aaa(x1)
U3_aaa(x1, x2, x3, x4, x5)  =  U3_aaa(x5)
.(x1, x2)  =  .(x2)
U2_ga(x1, x2, x3)  =  U2_ga(x1, x3)
app_in_gaa(x1, x2, x3)  =  app_in_gaa(x1)
[]  =  []
app_out_gaa(x1, x2, x3)  =  app_out_gaa(x1)
U3_gaa(x1, x2, x3, x4, x5)  =  U3_gaa(x2, x5)
sublist_out_ga(x1, x2)  =  sublist_out_ga(x1)

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_ga(Xs, Ys) → U1_ga(Xs, Ys, app_in_aaa(X1, Zs, Ys))
app_in_aaa([], X, X) → app_out_aaa([], X, X)
app_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U3_aaa(X, Xs, Ys, Zs, app_in_aaa(Xs, Ys, Zs))
U3_aaa(X, Xs, Ys, Zs, app_out_aaa(Xs, Ys, Zs)) → app_out_aaa(.(X, Xs), Ys, .(X, Zs))
U1_ga(Xs, Ys, app_out_aaa(X1, Zs, Ys)) → U2_ga(Xs, Ys, app_in_gaa(Xs, X2, Zs))
app_in_gaa([], X, X) → app_out_gaa([], X, X)
app_in_gaa(.(X, Xs), Ys, .(X, Zs)) → U3_gaa(X, Xs, Ys, Zs, app_in_gaa(Xs, Ys, Zs))
U3_gaa(X, Xs, Ys, Zs, app_out_gaa(Xs, Ys, Zs)) → app_out_gaa(.(X, Xs), Ys, .(X, Zs))
U2_ga(Xs, Ys, app_out_gaa(Xs, X2, Zs)) → sublist_out_ga(Xs, Ys)

The argument filtering Pi contains the following mapping:
sublist_in_ga(x1, x2)  =  sublist_in_ga(x1)
U1_ga(x1, x2, x3)  =  U1_ga(x1, x3)
app_in_aaa(x1, x2, x3)  =  app_in_aaa
app_out_aaa(x1, x2, x3)  =  app_out_aaa(x1)
U3_aaa(x1, x2, x3, x4, x5)  =  U3_aaa(x5)
.(x1, x2)  =  .(x2)
U2_ga(x1, x2, x3)  =  U2_ga(x1, x3)
app_in_gaa(x1, x2, x3)  =  app_in_gaa(x1)
[]  =  []
app_out_gaa(x1, x2, x3)  =  app_out_gaa(x1)
U3_gaa(x1, x2, x3, x4, x5)  =  U3_gaa(x2, x5)
sublist_out_ga(x1, x2)  =  sublist_out_ga(x1)

(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_GA(Xs, Ys) → U1_GA(Xs, Ys, app_in_aaa(X1, Zs, Ys))
SUBLIST_IN_GA(Xs, Ys) → APP_IN_AAA(X1, Zs, Ys)
APP_IN_AAA(.(X, Xs), Ys, .(X, Zs)) → U3_AAA(X, Xs, Ys, Zs, app_in_aaa(Xs, Ys, Zs))
APP_IN_AAA(.(X, Xs), Ys, .(X, Zs)) → APP_IN_AAA(Xs, Ys, Zs)
U1_GA(Xs, Ys, app_out_aaa(X1, Zs, Ys)) → U2_GA(Xs, Ys, app_in_gaa(Xs, X2, Zs))
U1_GA(Xs, Ys, app_out_aaa(X1, Zs, Ys)) → APP_IN_GAA(Xs, X2, Zs)
APP_IN_GAA(.(X, Xs), Ys, .(X, Zs)) → U3_GAA(X, Xs, Ys, Zs, app_in_gaa(Xs, Ys, Zs))
APP_IN_GAA(.(X, Xs), Ys, .(X, Zs)) → APP_IN_GAA(Xs, Ys, Zs)

The TRS R consists of the following rules:

sublist_in_ga(Xs, Ys) → U1_ga(Xs, Ys, app_in_aaa(X1, Zs, Ys))
app_in_aaa([], X, X) → app_out_aaa([], X, X)
app_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U3_aaa(X, Xs, Ys, Zs, app_in_aaa(Xs, Ys, Zs))
U3_aaa(X, Xs, Ys, Zs, app_out_aaa(Xs, Ys, Zs)) → app_out_aaa(.(X, Xs), Ys, .(X, Zs))
U1_ga(Xs, Ys, app_out_aaa(X1, Zs, Ys)) → U2_ga(Xs, Ys, app_in_gaa(Xs, X2, Zs))
app_in_gaa([], X, X) → app_out_gaa([], X, X)
app_in_gaa(.(X, Xs), Ys, .(X, Zs)) → U3_gaa(X, Xs, Ys, Zs, app_in_gaa(Xs, Ys, Zs))
U3_gaa(X, Xs, Ys, Zs, app_out_gaa(Xs, Ys, Zs)) → app_out_gaa(.(X, Xs), Ys, .(X, Zs))
U2_ga(Xs, Ys, app_out_gaa(Xs, X2, Zs)) → sublist_out_ga(Xs, Ys)

The argument filtering Pi contains the following mapping:
sublist_in_ga(x1, x2)  =  sublist_in_ga(x1)
U1_ga(x1, x2, x3)  =  U1_ga(x1, x3)
app_in_aaa(x1, x2, x3)  =  app_in_aaa
app_out_aaa(x1, x2, x3)  =  app_out_aaa(x1)
U3_aaa(x1, x2, x3, x4, x5)  =  U3_aaa(x5)
.(x1, x2)  =  .(x2)
U2_ga(x1, x2, x3)  =  U2_ga(x1, x3)
app_in_gaa(x1, x2, x3)  =  app_in_gaa(x1)
[]  =  []
app_out_gaa(x1, x2, x3)  =  app_out_gaa(x1)
U3_gaa(x1, x2, x3, x4, x5)  =  U3_gaa(x2, x5)
sublist_out_ga(x1, x2)  =  sublist_out_ga(x1)
SUBLIST_IN_GA(x1, x2)  =  SUBLIST_IN_GA(x1)
U1_GA(x1, x2, x3)  =  U1_GA(x1, x3)
APP_IN_AAA(x1, x2, x3)  =  APP_IN_AAA
U3_AAA(x1, x2, x3, x4, x5)  =  U3_AAA(x5)
U2_GA(x1, x2, x3)  =  U2_GA(x1, x3)
APP_IN_GAA(x1, x2, x3)  =  APP_IN_GAA(x1)
U3_GAA(x1, x2, x3, x4, x5)  =  U3_GAA(x2, 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_GA(Xs, Ys) → U1_GA(Xs, Ys, app_in_aaa(X1, Zs, Ys))
SUBLIST_IN_GA(Xs, Ys) → APP_IN_AAA(X1, Zs, Ys)
APP_IN_AAA(.(X, Xs), Ys, .(X, Zs)) → U3_AAA(X, Xs, Ys, Zs, app_in_aaa(Xs, Ys, Zs))
APP_IN_AAA(.(X, Xs), Ys, .(X, Zs)) → APP_IN_AAA(Xs, Ys, Zs)
U1_GA(Xs, Ys, app_out_aaa(X1, Zs, Ys)) → U2_GA(Xs, Ys, app_in_gaa(Xs, X2, Zs))
U1_GA(Xs, Ys, app_out_aaa(X1, Zs, Ys)) → APP_IN_GAA(Xs, X2, Zs)
APP_IN_GAA(.(X, Xs), Ys, .(X, Zs)) → U3_GAA(X, Xs, Ys, Zs, app_in_gaa(Xs, Ys, Zs))
APP_IN_GAA(.(X, Xs), Ys, .(X, Zs)) → APP_IN_GAA(Xs, Ys, Zs)

The TRS R consists of the following rules:

sublist_in_ga(Xs, Ys) → U1_ga(Xs, Ys, app_in_aaa(X1, Zs, Ys))
app_in_aaa([], X, X) → app_out_aaa([], X, X)
app_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U3_aaa(X, Xs, Ys, Zs, app_in_aaa(Xs, Ys, Zs))
U3_aaa(X, Xs, Ys, Zs, app_out_aaa(Xs, Ys, Zs)) → app_out_aaa(.(X, Xs), Ys, .(X, Zs))
U1_ga(Xs, Ys, app_out_aaa(X1, Zs, Ys)) → U2_ga(Xs, Ys, app_in_gaa(Xs, X2, Zs))
app_in_gaa([], X, X) → app_out_gaa([], X, X)
app_in_gaa(.(X, Xs), Ys, .(X, Zs)) → U3_gaa(X, Xs, Ys, Zs, app_in_gaa(Xs, Ys, Zs))
U3_gaa(X, Xs, Ys, Zs, app_out_gaa(Xs, Ys, Zs)) → app_out_gaa(.(X, Xs), Ys, .(X, Zs))
U2_ga(Xs, Ys, app_out_gaa(Xs, X2, Zs)) → sublist_out_ga(Xs, Ys)

The argument filtering Pi contains the following mapping:
sublist_in_ga(x1, x2)  =  sublist_in_ga(x1)
U1_ga(x1, x2, x3)  =  U1_ga(x1, x3)
app_in_aaa(x1, x2, x3)  =  app_in_aaa
app_out_aaa(x1, x2, x3)  =  app_out_aaa(x1)
U3_aaa(x1, x2, x3, x4, x5)  =  U3_aaa(x5)
.(x1, x2)  =  .(x2)
U2_ga(x1, x2, x3)  =  U2_ga(x1, x3)
app_in_gaa(x1, x2, x3)  =  app_in_gaa(x1)
[]  =  []
app_out_gaa(x1, x2, x3)  =  app_out_gaa(x1)
U3_gaa(x1, x2, x3, x4, x5)  =  U3_gaa(x2, x5)
sublist_out_ga(x1, x2)  =  sublist_out_ga(x1)
SUBLIST_IN_GA(x1, x2)  =  SUBLIST_IN_GA(x1)
U1_GA(x1, x2, x3)  =  U1_GA(x1, x3)
APP_IN_AAA(x1, x2, x3)  =  APP_IN_AAA
U3_AAA(x1, x2, x3, x4, x5)  =  U3_AAA(x5)
U2_GA(x1, x2, x3)  =  U2_GA(x1, x3)
APP_IN_GAA(x1, x2, x3)  =  APP_IN_GAA(x1)
U3_GAA(x1, x2, x3, x4, x5)  =  U3_GAA(x2, 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:

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

The TRS R consists of the following rules:

sublist_in_ga(Xs, Ys) → U1_ga(Xs, Ys, app_in_aaa(X1, Zs, Ys))
app_in_aaa([], X, X) → app_out_aaa([], X, X)
app_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U3_aaa(X, Xs, Ys, Zs, app_in_aaa(Xs, Ys, Zs))
U3_aaa(X, Xs, Ys, Zs, app_out_aaa(Xs, Ys, Zs)) → app_out_aaa(.(X, Xs), Ys, .(X, Zs))
U1_ga(Xs, Ys, app_out_aaa(X1, Zs, Ys)) → U2_ga(Xs, Ys, app_in_gaa(Xs, X2, Zs))
app_in_gaa([], X, X) → app_out_gaa([], X, X)
app_in_gaa(.(X, Xs), Ys, .(X, Zs)) → U3_gaa(X, Xs, Ys, Zs, app_in_gaa(Xs, Ys, Zs))
U3_gaa(X, Xs, Ys, Zs, app_out_gaa(Xs, Ys, Zs)) → app_out_gaa(.(X, Xs), Ys, .(X, Zs))
U2_ga(Xs, Ys, app_out_gaa(Xs, X2, Zs)) → sublist_out_ga(Xs, Ys)

The argument filtering Pi contains the following mapping:
sublist_in_ga(x1, x2)  =  sublist_in_ga(x1)
U1_ga(x1, x2, x3)  =  U1_ga(x1, x3)
app_in_aaa(x1, x2, x3)  =  app_in_aaa
app_out_aaa(x1, x2, x3)  =  app_out_aaa(x1)
U3_aaa(x1, x2, x3, x4, x5)  =  U3_aaa(x5)
.(x1, x2)  =  .(x2)
U2_ga(x1, x2, x3)  =  U2_ga(x1, x3)
app_in_gaa(x1, x2, x3)  =  app_in_gaa(x1)
[]  =  []
app_out_gaa(x1, x2, x3)  =  app_out_gaa(x1)
U3_gaa(x1, x2, x3, x4, x5)  =  U3_gaa(x2, x5)
sublist_out_ga(x1, x2)  =  sublist_out_ga(x1)
APP_IN_GAA(x1, x2, x3)  =  APP_IN_GAA(x1)

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:

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

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

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:

APP_IN_GAA(.(Xs)) → APP_IN_GAA(Xs)

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:

  • APP_IN_GAA(.(Xs)) → APP_IN_GAA(Xs)
    The graph contains the following edges 1 > 1

(33) TRUE

(34) Obligation:

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

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

The TRS R consists of the following rules:

sublist_in_ga(Xs, Ys) → U1_ga(Xs, Ys, app_in_aaa(X1, Zs, Ys))
app_in_aaa([], X, X) → app_out_aaa([], X, X)
app_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U3_aaa(X, Xs, Ys, Zs, app_in_aaa(Xs, Ys, Zs))
U3_aaa(X, Xs, Ys, Zs, app_out_aaa(Xs, Ys, Zs)) → app_out_aaa(.(X, Xs), Ys, .(X, Zs))
U1_ga(Xs, Ys, app_out_aaa(X1, Zs, Ys)) → U2_ga(Xs, Ys, app_in_gaa(Xs, X2, Zs))
app_in_gaa([], X, X) → app_out_gaa([], X, X)
app_in_gaa(.(X, Xs), Ys, .(X, Zs)) → U3_gaa(X, Xs, Ys, Zs, app_in_gaa(Xs, Ys, Zs))
U3_gaa(X, Xs, Ys, Zs, app_out_gaa(Xs, Ys, Zs)) → app_out_gaa(.(X, Xs), Ys, .(X, Zs))
U2_ga(Xs, Ys, app_out_gaa(Xs, X2, Zs)) → sublist_out_ga(Xs, Ys)

The argument filtering Pi contains the following mapping:
sublist_in_ga(x1, x2)  =  sublist_in_ga(x1)
U1_ga(x1, x2, x3)  =  U1_ga(x1, x3)
app_in_aaa(x1, x2, x3)  =  app_in_aaa
app_out_aaa(x1, x2, x3)  =  app_out_aaa(x1)
U3_aaa(x1, x2, x3, x4, x5)  =  U3_aaa(x5)
.(x1, x2)  =  .(x2)
U2_ga(x1, x2, x3)  =  U2_ga(x1, x3)
app_in_gaa(x1, x2, x3)  =  app_in_gaa(x1)
[]  =  []
app_out_gaa(x1, x2, x3)  =  app_out_gaa(x1)
U3_gaa(x1, x2, x3, x4, x5)  =  U3_gaa(x2, x5)
sublist_out_ga(x1, x2)  =  sublist_out_ga(x1)
APP_IN_AAA(x1, x2, x3)  =  APP_IN_AAA

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:

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

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

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:

APP_IN_AAA → APP_IN_AAA

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

(39) NonTerminationProof (EQUIVALENT transformation)

We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.

s = APP_IN_AAA evaluates to t =APP_IN_AAA

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
  • Semiunifier: [ ]
  • Matcher: [ ]



Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from APP_IN_AAA to APP_IN_AAA.



(40) FALSE