Left Termination of the query pattern append3_in_4(a, a, a, g) 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:

append([], L, L).
append(.(H, L1), L2, .(H, L3)) :- append(L1, L2, L3).
append3(A, B, C, D) :- ','(append(A, B, E), append(E, C, D)).

Queries:

append3(a,a,a,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:
append3_in: (f,f,f,b)
append_in: (f,f,f) (f,f,b)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

append3_in_aaag(A, B, C, D) → U2_aaag(A, B, C, D, append_in_aaa(A, B, E))
append_in_aaa([], L, L) → append_out_aaa([], L, L)
append_in_aaa(.(H, L1), L2, .(H, L3)) → U1_aaa(H, L1, L2, L3, append_in_aaa(L1, L2, L3))
U1_aaa(H, L1, L2, L3, append_out_aaa(L1, L2, L3)) → append_out_aaa(.(H, L1), L2, .(H, L3))
U2_aaag(A, B, C, D, append_out_aaa(A, B, E)) → U3_aaag(A, B, C, D, append_in_aag(E, C, D))
append_in_aag([], L, L) → append_out_aag([], L, L)
append_in_aag(.(H, L1), L2, .(H, L3)) → U1_aag(H, L1, L2, L3, append_in_aag(L1, L2, L3))
U1_aag(H, L1, L2, L3, append_out_aag(L1, L2, L3)) → append_out_aag(.(H, L1), L2, .(H, L3))
U3_aaag(A, B, C, D, append_out_aag(E, C, D)) → append3_out_aaag(A, B, C, D)

The argument filtering Pi contains the following mapping:
append3_in_aaag(x1, x2, x3, x4)  =  append3_in_aaag(x4)
U2_aaag(x1, x2, x3, x4, x5)  =  U2_aaag(x4, x5)
append_in_aaa(x1, x2, x3)  =  append_in_aaa
append_out_aaa(x1, x2, x3)  =  append_out_aaa(x1)
U1_aaa(x1, x2, x3, x4, x5)  =  U1_aaa(x5)
.(x1, x2)  =  .(x2)
U3_aaag(x1, x2, x3, x4, x5)  =  U3_aaag(x1, x5)
append_in_aag(x1, x2, x3)  =  append_in_aag(x3)
append_out_aag(x1, x2, x3)  =  append_out_aag(x1, x2)
U1_aag(x1, x2, x3, x4, x5)  =  U1_aag(x5)
append3_out_aaag(x1, x2, x3, x4)  =  append3_out_aaag(x1, x3)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(2) Obligation:

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

append3_in_aaag(A, B, C, D) → U2_aaag(A, B, C, D, append_in_aaa(A, B, E))
append_in_aaa([], L, L) → append_out_aaa([], L, L)
append_in_aaa(.(H, L1), L2, .(H, L3)) → U1_aaa(H, L1, L2, L3, append_in_aaa(L1, L2, L3))
U1_aaa(H, L1, L2, L3, append_out_aaa(L1, L2, L3)) → append_out_aaa(.(H, L1), L2, .(H, L3))
U2_aaag(A, B, C, D, append_out_aaa(A, B, E)) → U3_aaag(A, B, C, D, append_in_aag(E, C, D))
append_in_aag([], L, L) → append_out_aag([], L, L)
append_in_aag(.(H, L1), L2, .(H, L3)) → U1_aag(H, L1, L2, L3, append_in_aag(L1, L2, L3))
U1_aag(H, L1, L2, L3, append_out_aag(L1, L2, L3)) → append_out_aag(.(H, L1), L2, .(H, L3))
U3_aaag(A, B, C, D, append_out_aag(E, C, D)) → append3_out_aaag(A, B, C, D)

The argument filtering Pi contains the following mapping:
append3_in_aaag(x1, x2, x3, x4)  =  append3_in_aaag(x4)
U2_aaag(x1, x2, x3, x4, x5)  =  U2_aaag(x4, x5)
append_in_aaa(x1, x2, x3)  =  append_in_aaa
append_out_aaa(x1, x2, x3)  =  append_out_aaa(x1)
U1_aaa(x1, x2, x3, x4, x5)  =  U1_aaa(x5)
.(x1, x2)  =  .(x2)
U3_aaag(x1, x2, x3, x4, x5)  =  U3_aaag(x1, x5)
append_in_aag(x1, x2, x3)  =  append_in_aag(x3)
append_out_aag(x1, x2, x3)  =  append_out_aag(x1, x2)
U1_aag(x1, x2, x3, x4, x5)  =  U1_aag(x5)
append3_out_aaag(x1, x2, x3, x4)  =  append3_out_aaag(x1, x3)

(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:

APPEND3_IN_AAAG(A, B, C, D) → U2_AAAG(A, B, C, D, append_in_aaa(A, B, E))
APPEND3_IN_AAAG(A, B, C, D) → APPEND_IN_AAA(A, B, E)
APPEND_IN_AAA(.(H, L1), L2, .(H, L3)) → U1_AAA(H, L1, L2, L3, append_in_aaa(L1, L2, L3))
APPEND_IN_AAA(.(H, L1), L2, .(H, L3)) → APPEND_IN_AAA(L1, L2, L3)
U2_AAAG(A, B, C, D, append_out_aaa(A, B, E)) → U3_AAAG(A, B, C, D, append_in_aag(E, C, D))
U2_AAAG(A, B, C, D, append_out_aaa(A, B, E)) → APPEND_IN_AAG(E, C, D)
APPEND_IN_AAG(.(H, L1), L2, .(H, L3)) → U1_AAG(H, L1, L2, L3, append_in_aag(L1, L2, L3))
APPEND_IN_AAG(.(H, L1), L2, .(H, L3)) → APPEND_IN_AAG(L1, L2, L3)

The TRS R consists of the following rules:

append3_in_aaag(A, B, C, D) → U2_aaag(A, B, C, D, append_in_aaa(A, B, E))
append_in_aaa([], L, L) → append_out_aaa([], L, L)
append_in_aaa(.(H, L1), L2, .(H, L3)) → U1_aaa(H, L1, L2, L3, append_in_aaa(L1, L2, L3))
U1_aaa(H, L1, L2, L3, append_out_aaa(L1, L2, L3)) → append_out_aaa(.(H, L1), L2, .(H, L3))
U2_aaag(A, B, C, D, append_out_aaa(A, B, E)) → U3_aaag(A, B, C, D, append_in_aag(E, C, D))
append_in_aag([], L, L) → append_out_aag([], L, L)
append_in_aag(.(H, L1), L2, .(H, L3)) → U1_aag(H, L1, L2, L3, append_in_aag(L1, L2, L3))
U1_aag(H, L1, L2, L3, append_out_aag(L1, L2, L3)) → append_out_aag(.(H, L1), L2, .(H, L3))
U3_aaag(A, B, C, D, append_out_aag(E, C, D)) → append3_out_aaag(A, B, C, D)

The argument filtering Pi contains the following mapping:
append3_in_aaag(x1, x2, x3, x4)  =  append3_in_aaag(x4)
U2_aaag(x1, x2, x3, x4, x5)  =  U2_aaag(x4, x5)
append_in_aaa(x1, x2, x3)  =  append_in_aaa
append_out_aaa(x1, x2, x3)  =  append_out_aaa(x1)
U1_aaa(x1, x2, x3, x4, x5)  =  U1_aaa(x5)
.(x1, x2)  =  .(x2)
U3_aaag(x1, x2, x3, x4, x5)  =  U3_aaag(x1, x5)
append_in_aag(x1, x2, x3)  =  append_in_aag(x3)
append_out_aag(x1, x2, x3)  =  append_out_aag(x1, x2)
U1_aag(x1, x2, x3, x4, x5)  =  U1_aag(x5)
append3_out_aaag(x1, x2, x3, x4)  =  append3_out_aaag(x1, x3)
APPEND3_IN_AAAG(x1, x2, x3, x4)  =  APPEND3_IN_AAAG(x4)
U2_AAAG(x1, x2, x3, x4, x5)  =  U2_AAAG(x4, x5)
APPEND_IN_AAA(x1, x2, x3)  =  APPEND_IN_AAA
U1_AAA(x1, x2, x3, x4, x5)  =  U1_AAA(x5)
U3_AAAG(x1, x2, x3, x4, x5)  =  U3_AAAG(x1, x5)
APPEND_IN_AAG(x1, x2, x3)  =  APPEND_IN_AAG(x3)
U1_AAG(x1, x2, x3, x4, x5)  =  U1_AAG(x5)

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

(4) Obligation:

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

APPEND3_IN_AAAG(A, B, C, D) → U2_AAAG(A, B, C, D, append_in_aaa(A, B, E))
APPEND3_IN_AAAG(A, B, C, D) → APPEND_IN_AAA(A, B, E)
APPEND_IN_AAA(.(H, L1), L2, .(H, L3)) → U1_AAA(H, L1, L2, L3, append_in_aaa(L1, L2, L3))
APPEND_IN_AAA(.(H, L1), L2, .(H, L3)) → APPEND_IN_AAA(L1, L2, L3)
U2_AAAG(A, B, C, D, append_out_aaa(A, B, E)) → U3_AAAG(A, B, C, D, append_in_aag(E, C, D))
U2_AAAG(A, B, C, D, append_out_aaa(A, B, E)) → APPEND_IN_AAG(E, C, D)
APPEND_IN_AAG(.(H, L1), L2, .(H, L3)) → U1_AAG(H, L1, L2, L3, append_in_aag(L1, L2, L3))
APPEND_IN_AAG(.(H, L1), L2, .(H, L3)) → APPEND_IN_AAG(L1, L2, L3)

The TRS R consists of the following rules:

append3_in_aaag(A, B, C, D) → U2_aaag(A, B, C, D, append_in_aaa(A, B, E))
append_in_aaa([], L, L) → append_out_aaa([], L, L)
append_in_aaa(.(H, L1), L2, .(H, L3)) → U1_aaa(H, L1, L2, L3, append_in_aaa(L1, L2, L3))
U1_aaa(H, L1, L2, L3, append_out_aaa(L1, L2, L3)) → append_out_aaa(.(H, L1), L2, .(H, L3))
U2_aaag(A, B, C, D, append_out_aaa(A, B, E)) → U3_aaag(A, B, C, D, append_in_aag(E, C, D))
append_in_aag([], L, L) → append_out_aag([], L, L)
append_in_aag(.(H, L1), L2, .(H, L3)) → U1_aag(H, L1, L2, L3, append_in_aag(L1, L2, L3))
U1_aag(H, L1, L2, L3, append_out_aag(L1, L2, L3)) → append_out_aag(.(H, L1), L2, .(H, L3))
U3_aaag(A, B, C, D, append_out_aag(E, C, D)) → append3_out_aaag(A, B, C, D)

The argument filtering Pi contains the following mapping:
append3_in_aaag(x1, x2, x3, x4)  =  append3_in_aaag(x4)
U2_aaag(x1, x2, x3, x4, x5)  =  U2_aaag(x4, x5)
append_in_aaa(x1, x2, x3)  =  append_in_aaa
append_out_aaa(x1, x2, x3)  =  append_out_aaa(x1)
U1_aaa(x1, x2, x3, x4, x5)  =  U1_aaa(x5)
.(x1, x2)  =  .(x2)
U3_aaag(x1, x2, x3, x4, x5)  =  U3_aaag(x1, x5)
append_in_aag(x1, x2, x3)  =  append_in_aag(x3)
append_out_aag(x1, x2, x3)  =  append_out_aag(x1, x2)
U1_aag(x1, x2, x3, x4, x5)  =  U1_aag(x5)
append3_out_aaag(x1, x2, x3, x4)  =  append3_out_aaag(x1, x3)
APPEND3_IN_AAAG(x1, x2, x3, x4)  =  APPEND3_IN_AAAG(x4)
U2_AAAG(x1, x2, x3, x4, x5)  =  U2_AAAG(x4, x5)
APPEND_IN_AAA(x1, x2, x3)  =  APPEND_IN_AAA
U1_AAA(x1, x2, x3, x4, x5)  =  U1_AAA(x5)
U3_AAAG(x1, x2, x3, x4, x5)  =  U3_AAAG(x1, x5)
APPEND_IN_AAG(x1, x2, x3)  =  APPEND_IN_AAG(x3)
U1_AAG(x1, x2, x3, x4, x5)  =  U1_AAG(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:

APPEND_IN_AAG(.(H, L1), L2, .(H, L3)) → APPEND_IN_AAG(L1, L2, L3)

The TRS R consists of the following rules:

append3_in_aaag(A, B, C, D) → U2_aaag(A, B, C, D, append_in_aaa(A, B, E))
append_in_aaa([], L, L) → append_out_aaa([], L, L)
append_in_aaa(.(H, L1), L2, .(H, L3)) → U1_aaa(H, L1, L2, L3, append_in_aaa(L1, L2, L3))
U1_aaa(H, L1, L2, L3, append_out_aaa(L1, L2, L3)) → append_out_aaa(.(H, L1), L2, .(H, L3))
U2_aaag(A, B, C, D, append_out_aaa(A, B, E)) → U3_aaag(A, B, C, D, append_in_aag(E, C, D))
append_in_aag([], L, L) → append_out_aag([], L, L)
append_in_aag(.(H, L1), L2, .(H, L3)) → U1_aag(H, L1, L2, L3, append_in_aag(L1, L2, L3))
U1_aag(H, L1, L2, L3, append_out_aag(L1, L2, L3)) → append_out_aag(.(H, L1), L2, .(H, L3))
U3_aaag(A, B, C, D, append_out_aag(E, C, D)) → append3_out_aaag(A, B, C, D)

The argument filtering Pi contains the following mapping:
append3_in_aaag(x1, x2, x3, x4)  =  append3_in_aaag(x4)
U2_aaag(x1, x2, x3, x4, x5)  =  U2_aaag(x4, x5)
append_in_aaa(x1, x2, x3)  =  append_in_aaa
append_out_aaa(x1, x2, x3)  =  append_out_aaa(x1)
U1_aaa(x1, x2, x3, x4, x5)  =  U1_aaa(x5)
.(x1, x2)  =  .(x2)
U3_aaag(x1, x2, x3, x4, x5)  =  U3_aaag(x1, x5)
append_in_aag(x1, x2, x3)  =  append_in_aag(x3)
append_out_aag(x1, x2, x3)  =  append_out_aag(x1, x2)
U1_aag(x1, x2, x3, x4, x5)  =  U1_aag(x5)
append3_out_aaag(x1, x2, x3, x4)  =  append3_out_aaag(x1, x3)
APPEND_IN_AAG(x1, x2, x3)  =  APPEND_IN_AAG(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:

APPEND_IN_AAG(.(H, L1), L2, .(H, L3)) → APPEND_IN_AAG(L1, L2, L3)

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

APPEND_IN_AAG(.(L3)) → APPEND_IN_AAG(L3)

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:

  • APPEND_IN_AAG(.(L3)) → APPEND_IN_AAG(L3)
    The graph contains the following edges 1 > 1

(13) TRUE

(14) Obligation:

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

APPEND_IN_AAA(.(H, L1), L2, .(H, L3)) → APPEND_IN_AAA(L1, L2, L3)

The TRS R consists of the following rules:

append3_in_aaag(A, B, C, D) → U2_aaag(A, B, C, D, append_in_aaa(A, B, E))
append_in_aaa([], L, L) → append_out_aaa([], L, L)
append_in_aaa(.(H, L1), L2, .(H, L3)) → U1_aaa(H, L1, L2, L3, append_in_aaa(L1, L2, L3))
U1_aaa(H, L1, L2, L3, append_out_aaa(L1, L2, L3)) → append_out_aaa(.(H, L1), L2, .(H, L3))
U2_aaag(A, B, C, D, append_out_aaa(A, B, E)) → U3_aaag(A, B, C, D, append_in_aag(E, C, D))
append_in_aag([], L, L) → append_out_aag([], L, L)
append_in_aag(.(H, L1), L2, .(H, L3)) → U1_aag(H, L1, L2, L3, append_in_aag(L1, L2, L3))
U1_aag(H, L1, L2, L3, append_out_aag(L1, L2, L3)) → append_out_aag(.(H, L1), L2, .(H, L3))
U3_aaag(A, B, C, D, append_out_aag(E, C, D)) → append3_out_aaag(A, B, C, D)

The argument filtering Pi contains the following mapping:
append3_in_aaag(x1, x2, x3, x4)  =  append3_in_aaag(x4)
U2_aaag(x1, x2, x3, x4, x5)  =  U2_aaag(x4, x5)
append_in_aaa(x1, x2, x3)  =  append_in_aaa
append_out_aaa(x1, x2, x3)  =  append_out_aaa(x1)
U1_aaa(x1, x2, x3, x4, x5)  =  U1_aaa(x5)
.(x1, x2)  =  .(x2)
U3_aaag(x1, x2, x3, x4, x5)  =  U3_aaag(x1, x5)
append_in_aag(x1, x2, x3)  =  append_in_aag(x3)
append_out_aag(x1, x2, x3)  =  append_out_aag(x1, x2)
U1_aag(x1, x2, x3, x4, x5)  =  U1_aag(x5)
append3_out_aaag(x1, x2, x3, x4)  =  append3_out_aaag(x1, x3)
APPEND_IN_AAA(x1, x2, x3)  =  APPEND_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:

APPEND_IN_AAA(.(H, L1), L2, .(H, L3)) → APPEND_IN_AAA(L1, L2, L3)

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

APPEND_IN_AAA → APPEND_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 = APPEND_IN_AAA evaluates to t =APPEND_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 APPEND_IN_AAA to APPEND_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:
append3_in: (f,f,f,b)
append_in: (f,f,f) (f,f,b)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

append3_in_aaag(A, B, C, D) → U2_aaag(A, B, C, D, append_in_aaa(A, B, E))
append_in_aaa([], L, L) → append_out_aaa([], L, L)
append_in_aaa(.(H, L1), L2, .(H, L3)) → U1_aaa(H, L1, L2, L3, append_in_aaa(L1, L2, L3))
U1_aaa(H, L1, L2, L3, append_out_aaa(L1, L2, L3)) → append_out_aaa(.(H, L1), L2, .(H, L3))
U2_aaag(A, B, C, D, append_out_aaa(A, B, E)) → U3_aaag(A, B, C, D, append_in_aag(E, C, D))
append_in_aag([], L, L) → append_out_aag([], L, L)
append_in_aag(.(H, L1), L2, .(H, L3)) → U1_aag(H, L1, L2, L3, append_in_aag(L1, L2, L3))
U1_aag(H, L1, L2, L3, append_out_aag(L1, L2, L3)) → append_out_aag(.(H, L1), L2, .(H, L3))
U3_aaag(A, B, C, D, append_out_aag(E, C, D)) → append3_out_aaag(A, B, C, D)

The argument filtering Pi contains the following mapping:
append3_in_aaag(x1, x2, x3, x4)  =  append3_in_aaag(x4)
U2_aaag(x1, x2, x3, x4, x5)  =  U2_aaag(x4, x5)
append_in_aaa(x1, x2, x3)  =  append_in_aaa
append_out_aaa(x1, x2, x3)  =  append_out_aaa(x1)
U1_aaa(x1, x2, x3, x4, x5)  =  U1_aaa(x5)
.(x1, x2)  =  .(x2)
U3_aaag(x1, x2, x3, x4, x5)  =  U3_aaag(x1, x4, x5)
append_in_aag(x1, x2, x3)  =  append_in_aag(x3)
append_out_aag(x1, x2, x3)  =  append_out_aag(x1, x2, x3)
U1_aag(x1, x2, x3, x4, x5)  =  U1_aag(x4, x5)
append3_out_aaag(x1, x2, x3, x4)  =  append3_out_aaag(x1, x3, x4)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(22) Obligation:

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

append3_in_aaag(A, B, C, D) → U2_aaag(A, B, C, D, append_in_aaa(A, B, E))
append_in_aaa([], L, L) → append_out_aaa([], L, L)
append_in_aaa(.(H, L1), L2, .(H, L3)) → U1_aaa(H, L1, L2, L3, append_in_aaa(L1, L2, L3))
U1_aaa(H, L1, L2, L3, append_out_aaa(L1, L2, L3)) → append_out_aaa(.(H, L1), L2, .(H, L3))
U2_aaag(A, B, C, D, append_out_aaa(A, B, E)) → U3_aaag(A, B, C, D, append_in_aag(E, C, D))
append_in_aag([], L, L) → append_out_aag([], L, L)
append_in_aag(.(H, L1), L2, .(H, L3)) → U1_aag(H, L1, L2, L3, append_in_aag(L1, L2, L3))
U1_aag(H, L1, L2, L3, append_out_aag(L1, L2, L3)) → append_out_aag(.(H, L1), L2, .(H, L3))
U3_aaag(A, B, C, D, append_out_aag(E, C, D)) → append3_out_aaag(A, B, C, D)

The argument filtering Pi contains the following mapping:
append3_in_aaag(x1, x2, x3, x4)  =  append3_in_aaag(x4)
U2_aaag(x1, x2, x3, x4, x5)  =  U2_aaag(x4, x5)
append_in_aaa(x1, x2, x3)  =  append_in_aaa
append_out_aaa(x1, x2, x3)  =  append_out_aaa(x1)
U1_aaa(x1, x2, x3, x4, x5)  =  U1_aaa(x5)
.(x1, x2)  =  .(x2)
U3_aaag(x1, x2, x3, x4, x5)  =  U3_aaag(x1, x4, x5)
append_in_aag(x1, x2, x3)  =  append_in_aag(x3)
append_out_aag(x1, x2, x3)  =  append_out_aag(x1, x2, x3)
U1_aag(x1, x2, x3, x4, x5)  =  U1_aag(x4, x5)
append3_out_aaag(x1, x2, x3, x4)  =  append3_out_aaag(x1, x3, x4)

(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:

APPEND3_IN_AAAG(A, B, C, D) → U2_AAAG(A, B, C, D, append_in_aaa(A, B, E))
APPEND3_IN_AAAG(A, B, C, D) → APPEND_IN_AAA(A, B, E)
APPEND_IN_AAA(.(H, L1), L2, .(H, L3)) → U1_AAA(H, L1, L2, L3, append_in_aaa(L1, L2, L3))
APPEND_IN_AAA(.(H, L1), L2, .(H, L3)) → APPEND_IN_AAA(L1, L2, L3)
U2_AAAG(A, B, C, D, append_out_aaa(A, B, E)) → U3_AAAG(A, B, C, D, append_in_aag(E, C, D))
U2_AAAG(A, B, C, D, append_out_aaa(A, B, E)) → APPEND_IN_AAG(E, C, D)
APPEND_IN_AAG(.(H, L1), L2, .(H, L3)) → U1_AAG(H, L1, L2, L3, append_in_aag(L1, L2, L3))
APPEND_IN_AAG(.(H, L1), L2, .(H, L3)) → APPEND_IN_AAG(L1, L2, L3)

The TRS R consists of the following rules:

append3_in_aaag(A, B, C, D) → U2_aaag(A, B, C, D, append_in_aaa(A, B, E))
append_in_aaa([], L, L) → append_out_aaa([], L, L)
append_in_aaa(.(H, L1), L2, .(H, L3)) → U1_aaa(H, L1, L2, L3, append_in_aaa(L1, L2, L3))
U1_aaa(H, L1, L2, L3, append_out_aaa(L1, L2, L3)) → append_out_aaa(.(H, L1), L2, .(H, L3))
U2_aaag(A, B, C, D, append_out_aaa(A, B, E)) → U3_aaag(A, B, C, D, append_in_aag(E, C, D))
append_in_aag([], L, L) → append_out_aag([], L, L)
append_in_aag(.(H, L1), L2, .(H, L3)) → U1_aag(H, L1, L2, L3, append_in_aag(L1, L2, L3))
U1_aag(H, L1, L2, L3, append_out_aag(L1, L2, L3)) → append_out_aag(.(H, L1), L2, .(H, L3))
U3_aaag(A, B, C, D, append_out_aag(E, C, D)) → append3_out_aaag(A, B, C, D)

The argument filtering Pi contains the following mapping:
append3_in_aaag(x1, x2, x3, x4)  =  append3_in_aaag(x4)
U2_aaag(x1, x2, x3, x4, x5)  =  U2_aaag(x4, x5)
append_in_aaa(x1, x2, x3)  =  append_in_aaa
append_out_aaa(x1, x2, x3)  =  append_out_aaa(x1)
U1_aaa(x1, x2, x3, x4, x5)  =  U1_aaa(x5)
.(x1, x2)  =  .(x2)
U3_aaag(x1, x2, x3, x4, x5)  =  U3_aaag(x1, x4, x5)
append_in_aag(x1, x2, x3)  =  append_in_aag(x3)
append_out_aag(x1, x2, x3)  =  append_out_aag(x1, x2, x3)
U1_aag(x1, x2, x3, x4, x5)  =  U1_aag(x4, x5)
append3_out_aaag(x1, x2, x3, x4)  =  append3_out_aaag(x1, x3, x4)
APPEND3_IN_AAAG(x1, x2, x3, x4)  =  APPEND3_IN_AAAG(x4)
U2_AAAG(x1, x2, x3, x4, x5)  =  U2_AAAG(x4, x5)
APPEND_IN_AAA(x1, x2, x3)  =  APPEND_IN_AAA
U1_AAA(x1, x2, x3, x4, x5)  =  U1_AAA(x5)
U3_AAAG(x1, x2, x3, x4, x5)  =  U3_AAAG(x1, x4, x5)
APPEND_IN_AAG(x1, x2, x3)  =  APPEND_IN_AAG(x3)
U1_AAG(x1, x2, x3, x4, x5)  =  U1_AAG(x4, x5)

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

(24) Obligation:

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

APPEND3_IN_AAAG(A, B, C, D) → U2_AAAG(A, B, C, D, append_in_aaa(A, B, E))
APPEND3_IN_AAAG(A, B, C, D) → APPEND_IN_AAA(A, B, E)
APPEND_IN_AAA(.(H, L1), L2, .(H, L3)) → U1_AAA(H, L1, L2, L3, append_in_aaa(L1, L2, L3))
APPEND_IN_AAA(.(H, L1), L2, .(H, L3)) → APPEND_IN_AAA(L1, L2, L3)
U2_AAAG(A, B, C, D, append_out_aaa(A, B, E)) → U3_AAAG(A, B, C, D, append_in_aag(E, C, D))
U2_AAAG(A, B, C, D, append_out_aaa(A, B, E)) → APPEND_IN_AAG(E, C, D)
APPEND_IN_AAG(.(H, L1), L2, .(H, L3)) → U1_AAG(H, L1, L2, L3, append_in_aag(L1, L2, L3))
APPEND_IN_AAG(.(H, L1), L2, .(H, L3)) → APPEND_IN_AAG(L1, L2, L3)

The TRS R consists of the following rules:

append3_in_aaag(A, B, C, D) → U2_aaag(A, B, C, D, append_in_aaa(A, B, E))
append_in_aaa([], L, L) → append_out_aaa([], L, L)
append_in_aaa(.(H, L1), L2, .(H, L3)) → U1_aaa(H, L1, L2, L3, append_in_aaa(L1, L2, L3))
U1_aaa(H, L1, L2, L3, append_out_aaa(L1, L2, L3)) → append_out_aaa(.(H, L1), L2, .(H, L3))
U2_aaag(A, B, C, D, append_out_aaa(A, B, E)) → U3_aaag(A, B, C, D, append_in_aag(E, C, D))
append_in_aag([], L, L) → append_out_aag([], L, L)
append_in_aag(.(H, L1), L2, .(H, L3)) → U1_aag(H, L1, L2, L3, append_in_aag(L1, L2, L3))
U1_aag(H, L1, L2, L3, append_out_aag(L1, L2, L3)) → append_out_aag(.(H, L1), L2, .(H, L3))
U3_aaag(A, B, C, D, append_out_aag(E, C, D)) → append3_out_aaag(A, B, C, D)

The argument filtering Pi contains the following mapping:
append3_in_aaag(x1, x2, x3, x4)  =  append3_in_aaag(x4)
U2_aaag(x1, x2, x3, x4, x5)  =  U2_aaag(x4, x5)
append_in_aaa(x1, x2, x3)  =  append_in_aaa
append_out_aaa(x1, x2, x3)  =  append_out_aaa(x1)
U1_aaa(x1, x2, x3, x4, x5)  =  U1_aaa(x5)
.(x1, x2)  =  .(x2)
U3_aaag(x1, x2, x3, x4, x5)  =  U3_aaag(x1, x4, x5)
append_in_aag(x1, x2, x3)  =  append_in_aag(x3)
append_out_aag(x1, x2, x3)  =  append_out_aag(x1, x2, x3)
U1_aag(x1, x2, x3, x4, x5)  =  U1_aag(x4, x5)
append3_out_aaag(x1, x2, x3, x4)  =  append3_out_aaag(x1, x3, x4)
APPEND3_IN_AAAG(x1, x2, x3, x4)  =  APPEND3_IN_AAAG(x4)
U2_AAAG(x1, x2, x3, x4, x5)  =  U2_AAAG(x4, x5)
APPEND_IN_AAA(x1, x2, x3)  =  APPEND_IN_AAA
U1_AAA(x1, x2, x3, x4, x5)  =  U1_AAA(x5)
U3_AAAG(x1, x2, x3, x4, x5)  =  U3_AAAG(x1, x4, x5)
APPEND_IN_AAG(x1, x2, x3)  =  APPEND_IN_AAG(x3)
U1_AAG(x1, x2, x3, x4, x5)  =  U1_AAG(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:

APPEND_IN_AAG(.(H, L1), L2, .(H, L3)) → APPEND_IN_AAG(L1, L2, L3)

The TRS R consists of the following rules:

append3_in_aaag(A, B, C, D) → U2_aaag(A, B, C, D, append_in_aaa(A, B, E))
append_in_aaa([], L, L) → append_out_aaa([], L, L)
append_in_aaa(.(H, L1), L2, .(H, L3)) → U1_aaa(H, L1, L2, L3, append_in_aaa(L1, L2, L3))
U1_aaa(H, L1, L2, L3, append_out_aaa(L1, L2, L3)) → append_out_aaa(.(H, L1), L2, .(H, L3))
U2_aaag(A, B, C, D, append_out_aaa(A, B, E)) → U3_aaag(A, B, C, D, append_in_aag(E, C, D))
append_in_aag([], L, L) → append_out_aag([], L, L)
append_in_aag(.(H, L1), L2, .(H, L3)) → U1_aag(H, L1, L2, L3, append_in_aag(L1, L2, L3))
U1_aag(H, L1, L2, L3, append_out_aag(L1, L2, L3)) → append_out_aag(.(H, L1), L2, .(H, L3))
U3_aaag(A, B, C, D, append_out_aag(E, C, D)) → append3_out_aaag(A, B, C, D)

The argument filtering Pi contains the following mapping:
append3_in_aaag(x1, x2, x3, x4)  =  append3_in_aaag(x4)
U2_aaag(x1, x2, x3, x4, x5)  =  U2_aaag(x4, x5)
append_in_aaa(x1, x2, x3)  =  append_in_aaa
append_out_aaa(x1, x2, x3)  =  append_out_aaa(x1)
U1_aaa(x1, x2, x3, x4, x5)  =  U1_aaa(x5)
.(x1, x2)  =  .(x2)
U3_aaag(x1, x2, x3, x4, x5)  =  U3_aaag(x1, x4, x5)
append_in_aag(x1, x2, x3)  =  append_in_aag(x3)
append_out_aag(x1, x2, x3)  =  append_out_aag(x1, x2, x3)
U1_aag(x1, x2, x3, x4, x5)  =  U1_aag(x4, x5)
append3_out_aaag(x1, x2, x3, x4)  =  append3_out_aaag(x1, x3, x4)
APPEND_IN_AAG(x1, x2, x3)  =  APPEND_IN_AAG(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:

APPEND_IN_AAG(.(H, L1), L2, .(H, L3)) → APPEND_IN_AAG(L1, L2, L3)

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

APPEND_IN_AAG(.(L3)) → APPEND_IN_AAG(L3)

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:

  • APPEND_IN_AAG(.(L3)) → APPEND_IN_AAG(L3)
    The graph contains the following edges 1 > 1

(33) TRUE

(34) Obligation:

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

APPEND_IN_AAA(.(H, L1), L2, .(H, L3)) → APPEND_IN_AAA(L1, L2, L3)

The TRS R consists of the following rules:

append3_in_aaag(A, B, C, D) → U2_aaag(A, B, C, D, append_in_aaa(A, B, E))
append_in_aaa([], L, L) → append_out_aaa([], L, L)
append_in_aaa(.(H, L1), L2, .(H, L3)) → U1_aaa(H, L1, L2, L3, append_in_aaa(L1, L2, L3))
U1_aaa(H, L1, L2, L3, append_out_aaa(L1, L2, L3)) → append_out_aaa(.(H, L1), L2, .(H, L3))
U2_aaag(A, B, C, D, append_out_aaa(A, B, E)) → U3_aaag(A, B, C, D, append_in_aag(E, C, D))
append_in_aag([], L, L) → append_out_aag([], L, L)
append_in_aag(.(H, L1), L2, .(H, L3)) → U1_aag(H, L1, L2, L3, append_in_aag(L1, L2, L3))
U1_aag(H, L1, L2, L3, append_out_aag(L1, L2, L3)) → append_out_aag(.(H, L1), L2, .(H, L3))
U3_aaag(A, B, C, D, append_out_aag(E, C, D)) → append3_out_aaag(A, B, C, D)

The argument filtering Pi contains the following mapping:
append3_in_aaag(x1, x2, x3, x4)  =  append3_in_aaag(x4)
U2_aaag(x1, x2, x3, x4, x5)  =  U2_aaag(x4, x5)
append_in_aaa(x1, x2, x3)  =  append_in_aaa
append_out_aaa(x1, x2, x3)  =  append_out_aaa(x1)
U1_aaa(x1, x2, x3, x4, x5)  =  U1_aaa(x5)
.(x1, x2)  =  .(x2)
U3_aaag(x1, x2, x3, x4, x5)  =  U3_aaag(x1, x4, x5)
append_in_aag(x1, x2, x3)  =  append_in_aag(x3)
append_out_aag(x1, x2, x3)  =  append_out_aag(x1, x2, x3)
U1_aag(x1, x2, x3, x4, x5)  =  U1_aag(x4, x5)
append3_out_aaag(x1, x2, x3, x4)  =  append3_out_aaag(x1, x3, x4)
APPEND_IN_AAA(x1, x2, x3)  =  APPEND_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:

APPEND_IN_AAA(.(H, L1), L2, .(H, L3)) → APPEND_IN_AAA(L1, L2, L3)

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

APPEND_IN_AAA → APPEND_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 = APPEND_IN_AAA evaluates to t =APPEND_IN_AAA

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



Rewriting sequence

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



(40) FALSE