Left Termination of the query pattern append_in_3(a, a, 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 PiDP
    ↳7 UsableRulesProof (⇔)
    ↳8 PiDP
    ↳9 PiDPToQDPProof (⇐)
    ↳10 QDP
    ↳11 NonTerminationProof (⇔)
    ↳12 FALSE
  ↳13 PrologToPiTRSProof (⇐)
    ↳14 PiTRS
    ↳15 DependencyPairsProof (⇔)
    ↳16 PiDP
    ↳17 DependencyGraphProof (⇔)
    ↳18 PiDP
    ↳19 UsableRulesProof (⇔)
    ↳20 PiDP
    ↳21 PiDPToQDPProof (⇐)
    ↳22 QDP
    ↳23 NonTerminationProof (⇔)
    ↳24 FALSE

(0) Obligation:

Clauses:

append([], L, L).
append(.(H, L1), L2, .(H, L3)) :- append(L1, L2, L3).
append1([], L, L).
append1(.(H, L1), L2, .(H, L3)) :- append1(L1, L2, L3).

Queries:

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

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))

The argument filtering Pi contains the following mapping:
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)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(2) Obligation:

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

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))

The argument filtering Pi contains the following mapping:
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)

(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:

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)

The TRS R consists of the following rules:

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))

The argument filtering Pi contains the following mapping:
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)
APPEND_IN_AAA(x1, x2, x3)  =  APPEND_IN_AAA
U1_AAA(x1, x2, x3, x4, x5)  =  U1_AAA(x5)

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

(4) Obligation:

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

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)

The TRS R consists of the following rules:

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))

The argument filtering Pi contains the following mapping:
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)
APPEND_IN_AAA(x1, x2, x3)  =  APPEND_IN_AAA
U1_AAA(x1, x2, x3, x4, x5)  =  U1_AAA(x5)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 1 less node.

(6) 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:

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))

The argument filtering Pi contains the following mapping:
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)
APPEND_IN_AAA(x1, x2, x3)  =  APPEND_IN_AAA

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

(7) UsableRulesProof (EQUIVALENT transformation)

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_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

(9) PiDPToQDPProof (SOUND transformation)

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_AAA → APPEND_IN_AAA

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

(11) 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.



(12) FALSE

(13) PrologToPiTRSProof (SOUND transformation)

We use the technique of [LOPSTR]. With regard to the inferred argument filtering the predicates were used in the following modes:
append_in: (f,f,f)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

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))

The argument filtering Pi contains the following mapping:
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)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(14) Obligation:

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

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))

The argument filtering Pi contains the following mapping:
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)

(15) 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:

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)

The TRS R consists of the following rules:

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))

The argument filtering Pi contains the following mapping:
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)
APPEND_IN_AAA(x1, x2, x3)  =  APPEND_IN_AAA
U1_AAA(x1, x2, x3, x4, x5)  =  U1_AAA(x5)

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

(16) Obligation:

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

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)

The TRS R consists of the following rules:

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))

The argument filtering Pi contains the following mapping:
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)
APPEND_IN_AAA(x1, x2, x3)  =  APPEND_IN_AAA
U1_AAA(x1, x2, x3, x4, x5)  =  U1_AAA(x5)

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

(17) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 1 less node.

(18) 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:

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))

The argument filtering Pi contains the following mapping:
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)
APPEND_IN_AAA(x1, x2, x3)  =  APPEND_IN_AAA

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

(19) UsableRulesProof (EQUIVALENT transformation)

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

(20) 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

(21) PiDPToQDPProof (SOUND transformation)

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

(22) 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.

(23) 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.



(24) FALSE