Left Termination of the query pattern p_in_3(a, a, a) w.r.t. the given Prolog program could not be shown:

(0) Obligation:

Clauses:

p(X, Y, Z) :- append(.(X, Y), Z, Y).
append([], Y, Y).
append(.(H, Xs), Ys, .(H, Zs)) :- append(Xs, Ys, Zs).

Queries:

p(a,a,a).

(1) PrologToDTProblemTransformerProof (SOUND transformation)

Built DT problem from termination graph.

(2) Obligation:

Triples:

append3(T27, .(T27, T26), T25) :- append3(T27, T26, T25).
p1(T12, T11, T10) :- append3(T12, T11, T10).

Clauses:

appendc3(T27, .(T27, T26), T25) :- appendc3(T27, T26, T25).

Afs:

p1(x1, x2, x3)  =  p1

(3) TriplesToPiDPProof (SOUND transformation)

We use the technique of [LOPSTR]. With regard to the inferred argument filtering the predicates were used in the following modes:
p1_in: (f,f,f)
append3_in: (f,f,f)
Transforming TRIPLES into the following Term Rewriting System:
Pi DP problem:
The TRS P consists of the following rules:

P1_IN_AAA(T12, T11, T10) → U2_AAA(T12, T11, T10, append3_in_aaa(T12, T11, T10))
P1_IN_AAA(T12, T11, T10) → APPEND3_IN_AAA(T12, T11, T10)
APPEND3_IN_AAA(T27, .(T27, T26), T25) → U1_AAA(T27, T26, T25, append3_in_aaa(T27, T26, T25))
APPEND3_IN_AAA(T27, .(T27, T26), T25) → APPEND3_IN_AAA(T27, T26, T25)

R is empty.
The argument filtering Pi contains the following mapping:
append3_in_aaa(x1, x2, x3)  =  append3_in_aaa
.(x1, x2)  =  .(x1, x2)
P1_IN_AAA(x1, x2, x3)  =  P1_IN_AAA
U2_AAA(x1, x2, x3, x4)  =  U2_AAA(x4)
APPEND3_IN_AAA(x1, x2, x3)  =  APPEND3_IN_AAA
U1_AAA(x1, x2, x3, x4)  =  U1_AAA(x4)

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

Infinitary Constructor Rewriting Termination of PiDP implies Termination of TRIPLES

(4) Obligation:

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

P1_IN_AAA(T12, T11, T10) → U2_AAA(T12, T11, T10, append3_in_aaa(T12, T11, T10))
P1_IN_AAA(T12, T11, T10) → APPEND3_IN_AAA(T12, T11, T10)
APPEND3_IN_AAA(T27, .(T27, T26), T25) → U1_AAA(T27, T26, T25, append3_in_aaa(T27, T26, T25))
APPEND3_IN_AAA(T27, .(T27, T26), T25) → APPEND3_IN_AAA(T27, T26, T25)

R is empty.
The argument filtering Pi contains the following mapping:
append3_in_aaa(x1, x2, x3)  =  append3_in_aaa
.(x1, x2)  =  .(x1, x2)
P1_IN_AAA(x1, x2, x3)  =  P1_IN_AAA
U2_AAA(x1, x2, x3, x4)  =  U2_AAA(x4)
APPEND3_IN_AAA(x1, x2, x3)  =  APPEND3_IN_AAA
U1_AAA(x1, x2, x3, x4)  =  U1_AAA(x4)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 3 less nodes.

(6) Obligation:

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

APPEND3_IN_AAA(T27, .(T27, T26), T25) → APPEND3_IN_AAA(T27, T26, T25)

R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x1, x2)
APPEND3_IN_AAA(x1, x2, x3)  =  APPEND3_IN_AAA

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

(7) PiDPToQDPProof (SOUND transformation)

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

(8) Obligation:

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

APPEND3_IN_AAA → APPEND3_IN_AAA

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

(9) 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 = APPEND3_IN_AAA evaluates to t =APPEND3_IN_AAA

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:

• Matcher: [ ]
• Semiunifier: [ ]

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Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from APPEND3_IN_AAA to APPEND3_IN_AAA.

(10) NO