Left Termination of the query pattern append_in_3(g, a, a) w.r.t. the given Prolog program could successfully be proven:

(0) Obligation:

Clauses:

append(X, Y, Z) :- ','(=(X, []), ','(!, =(Y, Z))).
append(X, Y, Z) :- ','(=(X, .(H, Xs)), ','(!, ','(=(Z, .(H, Zs)), append(Xs, Y, Zs)))).
=(X, X).

Queries:

append(g,a,a).

(1) BuiltinConflictTransformerProof (SOUND transformation)

Renamed defined predicates conflicting with built-in predicates [PROLOG].

(2) Obligation:

Clauses:

append(X, Y, Z) :- ','(user_defined_=(X, []), ','(!, user_defined_=(Y, Z))).
append(X, Y, Z) :- ','(user_defined_=(X, .(H, Xs)), ','(!, ','(user_defined_=(Z, .(H, Zs)), append(Xs, Y, Zs)))).
user_defined_=(X, X).

Queries:

append(g,a,a).

(3) PrologToDTProblemTransformerProof (SOUND transformation)

Built DT problem from termination graph.

(4) Obligation:

Triples:

append1(.(T49, T39), T51, .(T49, T52)) :- append1(T39, T51, T52).

Clauses:

appendc1([], T17, T17).
appendc1(.(T49, T39), T51, .(T49, T52)) :- appendc1(T39, T51, T52).

Afs:

append1(x1, x2, x3)  =  append1(x1)

(5) TriplesToPiDPProof (SOUND transformation)

We use the technique of [LOPSTR]. With regard to the inferred argument filtering the predicates were used in the following modes:
append1_in: (b,f,f)
Transforming TRIPLES into the following Term Rewriting System:
Pi DP problem:
The TRS P consists of the following rules:

APPEND1_IN_GAA(.(T49, T39), T51, .(T49, T52)) → U1_GAA(T49, T39, T51, T52, append1_in_gaa(T39, T51, T52))
APPEND1_IN_GAA(.(T49, T39), T51, .(T49, T52)) → APPEND1_IN_GAA(T39, T51, T52)

R is empty.
The argument filtering Pi contains the following mapping:
append1_in_gaa(x1, x2, x3)  =  append1_in_gaa(x1)
.(x1, x2)  =  .(x1, x2)
APPEND1_IN_GAA(x1, x2, x3)  =  APPEND1_IN_GAA(x1)
U1_GAA(x1, x2, x3, x4, x5)  =  U1_GAA(x1, x2, x5)

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

Infinitary Constructor Rewriting Termination of PiDP implies Termination of TRIPLES

(6) Obligation:

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

APPEND1_IN_GAA(.(T49, T39), T51, .(T49, T52)) → U1_GAA(T49, T39, T51, T52, append1_in_gaa(T39, T51, T52))
APPEND1_IN_GAA(.(T49, T39), T51, .(T49, T52)) → APPEND1_IN_GAA(T39, T51, T52)

R is empty.
The argument filtering Pi contains the following mapping:
append1_in_gaa(x1, x2, x3)  =  append1_in_gaa(x1)
.(x1, x2)  =  .(x1, x2)
APPEND1_IN_GAA(x1, x2, x3)  =  APPEND1_IN_GAA(x1)
U1_GAA(x1, x2, x3, x4, x5)  =  U1_GAA(x1, x2, x5)

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

(7) DependencyGraphProof (EQUIVALENT transformation)

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

(8) Obligation:

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

APPEND1_IN_GAA(.(T49, T39), T51, .(T49, T52)) → APPEND1_IN_GAA(T39, T51, T52)

R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x1, x2)
APPEND1_IN_GAA(x1, x2, x3)  =  APPEND1_IN_GAA(x1)

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:

APPEND1_IN_GAA(.(T49, T39)) → APPEND1_IN_GAA(T39)

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

(11) 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:

• APPEND1_IN_GAA(.(T49, T39)) → APPEND1_IN_GAA(T39)
  The graph contains the following edges 1 > 1

(12) YES