Left Termination of the query pattern interleave_in_3(g, g, a) w.r.t. the given Prolog program could successfully be proven:

(0) Obligation:

Clauses:

interleave([], Xs, Xs).
interleave(.(X, Xs), Ys, .(X, Zs)) :- interleave(Ys, Xs, Zs).

Queries:

interleave(g,g,a).

(1) PrologToDTProblemTransformerProof (SOUND transformation)

Built DT problem from termination graph.

(2) Obligation:

Triples:

interleave1(.(T10, T30), .(T28, T29), .(T10, .(T28, T32))) :- interleave1(T30, T29, T32).

Clauses:

interleavec1([], T5, T5).
interleavec1(.(T10, T19), [], .(T10, T19)).
interleavec1(.(T10, T30), .(T28, T29), .(T10, .(T28, T32))) :- interleavec1(T30, T29, T32).

Afs:

interleave1(x1, x2, x3)  =  interleave1(x1, x2)

(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:
interleave1_in: (b,b,f)
Transforming TRIPLES into the following Term Rewriting System:
Pi DP problem:
The TRS P consists of the following rules:

INTERLEAVE1_IN_GGA(.(T10, T30), .(T28, T29), .(T10, .(T28, T32))) → U1_GGA(T10, T30, T28, T29, T32, interleave1_in_gga(T30, T29, T32))
INTERLEAVE1_IN_GGA(.(T10, T30), .(T28, T29), .(T10, .(T28, T32))) → INTERLEAVE1_IN_GGA(T30, T29, T32)

R is empty.
The argument filtering Pi contains the following mapping:
interleave1_in_gga(x1, x2, x3)  =  interleave1_in_gga(x1, x2)
.(x1, x2)  =  .(x1, x2)
INTERLEAVE1_IN_GGA(x1, x2, x3)  =  INTERLEAVE1_IN_GGA(x1, x2)
U1_GGA(x1, x2, x3, x4, x5, x6)  =  U1_GGA(x1, x2, x3, x4, x6)

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:

INTERLEAVE1_IN_GGA(.(T10, T30), .(T28, T29), .(T10, .(T28, T32))) → U1_GGA(T10, T30, T28, T29, T32, interleave1_in_gga(T30, T29, T32))
INTERLEAVE1_IN_GGA(.(T10, T30), .(T28, T29), .(T10, .(T28, T32))) → INTERLEAVE1_IN_GGA(T30, T29, T32)

R is empty.
The argument filtering Pi contains the following mapping:
interleave1_in_gga(x1, x2, x3)  =  interleave1_in_gga(x1, x2)
.(x1, x2)  =  .(x1, x2)
INTERLEAVE1_IN_GGA(x1, x2, x3)  =  INTERLEAVE1_IN_GGA(x1, x2)
U1_GGA(x1, x2, x3, x4, x5, x6)  =  U1_GGA(x1, x2, x3, x4, x6)

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:

INTERLEAVE1_IN_GGA(.(T10, T30), .(T28, T29), .(T10, .(T28, T32))) → INTERLEAVE1_IN_GGA(T30, T29, T32)

R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x1, x2)
INTERLEAVE1_IN_GGA(x1, x2, x3)  =  INTERLEAVE1_IN_GGA(x1, x2)

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:

INTERLEAVE1_IN_GGA(.(T10, T30), .(T28, T29)) → INTERLEAVE1_IN_GGA(T30, T29)

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

(9) 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:

• INTERLEAVE1_IN_GGA(.(T10, T30), .(T28, T29)) → INTERLEAVE1_IN_GGA(T30, T29)
  The graph contains the following edges 1 > 1, 2 > 2

(10) YES