Left Termination of the query pattern select_in_3(a, a, g) w.r.t. the given Prolog program could successfully be proven:

(0) Obligation:

Clauses:

select(X, .(X, Xs), Xs).
select(X, .(Y, Xs), .(Y, Zs)) :- select(X, Xs, Zs).

Queries:

select(a,a,g).

(1) PrologToPrologProblemTransformerProof (SOUND transformation)

Built Prolog problem from termination graph.

(2) Obligation:

Clauses:

select1(T6, .(T6, T7), T7).
select1(T26, .(T13, .(T26, T27)), .(T13, T27)).
select1(T40, .(T13, .(T37, T41)), .(T13, .(T37, T39))) :- select1(T40, T41, T39).
select1(T64, .(T51, .(T64, T65)), .(T51, T65)).
select1(T78, .(T51, .(T75, T79)), .(T51, .(T75, T77))) :- select1(T78, T79, T77).

Queries:

select1(a,a,g).

(3) PrologToPiTRSProof (SOUND transformation)

We use the technique of [LOPSTR]. With regard to the inferred argument filtering the predicates were used in the following modes:
select1_in: (f,f,b)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

select1_in_aag(T6, .(T6, T7), T7) → select1_out_aag(T6, .(T6, T7), T7)
select1_in_aag(T26, .(T13, .(T26, T27)), .(T13, T27)) → select1_out_aag(T26, .(T13, .(T26, T27)), .(T13, T27))
select1_in_aag(T40, .(T13, .(T37, T41)), .(T13, .(T37, T39))) → U1_aag(T40, T13, T37, T41, T39, select1_in_aag(T40, T41, T39))
select1_in_aag(T78, .(T51, .(T75, T79)), .(T51, .(T75, T77))) → U2_aag(T78, T51, T75, T79, T77, select1_in_aag(T78, T79, T77))
U2_aag(T78, T51, T75, T79, T77, select1_out_aag(T78, T79, T77)) → select1_out_aag(T78, .(T51, .(T75, T79)), .(T51, .(T75, T77)))
U1_aag(T40, T13, T37, T41, T39, select1_out_aag(T40, T41, T39)) → select1_out_aag(T40, .(T13, .(T37, T41)), .(T13, .(T37, T39)))

The argument filtering Pi contains the following mapping:
select1_in_aag(x1, x2, x3)  =  select1_in_aag(x3)
select1_out_aag(x1, x2, x3)  =  select1_out_aag(x2)
.(x1, x2)  =  .(x2)
U1_aag(x1, x2, x3, x4, x5, x6)  =  U1_aag(x6)
U2_aag(x1, x2, x3, x4, x5, x6)  =  U2_aag(x6)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(4) Obligation:

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

select1_in_aag(T6, .(T6, T7), T7) → select1_out_aag(T6, .(T6, T7), T7)
select1_in_aag(T26, .(T13, .(T26, T27)), .(T13, T27)) → select1_out_aag(T26, .(T13, .(T26, T27)), .(T13, T27))
select1_in_aag(T40, .(T13, .(T37, T41)), .(T13, .(T37, T39))) → U1_aag(T40, T13, T37, T41, T39, select1_in_aag(T40, T41, T39))
select1_in_aag(T78, .(T51, .(T75, T79)), .(T51, .(T75, T77))) → U2_aag(T78, T51, T75, T79, T77, select1_in_aag(T78, T79, T77))
U2_aag(T78, T51, T75, T79, T77, select1_out_aag(T78, T79, T77)) → select1_out_aag(T78, .(T51, .(T75, T79)), .(T51, .(T75, T77)))
U1_aag(T40, T13, T37, T41, T39, select1_out_aag(T40, T41, T39)) → select1_out_aag(T40, .(T13, .(T37, T41)), .(T13, .(T37, T39)))

The argument filtering Pi contains the following mapping:
select1_in_aag(x1, x2, x3)  =  select1_in_aag(x3)
select1_out_aag(x1, x2, x3)  =  select1_out_aag(x2)
.(x1, x2)  =  .(x2)
U1_aag(x1, x2, x3, x4, x5, x6)  =  U1_aag(x6)
U2_aag(x1, x2, x3, x4, x5, x6)  =  U2_aag(x6)

(5) 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:

SELECT1_IN_AAG(T40, .(T13, .(T37, T41)), .(T13, .(T37, T39))) → U1_AAG(T40, T13, T37, T41, T39, select1_in_aag(T40, T41, T39))
SELECT1_IN_AAG(T40, .(T13, .(T37, T41)), .(T13, .(T37, T39))) → SELECT1_IN_AAG(T40, T41, T39)
SELECT1_IN_AAG(T78, .(T51, .(T75, T79)), .(T51, .(T75, T77))) → U2_AAG(T78, T51, T75, T79, T77, select1_in_aag(T78, T79, T77))

The TRS R consists of the following rules:

select1_in_aag(T6, .(T6, T7), T7) → select1_out_aag(T6, .(T6, T7), T7)
select1_in_aag(T26, .(T13, .(T26, T27)), .(T13, T27)) → select1_out_aag(T26, .(T13, .(T26, T27)), .(T13, T27))
select1_in_aag(T40, .(T13, .(T37, T41)), .(T13, .(T37, T39))) → U1_aag(T40, T13, T37, T41, T39, select1_in_aag(T40, T41, T39))
select1_in_aag(T78, .(T51, .(T75, T79)), .(T51, .(T75, T77))) → U2_aag(T78, T51, T75, T79, T77, select1_in_aag(T78, T79, T77))
U2_aag(T78, T51, T75, T79, T77, select1_out_aag(T78, T79, T77)) → select1_out_aag(T78, .(T51, .(T75, T79)), .(T51, .(T75, T77)))
U1_aag(T40, T13, T37, T41, T39, select1_out_aag(T40, T41, T39)) → select1_out_aag(T40, .(T13, .(T37, T41)), .(T13, .(T37, T39)))

The argument filtering Pi contains the following mapping:
select1_in_aag(x1, x2, x3)  =  select1_in_aag(x3)
select1_out_aag(x1, x2, x3)  =  select1_out_aag(x2)
.(x1, x2)  =  .(x2)
U1_aag(x1, x2, x3, x4, x5, x6)  =  U1_aag(x6)
U2_aag(x1, x2, x3, x4, x5, x6)  =  U2_aag(x6)
SELECT1_IN_AAG(x1, x2, x3)  =  SELECT1_IN_AAG(x3)
U1_AAG(x1, x2, x3, x4, x5, x6)  =  U1_AAG(x6)
U2_AAG(x1, x2, x3, x4, x5, x6)  =  U2_AAG(x6)

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

(6) Obligation:

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

SELECT1_IN_AAG(T40, .(T13, .(T37, T41)), .(T13, .(T37, T39))) → U1_AAG(T40, T13, T37, T41, T39, select1_in_aag(T40, T41, T39))
SELECT1_IN_AAG(T40, .(T13, .(T37, T41)), .(T13, .(T37, T39))) → SELECT1_IN_AAG(T40, T41, T39)
SELECT1_IN_AAG(T78, .(T51, .(T75, T79)), .(T51, .(T75, T77))) → U2_AAG(T78, T51, T75, T79, T77, select1_in_aag(T78, T79, T77))

The TRS R consists of the following rules:

select1_in_aag(T6, .(T6, T7), T7) → select1_out_aag(T6, .(T6, T7), T7)
select1_in_aag(T26, .(T13, .(T26, T27)), .(T13, T27)) → select1_out_aag(T26, .(T13, .(T26, T27)), .(T13, T27))
select1_in_aag(T40, .(T13, .(T37, T41)), .(T13, .(T37, T39))) → U1_aag(T40, T13, T37, T41, T39, select1_in_aag(T40, T41, T39))
select1_in_aag(T78, .(T51, .(T75, T79)), .(T51, .(T75, T77))) → U2_aag(T78, T51, T75, T79, T77, select1_in_aag(T78, T79, T77))
U2_aag(T78, T51, T75, T79, T77, select1_out_aag(T78, T79, T77)) → select1_out_aag(T78, .(T51, .(T75, T79)), .(T51, .(T75, T77)))
U1_aag(T40, T13, T37, T41, T39, select1_out_aag(T40, T41, T39)) → select1_out_aag(T40, .(T13, .(T37, T41)), .(T13, .(T37, T39)))

The argument filtering Pi contains the following mapping:
select1_in_aag(x1, x2, x3)  =  select1_in_aag(x3)
select1_out_aag(x1, x2, x3)  =  select1_out_aag(x2)
.(x1, x2)  =  .(x2)
U1_aag(x1, x2, x3, x4, x5, x6)  =  U1_aag(x6)
U2_aag(x1, x2, x3, x4, x5, x6)  =  U2_aag(x6)
SELECT1_IN_AAG(x1, x2, x3)  =  SELECT1_IN_AAG(x3)
U1_AAG(x1, x2, x3, x4, x5, x6)  =  U1_AAG(x6)
U2_AAG(x1, x2, x3, x4, x5, x6)  =  U2_AAG(x6)

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

(7) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 2 less nodes.

(8) Obligation:

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

SELECT1_IN_AAG(T40, .(T13, .(T37, T41)), .(T13, .(T37, T39))) → SELECT1_IN_AAG(T40, T41, T39)

The TRS R consists of the following rules:

select1_in_aag(T6, .(T6, T7), T7) → select1_out_aag(T6, .(T6, T7), T7)
select1_in_aag(T26, .(T13, .(T26, T27)), .(T13, T27)) → select1_out_aag(T26, .(T13, .(T26, T27)), .(T13, T27))
select1_in_aag(T40, .(T13, .(T37, T41)), .(T13, .(T37, T39))) → U1_aag(T40, T13, T37, T41, T39, select1_in_aag(T40, T41, T39))
select1_in_aag(T78, .(T51, .(T75, T79)), .(T51, .(T75, T77))) → U2_aag(T78, T51, T75, T79, T77, select1_in_aag(T78, T79, T77))
U2_aag(T78, T51, T75, T79, T77, select1_out_aag(T78, T79, T77)) → select1_out_aag(T78, .(T51, .(T75, T79)), .(T51, .(T75, T77)))
U1_aag(T40, T13, T37, T41, T39, select1_out_aag(T40, T41, T39)) → select1_out_aag(T40, .(T13, .(T37, T41)), .(T13, .(T37, T39)))

The argument filtering Pi contains the following mapping:
select1_in_aag(x1, x2, x3)  =  select1_in_aag(x3)
select1_out_aag(x1, x2, x3)  =  select1_out_aag(x2)
.(x1, x2)  =  .(x2)
U1_aag(x1, x2, x3, x4, x5, x6)  =  U1_aag(x6)
U2_aag(x1, x2, x3, x4, x5, x6)  =  U2_aag(x6)
SELECT1_IN_AAG(x1, x2, x3)  =  SELECT1_IN_AAG(x3)

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

(9) UsableRulesProof (EQUIVALENT transformation)

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

(10) Obligation:

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

SELECT1_IN_AAG(T40, .(T13, .(T37, T41)), .(T13, .(T37, T39))) → SELECT1_IN_AAG(T40, T41, T39)

R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x2)
SELECT1_IN_AAG(x1, x2, x3)  =  SELECT1_IN_AAG(x3)

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

(11) PiDPToQDPProof (SOUND transformation)

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

(12) Obligation:

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

SELECT1_IN_AAG(.(.(T39))) → SELECT1_IN_AAG(T39)

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

(13) 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:

• SELECT1_IN_AAG(.(.(T39))) → SELECT1_IN_AAG(T39)
  The graph contains the following edges 1 > 1

(14) YES