Left Termination of the query pattern flat_in_2(g, a) w.r.t. the given Prolog program could successfully be proven:

0 Prolog
↳1 PrologToDTProblemTransformerProof (⇐)
↳2 TRIPLES
↳3 TriplesToPiDPProof (⇐)
↳4 PiDP
↳5 DependencyGraphProof (⇔)
↳6 PiDP
↳7 PiDPToQDPProof (⇐)
↳8 QDP
↳9 UsableRulesReductionPairsProof (⇔)
↳10 QDP
↳11 MRRProof (⇔)
↳12 QDP
↳13 PisEmptyProof (⇔)
↳14 YES

(0) Obligation:

Clauses:

right(tree(X, XS1, XS2), XS2).
flat(niltree, nil).
flat(tree(X, niltree, XS), cons(X, YS)) :- ','(right(tree(X, niltree, XS), ZS), flat(ZS, YS)).
flat(tree(X, tree(Y, YS1, YS2), XS), ZS) :- flat(tree(Y, YS1, tree(X, YS2, XS)), ZS).

Queries:

flat(g,a).

(1) PrologToDTProblemTransformerProof (SOUND transformation)

Built DT problem from termination graph.

(2) Obligation:

Triples:

flat1(tree(T18, niltree, T19), cons(T18, T9)) :- flat1(T19, T9).
flat1(tree(T72, tree(T71, niltree, T73), T74), cons(T71, T59)) :- flat1(tree(T72, T73, T74), T59).
flat1(tree(T98, tree(T94, tree(T95, T96, T97), T99), T100), T102) :- flat1(tree(T95, T96, tree(T94, T97, tree(T98, T99, T100))), T102).

Clauses:

flatc1(niltree, nil).
flatc1(tree(T18, niltree, T19), cons(T18, T9)) :- flatc1(T19, T9).
flatc1(tree(T72, tree(T71, niltree, T73), T74), cons(T71, T59)) :- flatc1(tree(T72, T73, T74), T59).
flatc1(tree(T98, tree(T94, tree(T95, T96, T97), T99), T100), T102) :- flatc1(tree(T95, T96, tree(T94, T97, tree(T98, T99, T100))), T102).

Afs:

flat1(x1, x2)  =  flat1(x1)

(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:
flat1_in: (b,f)
Transforming TRIPLES into the following Term Rewriting System:
Pi DP problem:
The TRS P consists of the following rules:

FLAT1_IN_GA(tree(T18, niltree, T19), cons(T18, T9)) → U1_GA(T18, T19, T9, flat1_in_ga(T19, T9))
FLAT1_IN_GA(tree(T18, niltree, T19), cons(T18, T9)) → FLAT1_IN_GA(T19, T9)
FLAT1_IN_GA(tree(T72, tree(T71, niltree, T73), T74), cons(T71, T59)) → U2_GA(T72, T71, T73, T74, T59, flat1_in_ga(tree(T72, T73, T74), T59))
FLAT1_IN_GA(tree(T72, tree(T71, niltree, T73), T74), cons(T71, T59)) → FLAT1_IN_GA(tree(T72, T73, T74), T59)
FLAT1_IN_GA(tree(T98, tree(T94, tree(T95, T96, T97), T99), T100), T102) → U3_GA(T98, T94, T95, T96, T97, T99, T100, T102, flat1_in_ga(tree(T95, T96, tree(T94, T97, tree(T98, T99, T100))), T102))
FLAT1_IN_GA(tree(T98, tree(T94, tree(T95, T96, T97), T99), T100), T102) → FLAT1_IN_GA(tree(T95, T96, tree(T94, T97, tree(T98, T99, T100))), T102)

R is empty.
The argument filtering Pi contains the following mapping:
flat1_in_ga(x1, x2)  =  flat1_in_ga(x1)
tree(x1, x2, x3)  =  tree(x1, x2, x3)
niltree  =  niltree
cons(x1, x2)  =  cons(x1, x2)
FLAT1_IN_GA(x1, x2)  =  FLAT1_IN_GA(x1)
U1_GA(x1, x2, x3, x4)  =  U1_GA(x1, x2, x4)
U2_GA(x1, x2, x3, x4, x5, x6)  =  U2_GA(x1, x2, x3, x4, x6)
U3_GA(x1, x2, x3, x4, x5, x6, x7, x8, x9)  =  U3_GA(x1, x2, x3, x4, x5, x6, x7, x9)

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:

FLAT1_IN_GA(tree(T18, niltree, T19), cons(T18, T9)) → U1_GA(T18, T19, T9, flat1_in_ga(T19, T9))
FLAT1_IN_GA(tree(T18, niltree, T19), cons(T18, T9)) → FLAT1_IN_GA(T19, T9)
FLAT1_IN_GA(tree(T72, tree(T71, niltree, T73), T74), cons(T71, T59)) → U2_GA(T72, T71, T73, T74, T59, flat1_in_ga(tree(T72, T73, T74), T59))
FLAT1_IN_GA(tree(T72, tree(T71, niltree, T73), T74), cons(T71, T59)) → FLAT1_IN_GA(tree(T72, T73, T74), T59)
FLAT1_IN_GA(tree(T98, tree(T94, tree(T95, T96, T97), T99), T100), T102) → U3_GA(T98, T94, T95, T96, T97, T99, T100, T102, flat1_in_ga(tree(T95, T96, tree(T94, T97, tree(T98, T99, T100))), T102))
FLAT1_IN_GA(tree(T98, tree(T94, tree(T95, T96, T97), T99), T100), T102) → FLAT1_IN_GA(tree(T95, T96, tree(T94, T97, tree(T98, T99, T100))), T102)

R is empty.
The argument filtering Pi contains the following mapping:
flat1_in_ga(x1, x2)  =  flat1_in_ga(x1)
tree(x1, x2, x3)  =  tree(x1, x2, x3)
niltree  =  niltree
cons(x1, x2)  =  cons(x1, x2)
FLAT1_IN_GA(x1, x2)  =  FLAT1_IN_GA(x1)
U1_GA(x1, x2, x3, x4)  =  U1_GA(x1, x2, x4)
U2_GA(x1, x2, x3, x4, x5, x6)  =  U2_GA(x1, x2, x3, x4, x6)
U3_GA(x1, x2, x3, x4, x5, x6, x7, x8, x9)  =  U3_GA(x1, x2, x3, x4, x5, x6, x7, x9)

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:

FLAT1_IN_GA(tree(T72, tree(T71, niltree, T73), T74), cons(T71, T59)) → FLAT1_IN_GA(tree(T72, T73, T74), T59)
FLAT1_IN_GA(tree(T18, niltree, T19), cons(T18, T9)) → FLAT1_IN_GA(T19, T9)
FLAT1_IN_GA(tree(T98, tree(T94, tree(T95, T96, T97), T99), T100), T102) → FLAT1_IN_GA(tree(T95, T96, tree(T94, T97, tree(T98, T99, T100))), T102)

R is empty.
The argument filtering Pi contains the following mapping:
tree(x1, x2, x3)  =  tree(x1, x2, x3)
niltree  =  niltree
cons(x1, x2)  =  cons(x1, x2)
FLAT1_IN_GA(x1, x2)  =  FLAT1_IN_GA(x1)

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:

FLAT1_IN_GA(tree(T72, tree(T71, niltree, T73), T74)) → FLAT1_IN_GA(tree(T72, T73, T74))
FLAT1_IN_GA(tree(T18, niltree, T19)) → FLAT1_IN_GA(T19)
FLAT1_IN_GA(tree(T98, tree(T94, tree(T95, T96, T97), T99), T100)) → FLAT1_IN_GA(tree(T95, T96, tree(T94, T97, tree(T98, T99, T100))))

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

(9) UsableRulesReductionPairsProof (EQUIVALENT transformation)

By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

The following dependency pairs can be deleted:

FLAT1_IN_GA(tree(T72, tree(T71, niltree, T73), T74)) → FLAT1_IN_GA(tree(T72, T73, T74))
FLAT1_IN_GA(tree(T18, niltree, T19)) → FLAT1_IN_GA(T19)
No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [POLO]:

POL(FLAT1_IN_GA(x1)) = 2·x1   
POL(niltree) = 0   
POL(tree(x1, x2, x3)) = 2·x1 + 2·x2 + x3   

(10) Obligation:

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

FLAT1_IN_GA(tree(T98, tree(T94, tree(T95, T96, T97), T99), T100)) → FLAT1_IN_GA(tree(T95, T96, tree(T94, T97, tree(T98, T99, T100))))

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

(11) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

FLAT1_IN_GA(tree(T98, tree(T94, tree(T95, T96, T97), T99), T100)) → FLAT1_IN_GA(tree(T95, T96, tree(T94, T97, tree(T98, T99, T100))))


Used ordering: Polynomial interpretation [POLO]:

POL(FLAT1_IN_GA(x1)) = 2·x1   
POL(tree(x1, x2, x3)) = 2 + 2·x1 + 2·x2 + x3   

(12) Obligation:

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

(13) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(14) YES