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 PisEmptyProof (⇔)
↳12 YES

(0) Obligation:

Clauses:

flat(niltree, nil).
flat(tree(X, niltree, T), cons(X, Xs)) :- flat(T, Xs).
flat(tree(X, tree(Y, T1, T2), T3), Xs) :- flat(tree(Y, T1, tree(X, T2, T3)), Xs).

Queries:

flat(g,a).

(1) PrologToDTProblemTransformerProof (SOUND transformation)

Built DT problem from termination graph.

(2) Obligation:

Triples:

flat1(tree(T9, niltree, tree(T25, niltree, T26)), cons(T9, cons(T25, T28))) :- flat1(T26, T28).
flat1(tree(T9, niltree, tree(T53, tree(T54, T55, T56), T57)), cons(T9, T59)) :- flat1(tree(T54, T55, tree(T53, T56, T57)), T59).
flat1(tree(T100, tree(T99, niltree, T101), T102), cons(T99, T104)) :- flat1(tree(T100, T101, T102), T104).
flat1(tree(T128, tree(T124, tree(T125, T126, T127), T129), T130), T132) :- flat1(tree(T125, T126, tree(T124, T127, tree(T128, T129, T130))), T132).

Clauses:

flatc1(niltree, nil).
flatc1(tree(T9, niltree, niltree), cons(T9, nil)).
flatc1(tree(T9, niltree, tree(T25, niltree, T26)), cons(T9, cons(T25, T28))) :- flatc1(T26, T28).
flatc1(tree(T9, niltree, tree(T53, tree(T54, T55, T56), T57)), cons(T9, T59)) :- flatc1(tree(T54, T55, tree(T53, T56, T57)), T59).
flatc1(tree(T100, tree(T99, niltree, T101), T102), cons(T99, T104)) :- flatc1(tree(T100, T101, T102), T104).
flatc1(tree(T128, tree(T124, tree(T125, T126, T127), T129), T130), T132) :- flatc1(tree(T125, T126, tree(T124, T127, tree(T128, T129, T130))), T132).

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(T9, niltree, tree(T25, niltree, T26)), cons(T9, cons(T25, T28))) → U1_GA(T9, T25, T26, T28, flat1_in_ga(T26, T28))
FLAT1_IN_GA(tree(T9, niltree, tree(T25, niltree, T26)), cons(T9, cons(T25, T28))) → FLAT1_IN_GA(T26, T28)
FLAT1_IN_GA(tree(T9, niltree, tree(T53, tree(T54, T55, T56), T57)), cons(T9, T59)) → U2_GA(T9, T53, T54, T55, T56, T57, T59, flat1_in_ga(tree(T54, T55, tree(T53, T56, T57)), T59))
FLAT1_IN_GA(tree(T9, niltree, tree(T53, tree(T54, T55, T56), T57)), cons(T9, T59)) → FLAT1_IN_GA(tree(T54, T55, tree(T53, T56, T57)), T59)
FLAT1_IN_GA(tree(T100, tree(T99, niltree, T101), T102), cons(T99, T104)) → U3_GA(T100, T99, T101, T102, T104, flat1_in_ga(tree(T100, T101, T102), T104))
FLAT1_IN_GA(tree(T100, tree(T99, niltree, T101), T102), cons(T99, T104)) → FLAT1_IN_GA(tree(T100, T101, T102), T104)
FLAT1_IN_GA(tree(T128, tree(T124, tree(T125, T126, T127), T129), T130), T132) → U4_GA(T128, T124, T125, T126, T127, T129, T130, T132, flat1_in_ga(tree(T125, T126, tree(T124, T127, tree(T128, T129, T130))), T132))
FLAT1_IN_GA(tree(T128, tree(T124, tree(T125, T126, T127), T129), T130), T132) → FLAT1_IN_GA(tree(T125, T126, tree(T124, T127, tree(T128, T129, T130))), T132)

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, x5)  =  U1_GA(x1, x2, x3, x5)
U2_GA(x1, x2, x3, x4, x5, x6, x7, x8)  =  U2_GA(x1, x2, x3, x4, x5, x6, x8)
U3_GA(x1, x2, x3, x4, x5, x6)  =  U3_GA(x1, x2, x3, x4, x6)
U4_GA(x1, x2, x3, x4, x5, x6, x7, x8, x9)  =  U4_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(T9, niltree, tree(T25, niltree, T26)), cons(T9, cons(T25, T28))) → U1_GA(T9, T25, T26, T28, flat1_in_ga(T26, T28))
FLAT1_IN_GA(tree(T9, niltree, tree(T25, niltree, T26)), cons(T9, cons(T25, T28))) → FLAT1_IN_GA(T26, T28)
FLAT1_IN_GA(tree(T9, niltree, tree(T53, tree(T54, T55, T56), T57)), cons(T9, T59)) → U2_GA(T9, T53, T54, T55, T56, T57, T59, flat1_in_ga(tree(T54, T55, tree(T53, T56, T57)), T59))
FLAT1_IN_GA(tree(T9, niltree, tree(T53, tree(T54, T55, T56), T57)), cons(T9, T59)) → FLAT1_IN_GA(tree(T54, T55, tree(T53, T56, T57)), T59)
FLAT1_IN_GA(tree(T100, tree(T99, niltree, T101), T102), cons(T99, T104)) → U3_GA(T100, T99, T101, T102, T104, flat1_in_ga(tree(T100, T101, T102), T104))
FLAT1_IN_GA(tree(T100, tree(T99, niltree, T101), T102), cons(T99, T104)) → FLAT1_IN_GA(tree(T100, T101, T102), T104)
FLAT1_IN_GA(tree(T128, tree(T124, tree(T125, T126, T127), T129), T130), T132) → U4_GA(T128, T124, T125, T126, T127, T129, T130, T132, flat1_in_ga(tree(T125, T126, tree(T124, T127, tree(T128, T129, T130))), T132))
FLAT1_IN_GA(tree(T128, tree(T124, tree(T125, T126, T127), T129), T130), T132) → FLAT1_IN_GA(tree(T125, T126, tree(T124, T127, tree(T128, T129, T130))), T132)

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, x5)  =  U1_GA(x1, x2, x3, x5)
U2_GA(x1, x2, x3, x4, x5, x6, x7, x8)  =  U2_GA(x1, x2, x3, x4, x5, x6, x8)
U3_GA(x1, x2, x3, x4, x5, x6)  =  U3_GA(x1, x2, x3, x4, x6)
U4_GA(x1, x2, x3, x4, x5, x6, x7, x8, x9)  =  U4_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 4 less nodes.

(6) Obligation:

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

FLAT1_IN_GA(tree(T9, niltree, tree(T53, tree(T54, T55, T56), T57)), cons(T9, T59)) → FLAT1_IN_GA(tree(T54, T55, tree(T53, T56, T57)), T59)
FLAT1_IN_GA(tree(T9, niltree, tree(T25, niltree, T26)), cons(T9, cons(T25, T28))) → FLAT1_IN_GA(T26, T28)
FLAT1_IN_GA(tree(T100, tree(T99, niltree, T101), T102), cons(T99, T104)) → FLAT1_IN_GA(tree(T100, T101, T102), T104)
FLAT1_IN_GA(tree(T128, tree(T124, tree(T125, T126, T127), T129), T130), T132) → FLAT1_IN_GA(tree(T125, T126, tree(T124, T127, tree(T128, T129, T130))), T132)

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(T9, niltree, tree(T53, tree(T54, T55, T56), T57))) → FLAT1_IN_GA(tree(T54, T55, tree(T53, T56, T57)))
FLAT1_IN_GA(tree(T9, niltree, tree(T25, niltree, T26))) → FLAT1_IN_GA(T26)
FLAT1_IN_GA(tree(T100, tree(T99, niltree, T101), T102)) → FLAT1_IN_GA(tree(T100, T101, T102))
FLAT1_IN_GA(tree(T128, tree(T124, tree(T125, T126, T127), T129), T130)) → FLAT1_IN_GA(tree(T125, T126, tree(T124, T127, tree(T128, T129, T130))))

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(T9, niltree, tree(T53, tree(T54, T55, T56), T57))) → FLAT1_IN_GA(tree(T54, T55, tree(T53, T56, T57)))
FLAT1_IN_GA(tree(T9, niltree, tree(T25, niltree, T26))) → FLAT1_IN_GA(T26)
FLAT1_IN_GA(tree(T100, tree(T99, niltree, T101), T102)) → FLAT1_IN_GA(tree(T100, T101, T102))
FLAT1_IN_GA(tree(T128, tree(T124, tree(T125, T126, T127), T129), T130)) → FLAT1_IN_GA(tree(T125, T126, tree(T124, T127, tree(T128, T129, T130))))
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 + 2·x1 + 2·x2 + x3   

(10) Obligation:

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

(11) PisEmptyProof (EQUIVALENT transformation)

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

(12) YES