Left Termination of the query pattern fl_in_3(a, a, g) w.r.t. the given Prolog program could not be shown:

(0) Obligation:

Clauses:

fl([], [], 0).
fl(.(E, X), R, s(Z)) :- ','(append(E, Y, R), fl(X, Y, Z)).
append([], X, X).
append(.(X, Xs), Ys, .(X, Zs)) :- append(Xs, Ys, Zs).

Queries:

fl(a,a,g).

(1) PrologToDTProblemTransformerProof (SOUND transformation)

Built DT problem from termination graph.

(2) Obligation:

Triples:

p7([], T24, T24, T23, T14) :- fl1(T23, T24, T14).
p7(.(T31, T34), X43, .(T31, T35), T36, T14) :- p7(T34, X43, T35, T36, T14).
fl1(.(T15, T17), T16, s(T14)) :- p7(T15, X13, T16, T17, T14).

Clauses:

flc1([], [], 0).
flc1(.(T15, T17), T16, s(T14)) :- qc7(T15, X13, T16, T17, T14).
qc7([], T24, T24, T23, T14) :- flc1(T23, T24, T14).
qc7(.(T31, T34), X43, .(T31, T35), T36, T14) :- qc7(T34, X43, T35, T36, T14).

Afs:

fl1(x1, x2, x3)  =  fl1(x3)

(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:
fl1_in: (f,f,b)
p7_in: (f,f,f,f,b)
Transforming TRIPLES into the following Term Rewriting System:
Pi DP problem:
The TRS P consists of the following rules:

FL1_IN_AAG(.(T15, T17), T16, s(T14)) → U3_AAG(T15, T17, T16, T14, p7_in_aaaag(T15, X13, T16, T17, T14))
FL1_IN_AAG(.(T15, T17), T16, s(T14)) → P7_IN_AAAAG(T15, X13, T16, T17, T14)
P7_IN_AAAAG([], T24, T24, T23, T14) → U1_AAAAG(T24, T23, T14, fl1_in_aag(T23, T24, T14))
P7_IN_AAAAG([], T24, T24, T23, T14) → FL1_IN_AAG(T23, T24, T14)
P7_IN_AAAAG(.(T31, T34), X43, .(T31, T35), T36, T14) → U2_AAAAG(T31, T34, X43, T35, T36, T14, p7_in_aaaag(T34, X43, T35, T36, T14))
P7_IN_AAAAG(.(T31, T34), X43, .(T31, T35), T36, T14) → P7_IN_AAAAG(T34, X43, T35, T36, T14)

R is empty.
The argument filtering Pi contains the following mapping:
fl1_in_aag(x1, x2, x3)  =  fl1_in_aag(x3)
s(x1)  =  s(x1)
p7_in_aaaag(x1, x2, x3, x4, x5)  =  p7_in_aaaag(x5)
FL1_IN_AAG(x1, x2, x3)  =  FL1_IN_AAG(x3)
U3_AAG(x1, x2, x3, x4, x5)  =  U3_AAG(x4, x5)
P7_IN_AAAAG(x1, x2, x3, x4, x5)  =  P7_IN_AAAAG(x5)
U1_AAAAG(x1, x2, x3, x4)  =  U1_AAAAG(x3, x4)
U2_AAAAG(x1, x2, x3, x4, x5, x6, x7)  =  U2_AAAAG(x6, x7)

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:

FL1_IN_AAG(.(T15, T17), T16, s(T14)) → U3_AAG(T15, T17, T16, T14, p7_in_aaaag(T15, X13, T16, T17, T14))
FL1_IN_AAG(.(T15, T17), T16, s(T14)) → P7_IN_AAAAG(T15, X13, T16, T17, T14)
P7_IN_AAAAG([], T24, T24, T23, T14) → U1_AAAAG(T24, T23, T14, fl1_in_aag(T23, T24, T14))
P7_IN_AAAAG([], T24, T24, T23, T14) → FL1_IN_AAG(T23, T24, T14)
P7_IN_AAAAG(.(T31, T34), X43, .(T31, T35), T36, T14) → U2_AAAAG(T31, T34, X43, T35, T36, T14, p7_in_aaaag(T34, X43, T35, T36, T14))
P7_IN_AAAAG(.(T31, T34), X43, .(T31, T35), T36, T14) → P7_IN_AAAAG(T34, X43, T35, T36, T14)

R is empty.
The argument filtering Pi contains the following mapping:
fl1_in_aag(x1, x2, x3)  =  fl1_in_aag(x3)
s(x1)  =  s(x1)
p7_in_aaaag(x1, x2, x3, x4, x5)  =  p7_in_aaaag(x5)
FL1_IN_AAG(x1, x2, x3)  =  FL1_IN_AAG(x3)
U3_AAG(x1, x2, x3, x4, x5)  =  U3_AAG(x4, x5)
P7_IN_AAAAG(x1, x2, x3, x4, x5)  =  P7_IN_AAAAG(x5)
U1_AAAAG(x1, x2, x3, x4)  =  U1_AAAAG(x3, x4)
U2_AAAAG(x1, x2, x3, x4, x5, x6, x7)  =  U2_AAAAG(x6, x7)

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:

FL1_IN_AAG(.(T15, T17), T16, s(T14)) → P7_IN_AAAAG(T15, X13, T16, T17, T14)
P7_IN_AAAAG([], T24, T24, T23, T14) → FL1_IN_AAG(T23, T24, T14)
P7_IN_AAAAG(.(T31, T34), X43, .(T31, T35), T36, T14) → P7_IN_AAAAG(T34, X43, T35, T36, T14)

R is empty.
The argument filtering Pi contains the following mapping:
s(x1)  =  s(x1)
FL1_IN_AAG(x1, x2, x3)  =  FL1_IN_AAG(x3)
P7_IN_AAAAG(x1, x2, x3, x4, x5)  =  P7_IN_AAAAG(x5)

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:

FL1_IN_AAG(s(T14)) → P7_IN_AAAAG(T14)
P7_IN_AAAAG(T14) → FL1_IN_AAG(T14)
P7_IN_AAAAG(T14) → P7_IN_AAAAG(T14)

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:

FL1_IN_AAG(s(T14)) → P7_IN_AAAAG(T14)
P7_IN_AAAAG(T14) → FL1_IN_AAG(T14)

No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [POLO]:

POL(FL1_IN_AAG(x₁)) = x₁   
POL(P7_IN_AAAAG(x₁)) = 1 + 2·x₁   
POL(s(x₁)) = 2 + 2·x₁   

(10) Obligation:

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

P7_IN_AAAAG(T14) → P7_IN_AAAAG(T14)

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

(11) NonTerminationProof (EQUIVALENT transformation)

We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.

s = P7_IN_AAAAG(T14) evaluates to t =P7_IN_AAAAG(T14)

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:

• Semiunifier: [ ]
• Matcher: [ ]

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Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from P7_IN_AAAAG(T14) to P7_IN_AAAAG(T14).

(12) NO