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

0 Prolog
↳1 PrologToPrologProblemTransformerProof (⇐)
↳2 Prolog
    ↳3 PrologToPiTRSProof (⇐)
        ↳4 PiTRS
        ↳5 DependencyPairsProof (⇔)
        ↳6 PiDP
        ↳7 DependencyGraphProof (⇔)
        ↳8 PiDP
        ↳9 UsableRulesProof (⇔)
        ↳10 PiDP
        ↳11 PiDPToQDPProof (⇐)
        ↳12 QDP
        ↳13 UsableRulesReductionPairsProof (⇔)
        ↳14 QDP
        ↳15 NonTerminationProof (⇔)
        ↳16 NO
    ↳17 PrologToPiTRSProof (⇐)
        ↳18 PiTRS
        ↳19 DependencyPairsProof (⇔)
        ↳20 PiDP
        ↳21 DependencyGraphProof (⇔)
        ↳22 PiDP
        ↳23 UsableRulesProof (⇔)
        ↳24 PiDP
        ↳25 PiDPToQDPProof (⇐)
        ↳26 QDP
        ↳27 UsableRulesReductionPairsProof (⇔)
        ↳28 QDP
        ↳29 NonTerminationProof (⇔)
        ↳30 NO

(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) PrologToPrologProblemTransformerProof (SOUND transformation)

Built Prolog problem from termination graph.

(2) Obligation:

Clauses:

p7([], T24, T24, T23, T14) :- fl1(T23, T24, T14).
p7(.(T31, T34), X43, .(T31, T35), T36, T14) :- p7(T34, X43, T35, T36, T14).
fl1([], [], 0).
fl1(.(T15, T17), T16, s(T14)) :- p7(T15, X13, T16, T17, T14).

Queries:

fl1(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:
fl1_in: (f,f,b)
p7_in: (f,f,f,f,b)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

fl1_in_aag([], [], 0) → fl1_out_aag([], [], 0)
fl1_in_aag(.(T15, T17), T16, s(T14)) → U3_aag(T15, T17, T16, 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))
U1_aaaag(T24, T23, T14, fl1_out_aag(T23, T24, T14)) → p7_out_aaaag([], T24, T24, T23, 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))
U2_aaaag(T31, T34, X43, T35, T36, T14, p7_out_aaaag(T34, X43, T35, T36, T14)) → p7_out_aaaag(.(T31, T34), X43, .(T31, T35), T36, T14)
U3_aag(T15, T17, T16, T14, p7_out_aaaag(T15, X13, T16, T17, T14)) → fl1_out_aag(.(T15, T17), T16, s(T14))

The argument filtering Pi contains the following mapping:
fl1_in_aag(x1, x2, x3)  =  fl1_in_aag(x3)
0  =  0
fl1_out_aag(x1, x2, x3)  =  fl1_out_aag
s(x1)  =  s(x1)
U3_aag(x1, x2, x3, x4, x5)  =  U3_aag(x5)
p7_in_aaaag(x1, x2, x3, x4, x5)  =  p7_in_aaaag(x5)
U1_aaaag(x1, x2, x3, x4)  =  U1_aaaag(x4)
p7_out_aaaag(x1, x2, x3, x4, x5)  =  p7_out_aaaag
U2_aaaag(x1, x2, x3, x4, x5, x6, x7)  =  U2_aaaag(x7)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(4) Obligation:

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

fl1_in_aag([], [], 0) → fl1_out_aag([], [], 0)
fl1_in_aag(.(T15, T17), T16, s(T14)) → U3_aag(T15, T17, T16, 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))
U1_aaaag(T24, T23, T14, fl1_out_aag(T23, T24, T14)) → p7_out_aaaag([], T24, T24, T23, 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))
U2_aaaag(T31, T34, X43, T35, T36, T14, p7_out_aaaag(T34, X43, T35, T36, T14)) → p7_out_aaaag(.(T31, T34), X43, .(T31, T35), T36, T14)
U3_aag(T15, T17, T16, T14, p7_out_aaaag(T15, X13, T16, T17, T14)) → fl1_out_aag(.(T15, T17), T16, s(T14))

The argument filtering Pi contains the following mapping:
fl1_in_aag(x1, x2, x3)  =  fl1_in_aag(x3)
0  =  0
fl1_out_aag(x1, x2, x3)  =  fl1_out_aag
s(x1)  =  s(x1)
U3_aag(x1, x2, x3, x4, x5)  =  U3_aag(x5)
p7_in_aaaag(x1, x2, x3, x4, x5)  =  p7_in_aaaag(x5)
U1_aaaag(x1, x2, x3, x4)  =  U1_aaaag(x4)
p7_out_aaaag(x1, x2, x3, x4, x5)  =  p7_out_aaaag
U2_aaaag(x1, x2, x3, x4, x5, x6, x7)  =  U2_aaaag(x7)

(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:

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)

The TRS R consists of the following rules:

fl1_in_aag([], [], 0) → fl1_out_aag([], [], 0)
fl1_in_aag(.(T15, T17), T16, s(T14)) → U3_aag(T15, T17, T16, 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))
U1_aaaag(T24, T23, T14, fl1_out_aag(T23, T24, T14)) → p7_out_aaaag([], T24, T24, T23, 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))
U2_aaaag(T31, T34, X43, T35, T36, T14, p7_out_aaaag(T34, X43, T35, T36, T14)) → p7_out_aaaag(.(T31, T34), X43, .(T31, T35), T36, T14)
U3_aag(T15, T17, T16, T14, p7_out_aaaag(T15, X13, T16, T17, T14)) → fl1_out_aag(.(T15, T17), T16, s(T14))

The argument filtering Pi contains the following mapping:
fl1_in_aag(x1, x2, x3)  =  fl1_in_aag(x3)
0  =  0
fl1_out_aag(x1, x2, x3)  =  fl1_out_aag
s(x1)  =  s(x1)
U3_aag(x1, x2, x3, x4, x5)  =  U3_aag(x5)
p7_in_aaaag(x1, x2, x3, x4, x5)  =  p7_in_aaaag(x5)
U1_aaaag(x1, x2, x3, x4)  =  U1_aaaag(x4)
p7_out_aaaag(x1, x2, x3, x4, x5)  =  p7_out_aaaag
U2_aaaag(x1, x2, x3, x4, x5, x6, x7)  =  U2_aaaag(x7)
FL1_IN_AAG(x1, x2, x3)  =  FL1_IN_AAG(x3)
U3_AAG(x1, x2, x3, x4, x5)  =  U3_AAG(x5)
P7_IN_AAAAG(x1, x2, x3, x4, x5)  =  P7_IN_AAAAG(x5)
U1_AAAAG(x1, x2, x3, x4)  =  U1_AAAAG(x4)
U2_AAAAG(x1, x2, x3, x4, x5, x6, x7)  =  U2_AAAAG(x7)

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

(6) 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)

The TRS R consists of the following rules:

fl1_in_aag([], [], 0) → fl1_out_aag([], [], 0)
fl1_in_aag(.(T15, T17), T16, s(T14)) → U3_aag(T15, T17, T16, 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))
U1_aaaag(T24, T23, T14, fl1_out_aag(T23, T24, T14)) → p7_out_aaaag([], T24, T24, T23, 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))
U2_aaaag(T31, T34, X43, T35, T36, T14, p7_out_aaaag(T34, X43, T35, T36, T14)) → p7_out_aaaag(.(T31, T34), X43, .(T31, T35), T36, T14)
U3_aag(T15, T17, T16, T14, p7_out_aaaag(T15, X13, T16, T17, T14)) → fl1_out_aag(.(T15, T17), T16, s(T14))

The argument filtering Pi contains the following mapping:
fl1_in_aag(x1, x2, x3)  =  fl1_in_aag(x3)
0  =  0
fl1_out_aag(x1, x2, x3)  =  fl1_out_aag
s(x1)  =  s(x1)
U3_aag(x1, x2, x3, x4, x5)  =  U3_aag(x5)
p7_in_aaaag(x1, x2, x3, x4, x5)  =  p7_in_aaaag(x5)
U1_aaaag(x1, x2, x3, x4)  =  U1_aaaag(x4)
p7_out_aaaag(x1, x2, x3, x4, x5)  =  p7_out_aaaag
U2_aaaag(x1, x2, x3, x4, x5, x6, x7)  =  U2_aaaag(x7)
FL1_IN_AAG(x1, x2, x3)  =  FL1_IN_AAG(x3)
U3_AAG(x1, x2, x3, x4, x5)  =  U3_AAG(x5)
P7_IN_AAAAG(x1, x2, x3, x4, x5)  =  P7_IN_AAAAG(x5)
U1_AAAAG(x1, x2, x3, x4)  =  U1_AAAAG(x4)
U2_AAAAG(x1, x2, x3, x4, x5, x6, x7)  =  U2_AAAAG(x7)

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

(7) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 3 less nodes.

(8) 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)

The TRS R consists of the following rules:

fl1_in_aag([], [], 0) → fl1_out_aag([], [], 0)
fl1_in_aag(.(T15, T17), T16, s(T14)) → U3_aag(T15, T17, T16, 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))
U1_aaaag(T24, T23, T14, fl1_out_aag(T23, T24, T14)) → p7_out_aaaag([], T24, T24, T23, 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))
U2_aaaag(T31, T34, X43, T35, T36, T14, p7_out_aaaag(T34, X43, T35, T36, T14)) → p7_out_aaaag(.(T31, T34), X43, .(T31, T35), T36, T14)
U3_aag(T15, T17, T16, T14, p7_out_aaaag(T15, X13, T16, T17, T14)) → fl1_out_aag(.(T15, T17), T16, s(T14))

The argument filtering Pi contains the following mapping:
fl1_in_aag(x1, x2, x3)  =  fl1_in_aag(x3)
0  =  0
fl1_out_aag(x1, x2, x3)  =  fl1_out_aag
s(x1)  =  s(x1)
U3_aag(x1, x2, x3, x4, x5)  =  U3_aag(x5)
p7_in_aaaag(x1, x2, x3, x4, x5)  =  p7_in_aaaag(x5)
U1_aaaag(x1, x2, x3, x4)  =  U1_aaaag(x4)
p7_out_aaaag(x1, x2, x3, x4, x5)  =  p7_out_aaaag
U2_aaaag(x1, x2, x3, x4, x5, x6, x7)  =  U2_aaaag(x7)
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

(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:

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

(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:

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.

(13) 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   

(14) 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.

(15) 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: [ ]



Rewriting sequence

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



(16) NO

(17) PrologToPiTRSProof (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 Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

fl1_in_aag([], [], 0) → fl1_out_aag([], [], 0)
fl1_in_aag(.(T15, T17), T16, s(T14)) → U3_aag(T15, T17, T16, 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))
U1_aaaag(T24, T23, T14, fl1_out_aag(T23, T24, T14)) → p7_out_aaaag([], T24, T24, T23, 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))
U2_aaaag(T31, T34, X43, T35, T36, T14, p7_out_aaaag(T34, X43, T35, T36, T14)) → p7_out_aaaag(.(T31, T34), X43, .(T31, T35), T36, T14)
U3_aag(T15, T17, T16, T14, p7_out_aaaag(T15, X13, T16, T17, T14)) → fl1_out_aag(.(T15, T17), T16, s(T14))

The argument filtering Pi contains the following mapping:
fl1_in_aag(x1, x2, x3)  =  fl1_in_aag(x3)
0  =  0
fl1_out_aag(x1, x2, x3)  =  fl1_out_aag(x3)
s(x1)  =  s(x1)
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)
p7_out_aaaag(x1, x2, x3, x4, x5)  =  p7_out_aaaag(x5)
U2_aaaag(x1, x2, x3, x4, x5, x6, x7)  =  U2_aaaag(x6, x7)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(18) Obligation:

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

fl1_in_aag([], [], 0) → fl1_out_aag([], [], 0)
fl1_in_aag(.(T15, T17), T16, s(T14)) → U3_aag(T15, T17, T16, 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))
U1_aaaag(T24, T23, T14, fl1_out_aag(T23, T24, T14)) → p7_out_aaaag([], T24, T24, T23, 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))
U2_aaaag(T31, T34, X43, T35, T36, T14, p7_out_aaaag(T34, X43, T35, T36, T14)) → p7_out_aaaag(.(T31, T34), X43, .(T31, T35), T36, T14)
U3_aag(T15, T17, T16, T14, p7_out_aaaag(T15, X13, T16, T17, T14)) → fl1_out_aag(.(T15, T17), T16, s(T14))

The argument filtering Pi contains the following mapping:
fl1_in_aag(x1, x2, x3)  =  fl1_in_aag(x3)
0  =  0
fl1_out_aag(x1, x2, x3)  =  fl1_out_aag(x3)
s(x1)  =  s(x1)
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)
p7_out_aaaag(x1, x2, x3, x4, x5)  =  p7_out_aaaag(x5)
U2_aaaag(x1, x2, x3, x4, x5, x6, x7)  =  U2_aaaag(x6, x7)

(19) 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:

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)

The TRS R consists of the following rules:

fl1_in_aag([], [], 0) → fl1_out_aag([], [], 0)
fl1_in_aag(.(T15, T17), T16, s(T14)) → U3_aag(T15, T17, T16, 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))
U1_aaaag(T24, T23, T14, fl1_out_aag(T23, T24, T14)) → p7_out_aaaag([], T24, T24, T23, 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))
U2_aaaag(T31, T34, X43, T35, T36, T14, p7_out_aaaag(T34, X43, T35, T36, T14)) → p7_out_aaaag(.(T31, T34), X43, .(T31, T35), T36, T14)
U3_aag(T15, T17, T16, T14, p7_out_aaaag(T15, X13, T16, T17, T14)) → fl1_out_aag(.(T15, T17), T16, s(T14))

The argument filtering Pi contains the following mapping:
fl1_in_aag(x1, x2, x3)  =  fl1_in_aag(x3)
0  =  0
fl1_out_aag(x1, x2, x3)  =  fl1_out_aag(x3)
s(x1)  =  s(x1)
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)
p7_out_aaaag(x1, x2, x3, x4, x5)  =  p7_out_aaaag(x5)
U2_aaaag(x1, x2, x3, x4, x5, x6, x7)  =  U2_aaaag(x6, x7)
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

(20) 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)

The TRS R consists of the following rules:

fl1_in_aag([], [], 0) → fl1_out_aag([], [], 0)
fl1_in_aag(.(T15, T17), T16, s(T14)) → U3_aag(T15, T17, T16, 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))
U1_aaaag(T24, T23, T14, fl1_out_aag(T23, T24, T14)) → p7_out_aaaag([], T24, T24, T23, 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))
U2_aaaag(T31, T34, X43, T35, T36, T14, p7_out_aaaag(T34, X43, T35, T36, T14)) → p7_out_aaaag(.(T31, T34), X43, .(T31, T35), T36, T14)
U3_aag(T15, T17, T16, T14, p7_out_aaaag(T15, X13, T16, T17, T14)) → fl1_out_aag(.(T15, T17), T16, s(T14))

The argument filtering Pi contains the following mapping:
fl1_in_aag(x1, x2, x3)  =  fl1_in_aag(x3)
0  =  0
fl1_out_aag(x1, x2, x3)  =  fl1_out_aag(x3)
s(x1)  =  s(x1)
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)
p7_out_aaaag(x1, x2, x3, x4, x5)  =  p7_out_aaaag(x5)
U2_aaaag(x1, x2, x3, x4, x5, x6, x7)  =  U2_aaaag(x6, x7)
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

(21) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 3 less nodes.

(22) 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)

The TRS R consists of the following rules:

fl1_in_aag([], [], 0) → fl1_out_aag([], [], 0)
fl1_in_aag(.(T15, T17), T16, s(T14)) → U3_aag(T15, T17, T16, 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))
U1_aaaag(T24, T23, T14, fl1_out_aag(T23, T24, T14)) → p7_out_aaaag([], T24, T24, T23, 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))
U2_aaaag(T31, T34, X43, T35, T36, T14, p7_out_aaaag(T34, X43, T35, T36, T14)) → p7_out_aaaag(.(T31, T34), X43, .(T31, T35), T36, T14)
U3_aag(T15, T17, T16, T14, p7_out_aaaag(T15, X13, T16, T17, T14)) → fl1_out_aag(.(T15, T17), T16, s(T14))

The argument filtering Pi contains the following mapping:
fl1_in_aag(x1, x2, x3)  =  fl1_in_aag(x3)
0  =  0
fl1_out_aag(x1, x2, x3)  =  fl1_out_aag(x3)
s(x1)  =  s(x1)
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)
p7_out_aaaag(x1, x2, x3, x4, x5)  =  p7_out_aaaag(x5)
U2_aaaag(x1, x2, x3, x4, x5, x6, x7)  =  U2_aaaag(x6, x7)
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

(23) UsableRulesProof (EQUIVALENT transformation)

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

(24) 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

(25) PiDPToQDPProof (SOUND transformation)

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

(26) 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.

(27) 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   

(28) 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.

(29) 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: [ ]



Rewriting sequence

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



(30) NO