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 PrologToPiTRSProof (⇐)
    ↳2 PiTRS
    ↳3 DependencyPairsProof (⇔)
    ↳4 PiDP
    ↳5 DependencyGraphProof (⇔)
    ↳6 AND
      ↳7 PiDP
        ↳8 UsableRulesProof (⇔)
        ↳9 PiDP
        ↳10 PiDPToQDPProof (⇐)
        ↳11 QDP
        ↳12 NonTerminationProof (⇔)
        ↳13 FALSE
      ↳14 PiDP
        ↳15 UsableRulesProof (⇔)
        ↳16 PiDP
        ↳17 PiDPToQDPProof (⇐)
        ↳18 QDP
        ↳19 QDPSizeChangeProof (⇔)
        ↳20 TRUE
  ↳21 PrologToPiTRSProof (⇐)
    ↳22 PiTRS
    ↳23 DependencyPairsProof (⇔)
    ↳24 PiDP
    ↳25 DependencyGraphProof (⇔)
    ↳26 AND
      ↳27 PiDP
        ↳28 UsableRulesProof (⇔)
        ↳29 PiDP
        ↳30 PiDPToQDPProof (⇐)
        ↳31 QDP
        ↳32 NonTerminationProof (⇔)
        ↳33 FALSE
      ↳34 PiDP
        ↳35 UsableRulesProof (⇔)
        ↳36 PiDP
        ↳37 PiDPToQDPProof (⇐)
        ↳38 QDP
        ↳39 QDPSizeChangeProof (⇔)
        ↳40 TRUE

(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) PrologToPiTRSProof (SOUND transformation)

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

fl_in_aag([], [], 0) → fl_out_aag([], [], 0)
fl_in_aag(.(E, X), R, s(Z)) → U1_aag(E, X, R, Z, append_in_aaa(E, Y, R))
append_in_aaa([], X, X) → append_out_aaa([], X, X)
append_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U3_aaa(X, Xs, Ys, Zs, append_in_aaa(Xs, Ys, Zs))
U3_aaa(X, Xs, Ys, Zs, append_out_aaa(Xs, Ys, Zs)) → append_out_aaa(.(X, Xs), Ys, .(X, Zs))
U1_aag(E, X, R, Z, append_out_aaa(E, Y, R)) → U2_aag(E, X, R, Z, fl_in_aag(X, Y, Z))
U2_aag(E, X, R, Z, fl_out_aag(X, Y, Z)) → fl_out_aag(.(E, X), R, s(Z))

The argument filtering Pi contains the following mapping:
fl_in_aag(x1, x2, x3)  =  fl_in_aag(x3)
0  =  0
fl_out_aag(x1, x2, x3)  =  fl_out_aag
s(x1)  =  s(x1)
U1_aag(x1, x2, x3, x4, x5)  =  U1_aag(x4, x5)
append_in_aaa(x1, x2, x3)  =  append_in_aaa
append_out_aaa(x1, x2, x3)  =  append_out_aaa
U3_aaa(x1, x2, x3, x4, x5)  =  U3_aaa(x5)
U2_aag(x1, x2, x3, x4, x5)  =  U2_aag(x5)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(2) Obligation:

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

fl_in_aag([], [], 0) → fl_out_aag([], [], 0)
fl_in_aag(.(E, X), R, s(Z)) → U1_aag(E, X, R, Z, append_in_aaa(E, Y, R))
append_in_aaa([], X, X) → append_out_aaa([], X, X)
append_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U3_aaa(X, Xs, Ys, Zs, append_in_aaa(Xs, Ys, Zs))
U3_aaa(X, Xs, Ys, Zs, append_out_aaa(Xs, Ys, Zs)) → append_out_aaa(.(X, Xs), Ys, .(X, Zs))
U1_aag(E, X, R, Z, append_out_aaa(E, Y, R)) → U2_aag(E, X, R, Z, fl_in_aag(X, Y, Z))
U2_aag(E, X, R, Z, fl_out_aag(X, Y, Z)) → fl_out_aag(.(E, X), R, s(Z))

The argument filtering Pi contains the following mapping:
fl_in_aag(x1, x2, x3)  =  fl_in_aag(x3)
0  =  0
fl_out_aag(x1, x2, x3)  =  fl_out_aag
s(x1)  =  s(x1)
U1_aag(x1, x2, x3, x4, x5)  =  U1_aag(x4, x5)
append_in_aaa(x1, x2, x3)  =  append_in_aaa
append_out_aaa(x1, x2, x3)  =  append_out_aaa
U3_aaa(x1, x2, x3, x4, x5)  =  U3_aaa(x5)
U2_aag(x1, x2, x3, x4, x5)  =  U2_aag(x5)

(3) 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:

FL_IN_AAG(.(E, X), R, s(Z)) → U1_AAG(E, X, R, Z, append_in_aaa(E, Y, R))
FL_IN_AAG(.(E, X), R, s(Z)) → APPEND_IN_AAA(E, Y, R)
APPEND_IN_AAA(.(X, Xs), Ys, .(X, Zs)) → U3_AAA(X, Xs, Ys, Zs, append_in_aaa(Xs, Ys, Zs))
APPEND_IN_AAA(.(X, Xs), Ys, .(X, Zs)) → APPEND_IN_AAA(Xs, Ys, Zs)
U1_AAG(E, X, R, Z, append_out_aaa(E, Y, R)) → U2_AAG(E, X, R, Z, fl_in_aag(X, Y, Z))
U1_AAG(E, X, R, Z, append_out_aaa(E, Y, R)) → FL_IN_AAG(X, Y, Z)

The TRS R consists of the following rules:

fl_in_aag([], [], 0) → fl_out_aag([], [], 0)
fl_in_aag(.(E, X), R, s(Z)) → U1_aag(E, X, R, Z, append_in_aaa(E, Y, R))
append_in_aaa([], X, X) → append_out_aaa([], X, X)
append_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U3_aaa(X, Xs, Ys, Zs, append_in_aaa(Xs, Ys, Zs))
U3_aaa(X, Xs, Ys, Zs, append_out_aaa(Xs, Ys, Zs)) → append_out_aaa(.(X, Xs), Ys, .(X, Zs))
U1_aag(E, X, R, Z, append_out_aaa(E, Y, R)) → U2_aag(E, X, R, Z, fl_in_aag(X, Y, Z))
U2_aag(E, X, R, Z, fl_out_aag(X, Y, Z)) → fl_out_aag(.(E, X), R, s(Z))

The argument filtering Pi contains the following mapping:
fl_in_aag(x1, x2, x3)  =  fl_in_aag(x3)
0  =  0
fl_out_aag(x1, x2, x3)  =  fl_out_aag
s(x1)  =  s(x1)
U1_aag(x1, x2, x3, x4, x5)  =  U1_aag(x4, x5)
append_in_aaa(x1, x2, x3)  =  append_in_aaa
append_out_aaa(x1, x2, x3)  =  append_out_aaa
U3_aaa(x1, x2, x3, x4, x5)  =  U3_aaa(x5)
U2_aag(x1, x2, x3, x4, x5)  =  U2_aag(x5)
FL_IN_AAG(x1, x2, x3)  =  FL_IN_AAG(x3)
U1_AAG(x1, x2, x3, x4, x5)  =  U1_AAG(x4, x5)
APPEND_IN_AAA(x1, x2, x3)  =  APPEND_IN_AAA
U3_AAA(x1, x2, x3, x4, x5)  =  U3_AAA(x5)
U2_AAG(x1, x2, x3, x4, x5)  =  U2_AAG(x5)

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

(4) Obligation:

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

FL_IN_AAG(.(E, X), R, s(Z)) → U1_AAG(E, X, R, Z, append_in_aaa(E, Y, R))
FL_IN_AAG(.(E, X), R, s(Z)) → APPEND_IN_AAA(E, Y, R)
APPEND_IN_AAA(.(X, Xs), Ys, .(X, Zs)) → U3_AAA(X, Xs, Ys, Zs, append_in_aaa(Xs, Ys, Zs))
APPEND_IN_AAA(.(X, Xs), Ys, .(X, Zs)) → APPEND_IN_AAA(Xs, Ys, Zs)
U1_AAG(E, X, R, Z, append_out_aaa(E, Y, R)) → U2_AAG(E, X, R, Z, fl_in_aag(X, Y, Z))
U1_AAG(E, X, R, Z, append_out_aaa(E, Y, R)) → FL_IN_AAG(X, Y, Z)

The TRS R consists of the following rules:

fl_in_aag([], [], 0) → fl_out_aag([], [], 0)
fl_in_aag(.(E, X), R, s(Z)) → U1_aag(E, X, R, Z, append_in_aaa(E, Y, R))
append_in_aaa([], X, X) → append_out_aaa([], X, X)
append_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U3_aaa(X, Xs, Ys, Zs, append_in_aaa(Xs, Ys, Zs))
U3_aaa(X, Xs, Ys, Zs, append_out_aaa(Xs, Ys, Zs)) → append_out_aaa(.(X, Xs), Ys, .(X, Zs))
U1_aag(E, X, R, Z, append_out_aaa(E, Y, R)) → U2_aag(E, X, R, Z, fl_in_aag(X, Y, Z))
U2_aag(E, X, R, Z, fl_out_aag(X, Y, Z)) → fl_out_aag(.(E, X), R, s(Z))

The argument filtering Pi contains the following mapping:
fl_in_aag(x1, x2, x3)  =  fl_in_aag(x3)
0  =  0
fl_out_aag(x1, x2, x3)  =  fl_out_aag
s(x1)  =  s(x1)
U1_aag(x1, x2, x3, x4, x5)  =  U1_aag(x4, x5)
append_in_aaa(x1, x2, x3)  =  append_in_aaa
append_out_aaa(x1, x2, x3)  =  append_out_aaa
U3_aaa(x1, x2, x3, x4, x5)  =  U3_aaa(x5)
U2_aag(x1, x2, x3, x4, x5)  =  U2_aag(x5)
FL_IN_AAG(x1, x2, x3)  =  FL_IN_AAG(x3)
U1_AAG(x1, x2, x3, x4, x5)  =  U1_AAG(x4, x5)
APPEND_IN_AAA(x1, x2, x3)  =  APPEND_IN_AAA
U3_AAA(x1, x2, x3, x4, x5)  =  U3_AAA(x5)
U2_AAG(x1, x2, x3, x4, x5)  =  U2_AAG(x5)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 2 SCCs with 3 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

APPEND_IN_AAA(.(X, Xs), Ys, .(X, Zs)) → APPEND_IN_AAA(Xs, Ys, Zs)

The TRS R consists of the following rules:

fl_in_aag([], [], 0) → fl_out_aag([], [], 0)
fl_in_aag(.(E, X), R, s(Z)) → U1_aag(E, X, R, Z, append_in_aaa(E, Y, R))
append_in_aaa([], X, X) → append_out_aaa([], X, X)
append_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U3_aaa(X, Xs, Ys, Zs, append_in_aaa(Xs, Ys, Zs))
U3_aaa(X, Xs, Ys, Zs, append_out_aaa(Xs, Ys, Zs)) → append_out_aaa(.(X, Xs), Ys, .(X, Zs))
U1_aag(E, X, R, Z, append_out_aaa(E, Y, R)) → U2_aag(E, X, R, Z, fl_in_aag(X, Y, Z))
U2_aag(E, X, R, Z, fl_out_aag(X, Y, Z)) → fl_out_aag(.(E, X), R, s(Z))

The argument filtering Pi contains the following mapping:
fl_in_aag(x1, x2, x3)  =  fl_in_aag(x3)
0  =  0
fl_out_aag(x1, x2, x3)  =  fl_out_aag
s(x1)  =  s(x1)
U1_aag(x1, x2, x3, x4, x5)  =  U1_aag(x4, x5)
append_in_aaa(x1, x2, x3)  =  append_in_aaa
append_out_aaa(x1, x2, x3)  =  append_out_aaa
U3_aaa(x1, x2, x3, x4, x5)  =  U3_aaa(x5)
U2_aag(x1, x2, x3, x4, x5)  =  U2_aag(x5)
APPEND_IN_AAA(x1, x2, x3)  =  APPEND_IN_AAA

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

(8) UsableRulesProof (EQUIVALENT transformation)

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

(9) Obligation:

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

APPEND_IN_AAA(.(X, Xs), Ys, .(X, Zs)) → APPEND_IN_AAA(Xs, Ys, Zs)

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

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

(10) PiDPToQDPProof (SOUND transformation)

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

(11) Obligation:

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

APPEND_IN_AAA → APPEND_IN_AAA

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

(12) 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 = APPEND_IN_AAA evaluates to t =APPEND_IN_AAA

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 APPEND_IN_AAA to APPEND_IN_AAA.



(13) FALSE

(14) Obligation:

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

U1_AAG(E, X, R, Z, append_out_aaa(E, Y, R)) → FL_IN_AAG(X, Y, Z)
FL_IN_AAG(.(E, X), R, s(Z)) → U1_AAG(E, X, R, Z, append_in_aaa(E, Y, R))

The TRS R consists of the following rules:

fl_in_aag([], [], 0) → fl_out_aag([], [], 0)
fl_in_aag(.(E, X), R, s(Z)) → U1_aag(E, X, R, Z, append_in_aaa(E, Y, R))
append_in_aaa([], X, X) → append_out_aaa([], X, X)
append_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U3_aaa(X, Xs, Ys, Zs, append_in_aaa(Xs, Ys, Zs))
U3_aaa(X, Xs, Ys, Zs, append_out_aaa(Xs, Ys, Zs)) → append_out_aaa(.(X, Xs), Ys, .(X, Zs))
U1_aag(E, X, R, Z, append_out_aaa(E, Y, R)) → U2_aag(E, X, R, Z, fl_in_aag(X, Y, Z))
U2_aag(E, X, R, Z, fl_out_aag(X, Y, Z)) → fl_out_aag(.(E, X), R, s(Z))

The argument filtering Pi contains the following mapping:
fl_in_aag(x1, x2, x3)  =  fl_in_aag(x3)
0  =  0
fl_out_aag(x1, x2, x3)  =  fl_out_aag
s(x1)  =  s(x1)
U1_aag(x1, x2, x3, x4, x5)  =  U1_aag(x4, x5)
append_in_aaa(x1, x2, x3)  =  append_in_aaa
append_out_aaa(x1, x2, x3)  =  append_out_aaa
U3_aaa(x1, x2, x3, x4, x5)  =  U3_aaa(x5)
U2_aag(x1, x2, x3, x4, x5)  =  U2_aag(x5)
FL_IN_AAG(x1, x2, x3)  =  FL_IN_AAG(x3)
U1_AAG(x1, x2, x3, x4, x5)  =  U1_AAG(x4, x5)

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

(15) UsableRulesProof (EQUIVALENT transformation)

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

(16) Obligation:

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

U1_AAG(E, X, R, Z, append_out_aaa(E, Y, R)) → FL_IN_AAG(X, Y, Z)
FL_IN_AAG(.(E, X), R, s(Z)) → U1_AAG(E, X, R, Z, append_in_aaa(E, Y, R))

The TRS R consists of the following rules:

append_in_aaa([], X, X) → append_out_aaa([], X, X)
append_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U3_aaa(X, Xs, Ys, Zs, append_in_aaa(Xs, Ys, Zs))
U3_aaa(X, Xs, Ys, Zs, append_out_aaa(Xs, Ys, Zs)) → append_out_aaa(.(X, Xs), Ys, .(X, Zs))

The argument filtering Pi contains the following mapping:
s(x1)  =  s(x1)
append_in_aaa(x1, x2, x3)  =  append_in_aaa
append_out_aaa(x1, x2, x3)  =  append_out_aaa
U3_aaa(x1, x2, x3, x4, x5)  =  U3_aaa(x5)
FL_IN_AAG(x1, x2, x3)  =  FL_IN_AAG(x3)
U1_AAG(x1, x2, x3, x4, x5)  =  U1_AAG(x4, x5)

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

(17) PiDPToQDPProof (SOUND transformation)

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

(18) Obligation:

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

U1_AAG(Z, append_out_aaa) → FL_IN_AAG(Z)
FL_IN_AAG(s(Z)) → U1_AAG(Z, append_in_aaa)

The TRS R consists of the following rules:

append_in_aaa → append_out_aaa
append_in_aaa → U3_aaa(append_in_aaa)
U3_aaa(append_out_aaa) → append_out_aaa

The set Q consists of the following terms:

append_in_aaa
U3_aaa(x0)

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

(19) 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:

  • FL_IN_AAG(s(Z)) → U1_AAG(Z, append_in_aaa)
    The graph contains the following edges 1 > 1

  • U1_AAG(Z, append_out_aaa) → FL_IN_AAG(Z)
    The graph contains the following edges 1 >= 1

(20) TRUE

(21) PrologToPiTRSProof (SOUND transformation)

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

fl_in_aag([], [], 0) → fl_out_aag([], [], 0)
fl_in_aag(.(E, X), R, s(Z)) → U1_aag(E, X, R, Z, append_in_aaa(E, Y, R))
append_in_aaa([], X, X) → append_out_aaa([], X, X)
append_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U3_aaa(X, Xs, Ys, Zs, append_in_aaa(Xs, Ys, Zs))
U3_aaa(X, Xs, Ys, Zs, append_out_aaa(Xs, Ys, Zs)) → append_out_aaa(.(X, Xs), Ys, .(X, Zs))
U1_aag(E, X, R, Z, append_out_aaa(E, Y, R)) → U2_aag(E, X, R, Z, fl_in_aag(X, Y, Z))
U2_aag(E, X, R, Z, fl_out_aag(X, Y, Z)) → fl_out_aag(.(E, X), R, s(Z))

The argument filtering Pi contains the following mapping:
fl_in_aag(x1, x2, x3)  =  fl_in_aag(x3)
0  =  0
fl_out_aag(x1, x2, x3)  =  fl_out_aag(x3)
s(x1)  =  s(x1)
U1_aag(x1, x2, x3, x4, x5)  =  U1_aag(x4, x5)
append_in_aaa(x1, x2, x3)  =  append_in_aaa
append_out_aaa(x1, x2, x3)  =  append_out_aaa
U3_aaa(x1, x2, x3, x4, x5)  =  U3_aaa(x5)
U2_aag(x1, x2, x3, x4, x5)  =  U2_aag(x4, x5)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(22) Obligation:

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

fl_in_aag([], [], 0) → fl_out_aag([], [], 0)
fl_in_aag(.(E, X), R, s(Z)) → U1_aag(E, X, R, Z, append_in_aaa(E, Y, R))
append_in_aaa([], X, X) → append_out_aaa([], X, X)
append_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U3_aaa(X, Xs, Ys, Zs, append_in_aaa(Xs, Ys, Zs))
U3_aaa(X, Xs, Ys, Zs, append_out_aaa(Xs, Ys, Zs)) → append_out_aaa(.(X, Xs), Ys, .(X, Zs))
U1_aag(E, X, R, Z, append_out_aaa(E, Y, R)) → U2_aag(E, X, R, Z, fl_in_aag(X, Y, Z))
U2_aag(E, X, R, Z, fl_out_aag(X, Y, Z)) → fl_out_aag(.(E, X), R, s(Z))

The argument filtering Pi contains the following mapping:
fl_in_aag(x1, x2, x3)  =  fl_in_aag(x3)
0  =  0
fl_out_aag(x1, x2, x3)  =  fl_out_aag(x3)
s(x1)  =  s(x1)
U1_aag(x1, x2, x3, x4, x5)  =  U1_aag(x4, x5)
append_in_aaa(x1, x2, x3)  =  append_in_aaa
append_out_aaa(x1, x2, x3)  =  append_out_aaa
U3_aaa(x1, x2, x3, x4, x5)  =  U3_aaa(x5)
U2_aag(x1, x2, x3, x4, x5)  =  U2_aag(x4, x5)

(23) 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:

FL_IN_AAG(.(E, X), R, s(Z)) → U1_AAG(E, X, R, Z, append_in_aaa(E, Y, R))
FL_IN_AAG(.(E, X), R, s(Z)) → APPEND_IN_AAA(E, Y, R)
APPEND_IN_AAA(.(X, Xs), Ys, .(X, Zs)) → U3_AAA(X, Xs, Ys, Zs, append_in_aaa(Xs, Ys, Zs))
APPEND_IN_AAA(.(X, Xs), Ys, .(X, Zs)) → APPEND_IN_AAA(Xs, Ys, Zs)
U1_AAG(E, X, R, Z, append_out_aaa(E, Y, R)) → U2_AAG(E, X, R, Z, fl_in_aag(X, Y, Z))
U1_AAG(E, X, R, Z, append_out_aaa(E, Y, R)) → FL_IN_AAG(X, Y, Z)

The TRS R consists of the following rules:

fl_in_aag([], [], 0) → fl_out_aag([], [], 0)
fl_in_aag(.(E, X), R, s(Z)) → U1_aag(E, X, R, Z, append_in_aaa(E, Y, R))
append_in_aaa([], X, X) → append_out_aaa([], X, X)
append_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U3_aaa(X, Xs, Ys, Zs, append_in_aaa(Xs, Ys, Zs))
U3_aaa(X, Xs, Ys, Zs, append_out_aaa(Xs, Ys, Zs)) → append_out_aaa(.(X, Xs), Ys, .(X, Zs))
U1_aag(E, X, R, Z, append_out_aaa(E, Y, R)) → U2_aag(E, X, R, Z, fl_in_aag(X, Y, Z))
U2_aag(E, X, R, Z, fl_out_aag(X, Y, Z)) → fl_out_aag(.(E, X), R, s(Z))

The argument filtering Pi contains the following mapping:
fl_in_aag(x1, x2, x3)  =  fl_in_aag(x3)
0  =  0
fl_out_aag(x1, x2, x3)  =  fl_out_aag(x3)
s(x1)  =  s(x1)
U1_aag(x1, x2, x3, x4, x5)  =  U1_aag(x4, x5)
append_in_aaa(x1, x2, x3)  =  append_in_aaa
append_out_aaa(x1, x2, x3)  =  append_out_aaa
U3_aaa(x1, x2, x3, x4, x5)  =  U3_aaa(x5)
U2_aag(x1, x2, x3, x4, x5)  =  U2_aag(x4, x5)
FL_IN_AAG(x1, x2, x3)  =  FL_IN_AAG(x3)
U1_AAG(x1, x2, x3, x4, x5)  =  U1_AAG(x4, x5)
APPEND_IN_AAA(x1, x2, x3)  =  APPEND_IN_AAA
U3_AAA(x1, x2, x3, x4, x5)  =  U3_AAA(x5)
U2_AAG(x1, x2, x3, x4, x5)  =  U2_AAG(x4, x5)

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

(24) Obligation:

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

FL_IN_AAG(.(E, X), R, s(Z)) → U1_AAG(E, X, R, Z, append_in_aaa(E, Y, R))
FL_IN_AAG(.(E, X), R, s(Z)) → APPEND_IN_AAA(E, Y, R)
APPEND_IN_AAA(.(X, Xs), Ys, .(X, Zs)) → U3_AAA(X, Xs, Ys, Zs, append_in_aaa(Xs, Ys, Zs))
APPEND_IN_AAA(.(X, Xs), Ys, .(X, Zs)) → APPEND_IN_AAA(Xs, Ys, Zs)
U1_AAG(E, X, R, Z, append_out_aaa(E, Y, R)) → U2_AAG(E, X, R, Z, fl_in_aag(X, Y, Z))
U1_AAG(E, X, R, Z, append_out_aaa(E, Y, R)) → FL_IN_AAG(X, Y, Z)

The TRS R consists of the following rules:

fl_in_aag([], [], 0) → fl_out_aag([], [], 0)
fl_in_aag(.(E, X), R, s(Z)) → U1_aag(E, X, R, Z, append_in_aaa(E, Y, R))
append_in_aaa([], X, X) → append_out_aaa([], X, X)
append_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U3_aaa(X, Xs, Ys, Zs, append_in_aaa(Xs, Ys, Zs))
U3_aaa(X, Xs, Ys, Zs, append_out_aaa(Xs, Ys, Zs)) → append_out_aaa(.(X, Xs), Ys, .(X, Zs))
U1_aag(E, X, R, Z, append_out_aaa(E, Y, R)) → U2_aag(E, X, R, Z, fl_in_aag(X, Y, Z))
U2_aag(E, X, R, Z, fl_out_aag(X, Y, Z)) → fl_out_aag(.(E, X), R, s(Z))

The argument filtering Pi contains the following mapping:
fl_in_aag(x1, x2, x3)  =  fl_in_aag(x3)
0  =  0
fl_out_aag(x1, x2, x3)  =  fl_out_aag(x3)
s(x1)  =  s(x1)
U1_aag(x1, x2, x3, x4, x5)  =  U1_aag(x4, x5)
append_in_aaa(x1, x2, x3)  =  append_in_aaa
append_out_aaa(x1, x2, x3)  =  append_out_aaa
U3_aaa(x1, x2, x3, x4, x5)  =  U3_aaa(x5)
U2_aag(x1, x2, x3, x4, x5)  =  U2_aag(x4, x5)
FL_IN_AAG(x1, x2, x3)  =  FL_IN_AAG(x3)
U1_AAG(x1, x2, x3, x4, x5)  =  U1_AAG(x4, x5)
APPEND_IN_AAA(x1, x2, x3)  =  APPEND_IN_AAA
U3_AAA(x1, x2, x3, x4, x5)  =  U3_AAA(x5)
U2_AAG(x1, x2, x3, x4, x5)  =  U2_AAG(x4, x5)

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

(25) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 2 SCCs with 3 less nodes.

(26) Complex Obligation (AND)

(27) Obligation:

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

APPEND_IN_AAA(.(X, Xs), Ys, .(X, Zs)) → APPEND_IN_AAA(Xs, Ys, Zs)

The TRS R consists of the following rules:

fl_in_aag([], [], 0) → fl_out_aag([], [], 0)
fl_in_aag(.(E, X), R, s(Z)) → U1_aag(E, X, R, Z, append_in_aaa(E, Y, R))
append_in_aaa([], X, X) → append_out_aaa([], X, X)
append_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U3_aaa(X, Xs, Ys, Zs, append_in_aaa(Xs, Ys, Zs))
U3_aaa(X, Xs, Ys, Zs, append_out_aaa(Xs, Ys, Zs)) → append_out_aaa(.(X, Xs), Ys, .(X, Zs))
U1_aag(E, X, R, Z, append_out_aaa(E, Y, R)) → U2_aag(E, X, R, Z, fl_in_aag(X, Y, Z))
U2_aag(E, X, R, Z, fl_out_aag(X, Y, Z)) → fl_out_aag(.(E, X), R, s(Z))

The argument filtering Pi contains the following mapping:
fl_in_aag(x1, x2, x3)  =  fl_in_aag(x3)
0  =  0
fl_out_aag(x1, x2, x3)  =  fl_out_aag(x3)
s(x1)  =  s(x1)
U1_aag(x1, x2, x3, x4, x5)  =  U1_aag(x4, x5)
append_in_aaa(x1, x2, x3)  =  append_in_aaa
append_out_aaa(x1, x2, x3)  =  append_out_aaa
U3_aaa(x1, x2, x3, x4, x5)  =  U3_aaa(x5)
U2_aag(x1, x2, x3, x4, x5)  =  U2_aag(x4, x5)
APPEND_IN_AAA(x1, x2, x3)  =  APPEND_IN_AAA

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

(28) UsableRulesProof (EQUIVALENT transformation)

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

(29) Obligation:

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

APPEND_IN_AAA(.(X, Xs), Ys, .(X, Zs)) → APPEND_IN_AAA(Xs, Ys, Zs)

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

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

(30) PiDPToQDPProof (SOUND transformation)

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

(31) Obligation:

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

APPEND_IN_AAA → APPEND_IN_AAA

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

(32) 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 = APPEND_IN_AAA evaluates to t =APPEND_IN_AAA

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
  • Matcher: [ ]
  • Semiunifier: [ ]



Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from APPEND_IN_AAA to APPEND_IN_AAA.



(33) FALSE

(34) Obligation:

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

U1_AAG(E, X, R, Z, append_out_aaa(E, Y, R)) → FL_IN_AAG(X, Y, Z)
FL_IN_AAG(.(E, X), R, s(Z)) → U1_AAG(E, X, R, Z, append_in_aaa(E, Y, R))

The TRS R consists of the following rules:

fl_in_aag([], [], 0) → fl_out_aag([], [], 0)
fl_in_aag(.(E, X), R, s(Z)) → U1_aag(E, X, R, Z, append_in_aaa(E, Y, R))
append_in_aaa([], X, X) → append_out_aaa([], X, X)
append_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U3_aaa(X, Xs, Ys, Zs, append_in_aaa(Xs, Ys, Zs))
U3_aaa(X, Xs, Ys, Zs, append_out_aaa(Xs, Ys, Zs)) → append_out_aaa(.(X, Xs), Ys, .(X, Zs))
U1_aag(E, X, R, Z, append_out_aaa(E, Y, R)) → U2_aag(E, X, R, Z, fl_in_aag(X, Y, Z))
U2_aag(E, X, R, Z, fl_out_aag(X, Y, Z)) → fl_out_aag(.(E, X), R, s(Z))

The argument filtering Pi contains the following mapping:
fl_in_aag(x1, x2, x3)  =  fl_in_aag(x3)
0  =  0
fl_out_aag(x1, x2, x3)  =  fl_out_aag(x3)
s(x1)  =  s(x1)
U1_aag(x1, x2, x3, x4, x5)  =  U1_aag(x4, x5)
append_in_aaa(x1, x2, x3)  =  append_in_aaa
append_out_aaa(x1, x2, x3)  =  append_out_aaa
U3_aaa(x1, x2, x3, x4, x5)  =  U3_aaa(x5)
U2_aag(x1, x2, x3, x4, x5)  =  U2_aag(x4, x5)
FL_IN_AAG(x1, x2, x3)  =  FL_IN_AAG(x3)
U1_AAG(x1, x2, x3, x4, x5)  =  U1_AAG(x4, x5)

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

(35) UsableRulesProof (EQUIVALENT transformation)

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

(36) Obligation:

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

U1_AAG(E, X, R, Z, append_out_aaa(E, Y, R)) → FL_IN_AAG(X, Y, Z)
FL_IN_AAG(.(E, X), R, s(Z)) → U1_AAG(E, X, R, Z, append_in_aaa(E, Y, R))

The TRS R consists of the following rules:

append_in_aaa([], X, X) → append_out_aaa([], X, X)
append_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U3_aaa(X, Xs, Ys, Zs, append_in_aaa(Xs, Ys, Zs))
U3_aaa(X, Xs, Ys, Zs, append_out_aaa(Xs, Ys, Zs)) → append_out_aaa(.(X, Xs), Ys, .(X, Zs))

The argument filtering Pi contains the following mapping:
s(x1)  =  s(x1)
append_in_aaa(x1, x2, x3)  =  append_in_aaa
append_out_aaa(x1, x2, x3)  =  append_out_aaa
U3_aaa(x1, x2, x3, x4, x5)  =  U3_aaa(x5)
FL_IN_AAG(x1, x2, x3)  =  FL_IN_AAG(x3)
U1_AAG(x1, x2, x3, x4, x5)  =  U1_AAG(x4, x5)

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

(37) PiDPToQDPProof (SOUND transformation)

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

(38) Obligation:

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

U1_AAG(Z, append_out_aaa) → FL_IN_AAG(Z)
FL_IN_AAG(s(Z)) → U1_AAG(Z, append_in_aaa)

The TRS R consists of the following rules:

append_in_aaa → append_out_aaa
append_in_aaa → U3_aaa(append_in_aaa)
U3_aaa(append_out_aaa) → append_out_aaa

The set Q consists of the following terms:

append_in_aaa
U3_aaa(x0)

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

(39) 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:

  • FL_IN_AAG(s(Z)) → U1_AAG(Z, append_in_aaa)
    The graph contains the following edges 1 > 1

  • U1_AAG(Z, append_out_aaa) → FL_IN_AAG(Z)
    The graph contains the following edges 1 >= 1

(40) TRUE