(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(g,a,a).

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

fl_in_gaa([], [], 0) → fl_out_gaa([], [], 0)
fl_in_gaa(.(E, X), R, s(Z)) → U1_gaa(E, X, R, Z, append_in_gaa(E, Y, R))
append_in_gaa([], X, X) → append_out_gaa([], X, X)
append_in_gaa(.(X, Xs), Ys, .(X, Zs)) → U3_gaa(X, Xs, Ys, Zs, append_in_gaa(Xs, Ys, Zs))
U3_gaa(X, Xs, Ys, Zs, append_out_gaa(Xs, Ys, Zs)) → append_out_gaa(.(X, Xs), Ys, .(X, Zs))
U1_gaa(E, X, R, Z, append_out_gaa(E, Y, R)) → U2_gaa(E, X, R, Z, fl_in_gaa(X, Y, Z))
U2_gaa(E, X, R, Z, fl_out_gaa(X, Y, Z)) → fl_out_gaa(.(E, X), R, s(Z))

The argument filtering Pi contains the following mapping:
fl_in_gaa(x1, x2, x3)  =  fl_in_gaa(x1)
[]  =  []
fl_out_gaa(x1, x2, x3)  =  fl_out_gaa(x1, x3)
.(x1, x2)  =  .(x1, x2)
U1_gaa(x1, x2, x3, x4, x5)  =  U1_gaa(x1, x2, x5)
append_in_gaa(x1, x2, x3)  =  append_in_gaa(x1)
append_out_gaa(x1, x2, x3)  =  append_out_gaa(x1)
U3_gaa(x1, x2, x3, x4, x5)  =  U3_gaa(x1, x2, x5)
U2_gaa(x1, x2, x3, x4, x5)  =  U2_gaa(x1, x2, x5)
s(x1)  =  s(x1)

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_gaa([], [], 0) → fl_out_gaa([], [], 0)
fl_in_gaa(.(E, X), R, s(Z)) → U1_gaa(E, X, R, Z, append_in_gaa(E, Y, R))
append_in_gaa([], X, X) → append_out_gaa([], X, X)
append_in_gaa(.(X, Xs), Ys, .(X, Zs)) → U3_gaa(X, Xs, Ys, Zs, append_in_gaa(Xs, Ys, Zs))
U3_gaa(X, Xs, Ys, Zs, append_out_gaa(Xs, Ys, Zs)) → append_out_gaa(.(X, Xs), Ys, .(X, Zs))
U1_gaa(E, X, R, Z, append_out_gaa(E, Y, R)) → U2_gaa(E, X, R, Z, fl_in_gaa(X, Y, Z))
U2_gaa(E, X, R, Z, fl_out_gaa(X, Y, Z)) → fl_out_gaa(.(E, X), R, s(Z))

The argument filtering Pi contains the following mapping:
fl_in_gaa(x1, x2, x3)  =  fl_in_gaa(x1)
[]  =  []
fl_out_gaa(x1, x2, x3)  =  fl_out_gaa(x1, x3)
.(x1, x2)  =  .(x1, x2)
U1_gaa(x1, x2, x3, x4, x5)  =  U1_gaa(x1, x2, x5)
append_in_gaa(x1, x2, x3)  =  append_in_gaa(x1)
append_out_gaa(x1, x2, x3)  =  append_out_gaa(x1)
U3_gaa(x1, x2, x3, x4, x5)  =  U3_gaa(x1, x2, x5)
U2_gaa(x1, x2, x3, x4, x5)  =  U2_gaa(x1, x2, x5)
s(x1)  =  s(x1)

(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_GAA(.(E, X), R, s(Z)) → U1_GAA(E, X, R, Z, append_in_gaa(E, Y, R))
FL_IN_GAA(.(E, X), R, s(Z)) → APPEND_IN_GAA(E, Y, R)
APPEND_IN_GAA(.(X, Xs), Ys, .(X, Zs)) → U3_GAA(X, Xs, Ys, Zs, append_in_gaa(Xs, Ys, Zs))
APPEND_IN_GAA(.(X, Xs), Ys, .(X, Zs)) → APPEND_IN_GAA(Xs, Ys, Zs)
U1_GAA(E, X, R, Z, append_out_gaa(E, Y, R)) → U2_GAA(E, X, R, Z, fl_in_gaa(X, Y, Z))
U1_GAA(E, X, R, Z, append_out_gaa(E, Y, R)) → FL_IN_GAA(X, Y, Z)

The TRS R consists of the following rules:

fl_in_gaa([], [], 0) → fl_out_gaa([], [], 0)
fl_in_gaa(.(E, X), R, s(Z)) → U1_gaa(E, X, R, Z, append_in_gaa(E, Y, R))
append_in_gaa([], X, X) → append_out_gaa([], X, X)
append_in_gaa(.(X, Xs), Ys, .(X, Zs)) → U3_gaa(X, Xs, Ys, Zs, append_in_gaa(Xs, Ys, Zs))
U3_gaa(X, Xs, Ys, Zs, append_out_gaa(Xs, Ys, Zs)) → append_out_gaa(.(X, Xs), Ys, .(X, Zs))
U1_gaa(E, X, R, Z, append_out_gaa(E, Y, R)) → U2_gaa(E, X, R, Z, fl_in_gaa(X, Y, Z))
U2_gaa(E, X, R, Z, fl_out_gaa(X, Y, Z)) → fl_out_gaa(.(E, X), R, s(Z))

The argument filtering Pi contains the following mapping:
fl_in_gaa(x1, x2, x3)  =  fl_in_gaa(x1)
[]  =  []
fl_out_gaa(x1, x2, x3)  =  fl_out_gaa(x1, x3)
.(x1, x2)  =  .(x1, x2)
U1_gaa(x1, x2, x3, x4, x5)  =  U1_gaa(x1, x2, x5)
append_in_gaa(x1, x2, x3)  =  append_in_gaa(x1)
append_out_gaa(x1, x2, x3)  =  append_out_gaa(x1)
U3_gaa(x1, x2, x3, x4, x5)  =  U3_gaa(x1, x2, x5)
U2_gaa(x1, x2, x3, x4, x5)  =  U2_gaa(x1, x2, x5)
s(x1)  =  s(x1)
FL_IN_GAA(x1, x2, x3)  =  FL_IN_GAA(x1)
U1_GAA(x1, x2, x3, x4, x5)  =  U1_GAA(x1, x2, x5)
APPEND_IN_GAA(x1, x2, x3)  =  APPEND_IN_GAA(x1)
U3_GAA(x1, x2, x3, x4, x5)  =  U3_GAA(x1, x2, x5)
U2_GAA(x1, x2, x3, x4, x5)  =  U2_GAA(x1, x2, 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_GAA(.(E, X), R, s(Z)) → U1_GAA(E, X, R, Z, append_in_gaa(E, Y, R))
FL_IN_GAA(.(E, X), R, s(Z)) → APPEND_IN_GAA(E, Y, R)
APPEND_IN_GAA(.(X, Xs), Ys, .(X, Zs)) → U3_GAA(X, Xs, Ys, Zs, append_in_gaa(Xs, Ys, Zs))
APPEND_IN_GAA(.(X, Xs), Ys, .(X, Zs)) → APPEND_IN_GAA(Xs, Ys, Zs)
U1_GAA(E, X, R, Z, append_out_gaa(E, Y, R)) → U2_GAA(E, X, R, Z, fl_in_gaa(X, Y, Z))
U1_GAA(E, X, R, Z, append_out_gaa(E, Y, R)) → FL_IN_GAA(X, Y, Z)

The TRS R consists of the following rules:

fl_in_gaa([], [], 0) → fl_out_gaa([], [], 0)
fl_in_gaa(.(E, X), R, s(Z)) → U1_gaa(E, X, R, Z, append_in_gaa(E, Y, R))
append_in_gaa([], X, X) → append_out_gaa([], X, X)
append_in_gaa(.(X, Xs), Ys, .(X, Zs)) → U3_gaa(X, Xs, Ys, Zs, append_in_gaa(Xs, Ys, Zs))
U3_gaa(X, Xs, Ys, Zs, append_out_gaa(Xs, Ys, Zs)) → append_out_gaa(.(X, Xs), Ys, .(X, Zs))
U1_gaa(E, X, R, Z, append_out_gaa(E, Y, R)) → U2_gaa(E, X, R, Z, fl_in_gaa(X, Y, Z))
U2_gaa(E, X, R, Z, fl_out_gaa(X, Y, Z)) → fl_out_gaa(.(E, X), R, s(Z))

The argument filtering Pi contains the following mapping:
fl_in_gaa(x1, x2, x3)  =  fl_in_gaa(x1)
[]  =  []
fl_out_gaa(x1, x2, x3)  =  fl_out_gaa(x1, x3)
.(x1, x2)  =  .(x1, x2)
U1_gaa(x1, x2, x3, x4, x5)  =  U1_gaa(x1, x2, x5)
append_in_gaa(x1, x2, x3)  =  append_in_gaa(x1)
append_out_gaa(x1, x2, x3)  =  append_out_gaa(x1)
U3_gaa(x1, x2, x3, x4, x5)  =  U3_gaa(x1, x2, x5)
U2_gaa(x1, x2, x3, x4, x5)  =  U2_gaa(x1, x2, x5)
s(x1)  =  s(x1)
FL_IN_GAA(x1, x2, x3)  =  FL_IN_GAA(x1)
U1_GAA(x1, x2, x3, x4, x5)  =  U1_GAA(x1, x2, x5)
APPEND_IN_GAA(x1, x2, x3)  =  APPEND_IN_GAA(x1)
U3_GAA(x1, x2, x3, x4, x5)  =  U3_GAA(x1, x2, x5)
U2_GAA(x1, x2, x3, x4, x5)  =  U2_GAA(x1, x2, 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_GAA(.(X, Xs), Ys, .(X, Zs)) → APPEND_IN_GAA(Xs, Ys, Zs)

The TRS R consists of the following rules:

fl_in_gaa([], [], 0) → fl_out_gaa([], [], 0)
fl_in_gaa(.(E, X), R, s(Z)) → U1_gaa(E, X, R, Z, append_in_gaa(E, Y, R))
append_in_gaa([], X, X) → append_out_gaa([], X, X)
append_in_gaa(.(X, Xs), Ys, .(X, Zs)) → U3_gaa(X, Xs, Ys, Zs, append_in_gaa(Xs, Ys, Zs))
U3_gaa(X, Xs, Ys, Zs, append_out_gaa(Xs, Ys, Zs)) → append_out_gaa(.(X, Xs), Ys, .(X, Zs))
U1_gaa(E, X, R, Z, append_out_gaa(E, Y, R)) → U2_gaa(E, X, R, Z, fl_in_gaa(X, Y, Z))
U2_gaa(E, X, R, Z, fl_out_gaa(X, Y, Z)) → fl_out_gaa(.(E, X), R, s(Z))

The argument filtering Pi contains the following mapping:
fl_in_gaa(x1, x2, x3)  =  fl_in_gaa(x1)
[]  =  []
fl_out_gaa(x1, x2, x3)  =  fl_out_gaa(x1, x3)
.(x1, x2)  =  .(x1, x2)
U1_gaa(x1, x2, x3, x4, x5)  =  U1_gaa(x1, x2, x5)
append_in_gaa(x1, x2, x3)  =  append_in_gaa(x1)
append_out_gaa(x1, x2, x3)  =  append_out_gaa(x1)
U3_gaa(x1, x2, x3, x4, x5)  =  U3_gaa(x1, x2, x5)
U2_gaa(x1, x2, x3, x4, x5)  =  U2_gaa(x1, x2, x5)
s(x1)  =  s(x1)
APPEND_IN_GAA(x1, x2, x3)  =  APPEND_IN_GAA(x1)

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_GAA(.(X, Xs), Ys, .(X, Zs)) → APPEND_IN_GAA(Xs, Ys, Zs)

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

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_GAA(.(X, Xs)) → APPEND_IN_GAA(Xs)

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

(12) 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:

  • APPEND_IN_GAA(.(X, Xs)) → APPEND_IN_GAA(Xs)
    The graph contains the following edges 1 > 1

(13) TRUE

(14) Obligation:

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

U1_GAA(E, X, R, Z, append_out_gaa(E, Y, R)) → FL_IN_GAA(X, Y, Z)
FL_IN_GAA(.(E, X), R, s(Z)) → U1_GAA(E, X, R, Z, append_in_gaa(E, Y, R))

The TRS R consists of the following rules:

fl_in_gaa([], [], 0) → fl_out_gaa([], [], 0)
fl_in_gaa(.(E, X), R, s(Z)) → U1_gaa(E, X, R, Z, append_in_gaa(E, Y, R))
append_in_gaa([], X, X) → append_out_gaa([], X, X)
append_in_gaa(.(X, Xs), Ys, .(X, Zs)) → U3_gaa(X, Xs, Ys, Zs, append_in_gaa(Xs, Ys, Zs))
U3_gaa(X, Xs, Ys, Zs, append_out_gaa(Xs, Ys, Zs)) → append_out_gaa(.(X, Xs), Ys, .(X, Zs))
U1_gaa(E, X, R, Z, append_out_gaa(E, Y, R)) → U2_gaa(E, X, R, Z, fl_in_gaa(X, Y, Z))
U2_gaa(E, X, R, Z, fl_out_gaa(X, Y, Z)) → fl_out_gaa(.(E, X), R, s(Z))

The argument filtering Pi contains the following mapping:
fl_in_gaa(x1, x2, x3)  =  fl_in_gaa(x1)
[]  =  []
fl_out_gaa(x1, x2, x3)  =  fl_out_gaa(x1, x3)
.(x1, x2)  =  .(x1, x2)
U1_gaa(x1, x2, x3, x4, x5)  =  U1_gaa(x1, x2, x5)
append_in_gaa(x1, x2, x3)  =  append_in_gaa(x1)
append_out_gaa(x1, x2, x3)  =  append_out_gaa(x1)
U3_gaa(x1, x2, x3, x4, x5)  =  U3_gaa(x1, x2, x5)
U2_gaa(x1, x2, x3, x4, x5)  =  U2_gaa(x1, x2, x5)
s(x1)  =  s(x1)
FL_IN_GAA(x1, x2, x3)  =  FL_IN_GAA(x1)
U1_GAA(x1, x2, x3, x4, x5)  =  U1_GAA(x1, x2, 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_GAA(E, X, R, Z, append_out_gaa(E, Y, R)) → FL_IN_GAA(X, Y, Z)
FL_IN_GAA(.(E, X), R, s(Z)) → U1_GAA(E, X, R, Z, append_in_gaa(E, Y, R))

The TRS R consists of the following rules:

append_in_gaa([], X, X) → append_out_gaa([], X, X)
append_in_gaa(.(X, Xs), Ys, .(X, Zs)) → U3_gaa(X, Xs, Ys, Zs, append_in_gaa(Xs, Ys, Zs))
U3_gaa(X, Xs, Ys, Zs, append_out_gaa(Xs, Ys, Zs)) → append_out_gaa(.(X, Xs), Ys, .(X, Zs))

The argument filtering Pi contains the following mapping:
[]  =  []
.(x1, x2)  =  .(x1, x2)
append_in_gaa(x1, x2, x3)  =  append_in_gaa(x1)
append_out_gaa(x1, x2, x3)  =  append_out_gaa(x1)
U3_gaa(x1, x2, x3, x4, x5)  =  U3_gaa(x1, x2, x5)
s(x1)  =  s(x1)
FL_IN_GAA(x1, x2, x3)  =  FL_IN_GAA(x1)
U1_GAA(x1, x2, x3, x4, x5)  =  U1_GAA(x1, x2, 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_GAA(E, X, append_out_gaa(E)) → FL_IN_GAA(X)
FL_IN_GAA(.(E, X)) → U1_GAA(E, X, append_in_gaa(E))

The TRS R consists of the following rules:

append_in_gaa([]) → append_out_gaa([])
append_in_gaa(.(X, Xs)) → U3_gaa(X, Xs, append_in_gaa(Xs))
U3_gaa(X, Xs, append_out_gaa(Xs)) → append_out_gaa(.(X, Xs))

The set Q consists of the following terms:

append_in_gaa(x0)
U3_gaa(x0, x1, x2)

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

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

fl_in_gaa([], [], 0) → fl_out_gaa([], [], 0)
fl_in_gaa(.(E, X), R, s(Z)) → U1_gaa(E, X, R, Z, append_in_gaa(E, Y, R))
append_in_gaa([], X, X) → append_out_gaa([], X, X)
append_in_gaa(.(X, Xs), Ys, .(X, Zs)) → U3_gaa(X, Xs, Ys, Zs, append_in_gaa(Xs, Ys, Zs))
U3_gaa(X, Xs, Ys, Zs, append_out_gaa(Xs, Ys, Zs)) → append_out_gaa(.(X, Xs), Ys, .(X, Zs))
U1_gaa(E, X, R, Z, append_out_gaa(E, Y, R)) → U2_gaa(E, X, R, Z, fl_in_gaa(X, Y, Z))
U2_gaa(E, X, R, Z, fl_out_gaa(X, Y, Z)) → fl_out_gaa(.(E, X), R, s(Z))

The argument filtering Pi contains the following mapping:
fl_in_gaa(x1, x2, x3)  =  fl_in_gaa(x1)
[]  =  []
fl_out_gaa(x1, x2, x3)  =  fl_out_gaa(x3)
.(x1, x2)  =  .(x1, x2)
U1_gaa(x1, x2, x3, x4, x5)  =  U1_gaa(x2, x5)
append_in_gaa(x1, x2, x3)  =  append_in_gaa(x1)
append_out_gaa(x1, x2, x3)  =  append_out_gaa
U3_gaa(x1, x2, x3, x4, x5)  =  U3_gaa(x5)
U2_gaa(x1, x2, x3, x4, x5)  =  U2_gaa(x5)
s(x1)  =  s(x1)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(20) Obligation:

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

fl_in_gaa([], [], 0) → fl_out_gaa([], [], 0)
fl_in_gaa(.(E, X), R, s(Z)) → U1_gaa(E, X, R, Z, append_in_gaa(E, Y, R))
append_in_gaa([], X, X) → append_out_gaa([], X, X)
append_in_gaa(.(X, Xs), Ys, .(X, Zs)) → U3_gaa(X, Xs, Ys, Zs, append_in_gaa(Xs, Ys, Zs))
U3_gaa(X, Xs, Ys, Zs, append_out_gaa(Xs, Ys, Zs)) → append_out_gaa(.(X, Xs), Ys, .(X, Zs))
U1_gaa(E, X, R, Z, append_out_gaa(E, Y, R)) → U2_gaa(E, X, R, Z, fl_in_gaa(X, Y, Z))
U2_gaa(E, X, R, Z, fl_out_gaa(X, Y, Z)) → fl_out_gaa(.(E, X), R, s(Z))

The argument filtering Pi contains the following mapping:
fl_in_gaa(x1, x2, x3)  =  fl_in_gaa(x1)
[]  =  []
fl_out_gaa(x1, x2, x3)  =  fl_out_gaa(x3)
.(x1, x2)  =  .(x1, x2)
U1_gaa(x1, x2, x3, x4, x5)  =  U1_gaa(x2, x5)
append_in_gaa(x1, x2, x3)  =  append_in_gaa(x1)
append_out_gaa(x1, x2, x3)  =  append_out_gaa
U3_gaa(x1, x2, x3, x4, x5)  =  U3_gaa(x5)
U2_gaa(x1, x2, x3, x4, x5)  =  U2_gaa(x5)
s(x1)  =  s(x1)

(21) 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_GAA(.(E, X), R, s(Z)) → U1_GAA(E, X, R, Z, append_in_gaa(E, Y, R))
FL_IN_GAA(.(E, X), R, s(Z)) → APPEND_IN_GAA(E, Y, R)
APPEND_IN_GAA(.(X, Xs), Ys, .(X, Zs)) → U3_GAA(X, Xs, Ys, Zs, append_in_gaa(Xs, Ys, Zs))
APPEND_IN_GAA(.(X, Xs), Ys, .(X, Zs)) → APPEND_IN_GAA(Xs, Ys, Zs)
U1_GAA(E, X, R, Z, append_out_gaa(E, Y, R)) → U2_GAA(E, X, R, Z, fl_in_gaa(X, Y, Z))
U1_GAA(E, X, R, Z, append_out_gaa(E, Y, R)) → FL_IN_GAA(X, Y, Z)

The TRS R consists of the following rules:

fl_in_gaa([], [], 0) → fl_out_gaa([], [], 0)
fl_in_gaa(.(E, X), R, s(Z)) → U1_gaa(E, X, R, Z, append_in_gaa(E, Y, R))
append_in_gaa([], X, X) → append_out_gaa([], X, X)
append_in_gaa(.(X, Xs), Ys, .(X, Zs)) → U3_gaa(X, Xs, Ys, Zs, append_in_gaa(Xs, Ys, Zs))
U3_gaa(X, Xs, Ys, Zs, append_out_gaa(Xs, Ys, Zs)) → append_out_gaa(.(X, Xs), Ys, .(X, Zs))
U1_gaa(E, X, R, Z, append_out_gaa(E, Y, R)) → U2_gaa(E, X, R, Z, fl_in_gaa(X, Y, Z))
U2_gaa(E, X, R, Z, fl_out_gaa(X, Y, Z)) → fl_out_gaa(.(E, X), R, s(Z))

The argument filtering Pi contains the following mapping:
fl_in_gaa(x1, x2, x3)  =  fl_in_gaa(x1)
[]  =  []
fl_out_gaa(x1, x2, x3)  =  fl_out_gaa(x3)
.(x1, x2)  =  .(x1, x2)
U1_gaa(x1, x2, x3, x4, x5)  =  U1_gaa(x2, x5)
append_in_gaa(x1, x2, x3)  =  append_in_gaa(x1)
append_out_gaa(x1, x2, x3)  =  append_out_gaa
U3_gaa(x1, x2, x3, x4, x5)  =  U3_gaa(x5)
U2_gaa(x1, x2, x3, x4, x5)  =  U2_gaa(x5)
s(x1)  =  s(x1)
FL_IN_GAA(x1, x2, x3)  =  FL_IN_GAA(x1)
U1_GAA(x1, x2, x3, x4, x5)  =  U1_GAA(x2, x5)
APPEND_IN_GAA(x1, x2, x3)  =  APPEND_IN_GAA(x1)
U3_GAA(x1, x2, x3, x4, x5)  =  U3_GAA(x5)
U2_GAA(x1, x2, x3, x4, x5)  =  U2_GAA(x5)

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

(22) Obligation:

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

FL_IN_GAA(.(E, X), R, s(Z)) → U1_GAA(E, X, R, Z, append_in_gaa(E, Y, R))
FL_IN_GAA(.(E, X), R, s(Z)) → APPEND_IN_GAA(E, Y, R)
APPEND_IN_GAA(.(X, Xs), Ys, .(X, Zs)) → U3_GAA(X, Xs, Ys, Zs, append_in_gaa(Xs, Ys, Zs))
APPEND_IN_GAA(.(X, Xs), Ys, .(X, Zs)) → APPEND_IN_GAA(Xs, Ys, Zs)
U1_GAA(E, X, R, Z, append_out_gaa(E, Y, R)) → U2_GAA(E, X, R, Z, fl_in_gaa(X, Y, Z))
U1_GAA(E, X, R, Z, append_out_gaa(E, Y, R)) → FL_IN_GAA(X, Y, Z)

The TRS R consists of the following rules:

fl_in_gaa([], [], 0) → fl_out_gaa([], [], 0)
fl_in_gaa(.(E, X), R, s(Z)) → U1_gaa(E, X, R, Z, append_in_gaa(E, Y, R))
append_in_gaa([], X, X) → append_out_gaa([], X, X)
append_in_gaa(.(X, Xs), Ys, .(X, Zs)) → U3_gaa(X, Xs, Ys, Zs, append_in_gaa(Xs, Ys, Zs))
U3_gaa(X, Xs, Ys, Zs, append_out_gaa(Xs, Ys, Zs)) → append_out_gaa(.(X, Xs), Ys, .(X, Zs))
U1_gaa(E, X, R, Z, append_out_gaa(E, Y, R)) → U2_gaa(E, X, R, Z, fl_in_gaa(X, Y, Z))
U2_gaa(E, X, R, Z, fl_out_gaa(X, Y, Z)) → fl_out_gaa(.(E, X), R, s(Z))

The argument filtering Pi contains the following mapping:
fl_in_gaa(x1, x2, x3)  =  fl_in_gaa(x1)
[]  =  []
fl_out_gaa(x1, x2, x3)  =  fl_out_gaa(x3)
.(x1, x2)  =  .(x1, x2)
U1_gaa(x1, x2, x3, x4, x5)  =  U1_gaa(x2, x5)
append_in_gaa(x1, x2, x3)  =  append_in_gaa(x1)
append_out_gaa(x1, x2, x3)  =  append_out_gaa
U3_gaa(x1, x2, x3, x4, x5)  =  U3_gaa(x5)
U2_gaa(x1, x2, x3, x4, x5)  =  U2_gaa(x5)
s(x1)  =  s(x1)
FL_IN_GAA(x1, x2, x3)  =  FL_IN_GAA(x1)
U1_GAA(x1, x2, x3, x4, x5)  =  U1_GAA(x2, x5)
APPEND_IN_GAA(x1, x2, x3)  =  APPEND_IN_GAA(x1)
U3_GAA(x1, x2, x3, x4, x5)  =  U3_GAA(x5)
U2_GAA(x1, x2, x3, x4, x5)  =  U2_GAA(x5)

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

(23) DependencyGraphProof (EQUIVALENT transformation)

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

(24) Complex Obligation (AND)

(25) Obligation:

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

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

The TRS R consists of the following rules:

fl_in_gaa([], [], 0) → fl_out_gaa([], [], 0)
fl_in_gaa(.(E, X), R, s(Z)) → U1_gaa(E, X, R, Z, append_in_gaa(E, Y, R))
append_in_gaa([], X, X) → append_out_gaa([], X, X)
append_in_gaa(.(X, Xs), Ys, .(X, Zs)) → U3_gaa(X, Xs, Ys, Zs, append_in_gaa(Xs, Ys, Zs))
U3_gaa(X, Xs, Ys, Zs, append_out_gaa(Xs, Ys, Zs)) → append_out_gaa(.(X, Xs), Ys, .(X, Zs))
U1_gaa(E, X, R, Z, append_out_gaa(E, Y, R)) → U2_gaa(E, X, R, Z, fl_in_gaa(X, Y, Z))
U2_gaa(E, X, R, Z, fl_out_gaa(X, Y, Z)) → fl_out_gaa(.(E, X), R, s(Z))

The argument filtering Pi contains the following mapping:
fl_in_gaa(x1, x2, x3)  =  fl_in_gaa(x1)
[]  =  []
fl_out_gaa(x1, x2, x3)  =  fl_out_gaa(x3)
.(x1, x2)  =  .(x1, x2)
U1_gaa(x1, x2, x3, x4, x5)  =  U1_gaa(x2, x5)
append_in_gaa(x1, x2, x3)  =  append_in_gaa(x1)
append_out_gaa(x1, x2, x3)  =  append_out_gaa
U3_gaa(x1, x2, x3, x4, x5)  =  U3_gaa(x5)
U2_gaa(x1, x2, x3, x4, x5)  =  U2_gaa(x5)
s(x1)  =  s(x1)
APPEND_IN_GAA(x1, x2, x3)  =  APPEND_IN_GAA(x1)

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

(26) UsableRulesProof (EQUIVALENT transformation)

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

(27) Obligation:

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

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

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

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

(28) PiDPToQDPProof (SOUND transformation)

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

(29) Obligation:

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

APPEND_IN_GAA(.(X, Xs)) → APPEND_IN_GAA(Xs)

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

(30) 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:

  • APPEND_IN_GAA(.(X, Xs)) → APPEND_IN_GAA(Xs)
    The graph contains the following edges 1 > 1

(31) TRUE

(32) Obligation:

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

U1_GAA(E, X, R, Z, append_out_gaa(E, Y, R)) → FL_IN_GAA(X, Y, Z)
FL_IN_GAA(.(E, X), R, s(Z)) → U1_GAA(E, X, R, Z, append_in_gaa(E, Y, R))

The TRS R consists of the following rules:

fl_in_gaa([], [], 0) → fl_out_gaa([], [], 0)
fl_in_gaa(.(E, X), R, s(Z)) → U1_gaa(E, X, R, Z, append_in_gaa(E, Y, R))
append_in_gaa([], X, X) → append_out_gaa([], X, X)
append_in_gaa(.(X, Xs), Ys, .(X, Zs)) → U3_gaa(X, Xs, Ys, Zs, append_in_gaa(Xs, Ys, Zs))
U3_gaa(X, Xs, Ys, Zs, append_out_gaa(Xs, Ys, Zs)) → append_out_gaa(.(X, Xs), Ys, .(X, Zs))
U1_gaa(E, X, R, Z, append_out_gaa(E, Y, R)) → U2_gaa(E, X, R, Z, fl_in_gaa(X, Y, Z))
U2_gaa(E, X, R, Z, fl_out_gaa(X, Y, Z)) → fl_out_gaa(.(E, X), R, s(Z))

The argument filtering Pi contains the following mapping:
fl_in_gaa(x1, x2, x3)  =  fl_in_gaa(x1)
[]  =  []
fl_out_gaa(x1, x2, x3)  =  fl_out_gaa(x3)
.(x1, x2)  =  .(x1, x2)
U1_gaa(x1, x2, x3, x4, x5)  =  U1_gaa(x2, x5)
append_in_gaa(x1, x2, x3)  =  append_in_gaa(x1)
append_out_gaa(x1, x2, x3)  =  append_out_gaa
U3_gaa(x1, x2, x3, x4, x5)  =  U3_gaa(x5)
U2_gaa(x1, x2, x3, x4, x5)  =  U2_gaa(x5)
s(x1)  =  s(x1)
FL_IN_GAA(x1, x2, x3)  =  FL_IN_GAA(x1)
U1_GAA(x1, x2, x3, x4, x5)  =  U1_GAA(x2, x5)

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

(33) UsableRulesProof (EQUIVALENT transformation)

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

(34) Obligation:

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

U1_GAA(E, X, R, Z, append_out_gaa(E, Y, R)) → FL_IN_GAA(X, Y, Z)
FL_IN_GAA(.(E, X), R, s(Z)) → U1_GAA(E, X, R, Z, append_in_gaa(E, Y, R))

The TRS R consists of the following rules:

append_in_gaa([], X, X) → append_out_gaa([], X, X)
append_in_gaa(.(X, Xs), Ys, .(X, Zs)) → U3_gaa(X, Xs, Ys, Zs, append_in_gaa(Xs, Ys, Zs))
U3_gaa(X, Xs, Ys, Zs, append_out_gaa(Xs, Ys, Zs)) → append_out_gaa(.(X, Xs), Ys, .(X, Zs))

The argument filtering Pi contains the following mapping:
[]  =  []
.(x1, x2)  =  .(x1, x2)
append_in_gaa(x1, x2, x3)  =  append_in_gaa(x1)
append_out_gaa(x1, x2, x3)  =  append_out_gaa
U3_gaa(x1, x2, x3, x4, x5)  =  U3_gaa(x5)
s(x1)  =  s(x1)
FL_IN_GAA(x1, x2, x3)  =  FL_IN_GAA(x1)
U1_GAA(x1, x2, x3, x4, x5)  =  U1_GAA(x2, x5)

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

(35) PiDPToQDPProof (SOUND transformation)

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

(36) Obligation:

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

U1_GAA(X, append_out_gaa) → FL_IN_GAA(X)
FL_IN_GAA(.(E, X)) → U1_GAA(X, append_in_gaa(E))

The TRS R consists of the following rules:

append_in_gaa([]) → append_out_gaa
append_in_gaa(.(X, Xs)) → U3_gaa(append_in_gaa(Xs))
U3_gaa(append_out_gaa) → append_out_gaa

The set Q consists of the following terms:

append_in_gaa(x0)
U3_gaa(x0)

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

(37) 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_GAA(.(E, X)) → U1_GAA(X, append_in_gaa(E))
    The graph contains the following edges 1 > 1

  • U1_GAA(X, append_out_gaa) → FL_IN_GAA(X)
    The graph contains the following edges 1 >= 1

(38) TRUE