Left Termination of the query pattern conf_in_1(g) w.r.t. the given Prolog program could successfully be proven:

(0) Obligation:

Clauses:

conf(X) :- ','(del2(X, Z), ','(del(U, Y, Z), conf(Y))).
del2(X, Y) :- ','(del(U, X, Z), del(V, Z, Y)).
del(X, cons(X, T), T).
del(X, cons(H, T), cons(H, T1)) :- del(X, T, T1).
s2l(s(X), cons(Y, Xs)) :- s2l(X, Xs).
s2l(0, nil).
goal(X) :- ','(s2l(X, XS), conf(XS)).

Queries:

conf(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:
conf_in: (b)
del2_in: (b,f)
del_in: (f,b,f) (f,f,b)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

conf_in_g(X) → U1_g(X, del2_in_ga(X, Z))
del2_in_ga(X, Y) → U4_ga(X, Y, del_in_aga(U, X, Z))
del_in_aga(X, cons(X, T), T) → del_out_aga(X, cons(X, T), T)
del_in_aga(X, cons(H, T), cons(H, T1)) → U6_aga(X, H, T, T1, del_in_aga(X, T, T1))
U6_aga(X, H, T, T1, del_out_aga(X, T, T1)) → del_out_aga(X, cons(H, T), cons(H, T1))
U4_ga(X, Y, del_out_aga(U, X, Z)) → U5_ga(X, Y, del_in_aga(V, Z, Y))
U5_ga(X, Y, del_out_aga(V, Z, Y)) → del2_out_ga(X, Y)
U1_g(X, del2_out_ga(X, Z)) → U2_g(X, del_in_aag(U, Y, Z))
del_in_aag(X, cons(X, T), T) → del_out_aag(X, cons(X, T), T)
del_in_aag(X, cons(H, T), cons(H, T1)) → U6_aag(X, H, T, T1, del_in_aag(X, T, T1))
U6_aag(X, H, T, T1, del_out_aag(X, T, T1)) → del_out_aag(X, cons(H, T), cons(H, T1))
U2_g(X, del_out_aag(U, Y, Z)) → U3_g(X, conf_in_g(Y))
U3_g(X, conf_out_g(Y)) → conf_out_g(X)

The argument filtering Pi contains the following mapping:
conf_in_g(x1)  =  conf_in_g(x1)
U1_g(x1, x2)  =  U1_g(x1, x2)
del2_in_ga(x1, x2)  =  del2_in_ga(x1)
U4_ga(x1, x2, x3)  =  U4_ga(x1, x3)
del_in_aga(x1, x2, x3)  =  del_in_aga(x2)
cons(x1, x2)  =  cons(x2)
del_out_aga(x1, x2, x3)  =  del_out_aga(x2, x3)
U6_aga(x1, x2, x3, x4, x5)  =  U6_aga(x3, x5)
U5_ga(x1, x2, x3)  =  U5_ga(x1, x3)
del2_out_ga(x1, x2)  =  del2_out_ga(x1, x2)
U2_g(x1, x2)  =  U2_g(x1, x2)
del_in_aag(x1, x2, x3)  =  del_in_aag(x3)
del_out_aag(x1, x2, x3)  =  del_out_aag(x2, x3)
U6_aag(x1, x2, x3, x4, x5)  =  U6_aag(x4, x5)
U3_g(x1, x2)  =  U3_g(x1, x2)
conf_out_g(x1)  =  conf_out_g(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:

conf_in_g(X) → U1_g(X, del2_in_ga(X, Z))
del2_in_ga(X, Y) → U4_ga(X, Y, del_in_aga(U, X, Z))
del_in_aga(X, cons(X, T), T) → del_out_aga(X, cons(X, T), T)
del_in_aga(X, cons(H, T), cons(H, T1)) → U6_aga(X, H, T, T1, del_in_aga(X, T, T1))
U6_aga(X, H, T, T1, del_out_aga(X, T, T1)) → del_out_aga(X, cons(H, T), cons(H, T1))
U4_ga(X, Y, del_out_aga(U, X, Z)) → U5_ga(X, Y, del_in_aga(V, Z, Y))
U5_ga(X, Y, del_out_aga(V, Z, Y)) → del2_out_ga(X, Y)
U1_g(X, del2_out_ga(X, Z)) → U2_g(X, del_in_aag(U, Y, Z))
del_in_aag(X, cons(X, T), T) → del_out_aag(X, cons(X, T), T)
del_in_aag(X, cons(H, T), cons(H, T1)) → U6_aag(X, H, T, T1, del_in_aag(X, T, T1))
U6_aag(X, H, T, T1, del_out_aag(X, T, T1)) → del_out_aag(X, cons(H, T), cons(H, T1))
U2_g(X, del_out_aag(U, Y, Z)) → U3_g(X, conf_in_g(Y))
U3_g(X, conf_out_g(Y)) → conf_out_g(X)

The argument filtering Pi contains the following mapping:
conf_in_g(x1)  =  conf_in_g(x1)
U1_g(x1, x2)  =  U1_g(x1, x2)
del2_in_ga(x1, x2)  =  del2_in_ga(x1)
U4_ga(x1, x2, x3)  =  U4_ga(x1, x3)
del_in_aga(x1, x2, x3)  =  del_in_aga(x2)
cons(x1, x2)  =  cons(x2)
del_out_aga(x1, x2, x3)  =  del_out_aga(x2, x3)
U6_aga(x1, x2, x3, x4, x5)  =  U6_aga(x3, x5)
U5_ga(x1, x2, x3)  =  U5_ga(x1, x3)
del2_out_ga(x1, x2)  =  del2_out_ga(x1, x2)
U2_g(x1, x2)  =  U2_g(x1, x2)
del_in_aag(x1, x2, x3)  =  del_in_aag(x3)
del_out_aag(x1, x2, x3)  =  del_out_aag(x2, x3)
U6_aag(x1, x2, x3, x4, x5)  =  U6_aag(x4, x5)
U3_g(x1, x2)  =  U3_g(x1, x2)
conf_out_g(x1)  =  conf_out_g(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:

CONF_IN_G(X) → U1_G(X, del2_in_ga(X, Z))
CONF_IN_G(X) → DEL2_IN_GA(X, Z)
DEL2_IN_GA(X, Y) → U4_GA(X, Y, del_in_aga(U, X, Z))
DEL2_IN_GA(X, Y) → DEL_IN_AGA(U, X, Z)
DEL_IN_AGA(X, cons(H, T), cons(H, T1)) → U6_AGA(X, H, T, T1, del_in_aga(X, T, T1))
DEL_IN_AGA(X, cons(H, T), cons(H, T1)) → DEL_IN_AGA(X, T, T1)
U4_GA(X, Y, del_out_aga(U, X, Z)) → U5_GA(X, Y, del_in_aga(V, Z, Y))
U4_GA(X, Y, del_out_aga(U, X, Z)) → DEL_IN_AGA(V, Z, Y)
U1_G(X, del2_out_ga(X, Z)) → U2_G(X, del_in_aag(U, Y, Z))
U1_G(X, del2_out_ga(X, Z)) → DEL_IN_AAG(U, Y, Z)
DEL_IN_AAG(X, cons(H, T), cons(H, T1)) → U6_AAG(X, H, T, T1, del_in_aag(X, T, T1))
DEL_IN_AAG(X, cons(H, T), cons(H, T1)) → DEL_IN_AAG(X, T, T1)
U2_G(X, del_out_aag(U, Y, Z)) → U3_G(X, conf_in_g(Y))
U2_G(X, del_out_aag(U, Y, Z)) → CONF_IN_G(Y)

The TRS R consists of the following rules:

conf_in_g(X) → U1_g(X, del2_in_ga(X, Z))
del2_in_ga(X, Y) → U4_ga(X, Y, del_in_aga(U, X, Z))
del_in_aga(X, cons(X, T), T) → del_out_aga(X, cons(X, T), T)
del_in_aga(X, cons(H, T), cons(H, T1)) → U6_aga(X, H, T, T1, del_in_aga(X, T, T1))
U6_aga(X, H, T, T1, del_out_aga(X, T, T1)) → del_out_aga(X, cons(H, T), cons(H, T1))
U4_ga(X, Y, del_out_aga(U, X, Z)) → U5_ga(X, Y, del_in_aga(V, Z, Y))
U5_ga(X, Y, del_out_aga(V, Z, Y)) → del2_out_ga(X, Y)
U1_g(X, del2_out_ga(X, Z)) → U2_g(X, del_in_aag(U, Y, Z))
del_in_aag(X, cons(X, T), T) → del_out_aag(X, cons(X, T), T)
del_in_aag(X, cons(H, T), cons(H, T1)) → U6_aag(X, H, T, T1, del_in_aag(X, T, T1))
U6_aag(X, H, T, T1, del_out_aag(X, T, T1)) → del_out_aag(X, cons(H, T), cons(H, T1))
U2_g(X, del_out_aag(U, Y, Z)) → U3_g(X, conf_in_g(Y))
U3_g(X, conf_out_g(Y)) → conf_out_g(X)

The argument filtering Pi contains the following mapping:
conf_in_g(x1)  =  conf_in_g(x1)
U1_g(x1, x2)  =  U1_g(x1, x2)
del2_in_ga(x1, x2)  =  del2_in_ga(x1)
U4_ga(x1, x2, x3)  =  U4_ga(x1, x3)
del_in_aga(x1, x2, x3)  =  del_in_aga(x2)
cons(x1, x2)  =  cons(x2)
del_out_aga(x1, x2, x3)  =  del_out_aga(x2, x3)
U6_aga(x1, x2, x3, x4, x5)  =  U6_aga(x3, x5)
U5_ga(x1, x2, x3)  =  U5_ga(x1, x3)
del2_out_ga(x1, x2)  =  del2_out_ga(x1, x2)
U2_g(x1, x2)  =  U2_g(x1, x2)
del_in_aag(x1, x2, x3)  =  del_in_aag(x3)
del_out_aag(x1, x2, x3)  =  del_out_aag(x2, x3)
U6_aag(x1, x2, x3, x4, x5)  =  U6_aag(x4, x5)
U3_g(x1, x2)  =  U3_g(x1, x2)
conf_out_g(x1)  =  conf_out_g(x1)
CONF_IN_G(x1)  =  CONF_IN_G(x1)
U1_G(x1, x2)  =  U1_G(x1, x2)
DEL2_IN_GA(x1, x2)  =  DEL2_IN_GA(x1)
U4_GA(x1, x2, x3)  =  U4_GA(x1, x3)
DEL_IN_AGA(x1, x2, x3)  =  DEL_IN_AGA(x2)
U6_AGA(x1, x2, x3, x4, x5)  =  U6_AGA(x3, x5)
U5_GA(x1, x2, x3)  =  U5_GA(x1, x3)
U2_G(x1, x2)  =  U2_G(x1, x2)
DEL_IN_AAG(x1, x2, x3)  =  DEL_IN_AAG(x3)
U6_AAG(x1, x2, x3, x4, x5)  =  U6_AAG(x4, x5)
U3_G(x1, x2)  =  U3_G(x1, x2)

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

(4) Obligation:

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

CONF_IN_G(X) → U1_G(X, del2_in_ga(X, Z))
CONF_IN_G(X) → DEL2_IN_GA(X, Z)
DEL2_IN_GA(X, Y) → U4_GA(X, Y, del_in_aga(U, X, Z))
DEL2_IN_GA(X, Y) → DEL_IN_AGA(U, X, Z)
DEL_IN_AGA(X, cons(H, T), cons(H, T1)) → U6_AGA(X, H, T, T1, del_in_aga(X, T, T1))
DEL_IN_AGA(X, cons(H, T), cons(H, T1)) → DEL_IN_AGA(X, T, T1)
U4_GA(X, Y, del_out_aga(U, X, Z)) → U5_GA(X, Y, del_in_aga(V, Z, Y))
U4_GA(X, Y, del_out_aga(U, X, Z)) → DEL_IN_AGA(V, Z, Y)
U1_G(X, del2_out_ga(X, Z)) → U2_G(X, del_in_aag(U, Y, Z))
U1_G(X, del2_out_ga(X, Z)) → DEL_IN_AAG(U, Y, Z)
DEL_IN_AAG(X, cons(H, T), cons(H, T1)) → U6_AAG(X, H, T, T1, del_in_aag(X, T, T1))
DEL_IN_AAG(X, cons(H, T), cons(H, T1)) → DEL_IN_AAG(X, T, T1)
U2_G(X, del_out_aag(U, Y, Z)) → U3_G(X, conf_in_g(Y))
U2_G(X, del_out_aag(U, Y, Z)) → CONF_IN_G(Y)

The TRS R consists of the following rules:

conf_in_g(X) → U1_g(X, del2_in_ga(X, Z))
del2_in_ga(X, Y) → U4_ga(X, Y, del_in_aga(U, X, Z))
del_in_aga(X, cons(X, T), T) → del_out_aga(X, cons(X, T), T)
del_in_aga(X, cons(H, T), cons(H, T1)) → U6_aga(X, H, T, T1, del_in_aga(X, T, T1))
U6_aga(X, H, T, T1, del_out_aga(X, T, T1)) → del_out_aga(X, cons(H, T), cons(H, T1))
U4_ga(X, Y, del_out_aga(U, X, Z)) → U5_ga(X, Y, del_in_aga(V, Z, Y))
U5_ga(X, Y, del_out_aga(V, Z, Y)) → del2_out_ga(X, Y)
U1_g(X, del2_out_ga(X, Z)) → U2_g(X, del_in_aag(U, Y, Z))
del_in_aag(X, cons(X, T), T) → del_out_aag(X, cons(X, T), T)
del_in_aag(X, cons(H, T), cons(H, T1)) → U6_aag(X, H, T, T1, del_in_aag(X, T, T1))
U6_aag(X, H, T, T1, del_out_aag(X, T, T1)) → del_out_aag(X, cons(H, T), cons(H, T1))
U2_g(X, del_out_aag(U, Y, Z)) → U3_g(X, conf_in_g(Y))
U3_g(X, conf_out_g(Y)) → conf_out_g(X)

The argument filtering Pi contains the following mapping:
conf_in_g(x1)  =  conf_in_g(x1)
U1_g(x1, x2)  =  U1_g(x1, x2)
del2_in_ga(x1, x2)  =  del2_in_ga(x1)
U4_ga(x1, x2, x3)  =  U4_ga(x1, x3)
del_in_aga(x1, x2, x3)  =  del_in_aga(x2)
cons(x1, x2)  =  cons(x2)
del_out_aga(x1, x2, x3)  =  del_out_aga(x2, x3)
U6_aga(x1, x2, x3, x4, x5)  =  U6_aga(x3, x5)
U5_ga(x1, x2, x3)  =  U5_ga(x1, x3)
del2_out_ga(x1, x2)  =  del2_out_ga(x1, x2)
U2_g(x1, x2)  =  U2_g(x1, x2)
del_in_aag(x1, x2, x3)  =  del_in_aag(x3)
del_out_aag(x1, x2, x3)  =  del_out_aag(x2, x3)
U6_aag(x1, x2, x3, x4, x5)  =  U6_aag(x4, x5)
U3_g(x1, x2)  =  U3_g(x1, x2)
conf_out_g(x1)  =  conf_out_g(x1)
CONF_IN_G(x1)  =  CONF_IN_G(x1)
U1_G(x1, x2)  =  U1_G(x1, x2)
DEL2_IN_GA(x1, x2)  =  DEL2_IN_GA(x1)
U4_GA(x1, x2, x3)  =  U4_GA(x1, x3)
DEL_IN_AGA(x1, x2, x3)  =  DEL_IN_AGA(x2)
U6_AGA(x1, x2, x3, x4, x5)  =  U6_AGA(x3, x5)
U5_GA(x1, x2, x3)  =  U5_GA(x1, x3)
U2_G(x1, x2)  =  U2_G(x1, x2)
DEL_IN_AAG(x1, x2, x3)  =  DEL_IN_AAG(x3)
U6_AAG(x1, x2, x3, x4, x5)  =  U6_AAG(x4, x5)
U3_G(x1, x2)  =  U3_G(x1, x2)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 3 SCCs with 9 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

DEL_IN_AAG(X, cons(H, T), cons(H, T1)) → DEL_IN_AAG(X, T, T1)

The TRS R consists of the following rules:

conf_in_g(X) → U1_g(X, del2_in_ga(X, Z))
del2_in_ga(X, Y) → U4_ga(X, Y, del_in_aga(U, X, Z))
del_in_aga(X, cons(X, T), T) → del_out_aga(X, cons(X, T), T)
del_in_aga(X, cons(H, T), cons(H, T1)) → U6_aga(X, H, T, T1, del_in_aga(X, T, T1))
U6_aga(X, H, T, T1, del_out_aga(X, T, T1)) → del_out_aga(X, cons(H, T), cons(H, T1))
U4_ga(X, Y, del_out_aga(U, X, Z)) → U5_ga(X, Y, del_in_aga(V, Z, Y))
U5_ga(X, Y, del_out_aga(V, Z, Y)) → del2_out_ga(X, Y)
U1_g(X, del2_out_ga(X, Z)) → U2_g(X, del_in_aag(U, Y, Z))
del_in_aag(X, cons(X, T), T) → del_out_aag(X, cons(X, T), T)
del_in_aag(X, cons(H, T), cons(H, T1)) → U6_aag(X, H, T, T1, del_in_aag(X, T, T1))
U6_aag(X, H, T, T1, del_out_aag(X, T, T1)) → del_out_aag(X, cons(H, T), cons(H, T1))
U2_g(X, del_out_aag(U, Y, Z)) → U3_g(X, conf_in_g(Y))
U3_g(X, conf_out_g(Y)) → conf_out_g(X)

The argument filtering Pi contains the following mapping:
conf_in_g(x1)  =  conf_in_g(x1)
U1_g(x1, x2)  =  U1_g(x1, x2)
del2_in_ga(x1, x2)  =  del2_in_ga(x1)
U4_ga(x1, x2, x3)  =  U4_ga(x1, x3)
del_in_aga(x1, x2, x3)  =  del_in_aga(x2)
cons(x1, x2)  =  cons(x2)
del_out_aga(x1, x2, x3)  =  del_out_aga(x2, x3)
U6_aga(x1, x2, x3, x4, x5)  =  U6_aga(x3, x5)
U5_ga(x1, x2, x3)  =  U5_ga(x1, x3)
del2_out_ga(x1, x2)  =  del2_out_ga(x1, x2)
U2_g(x1, x2)  =  U2_g(x1, x2)
del_in_aag(x1, x2, x3)  =  del_in_aag(x3)
del_out_aag(x1, x2, x3)  =  del_out_aag(x2, x3)
U6_aag(x1, x2, x3, x4, x5)  =  U6_aag(x4, x5)
U3_g(x1, x2)  =  U3_g(x1, x2)
conf_out_g(x1)  =  conf_out_g(x1)
DEL_IN_AAG(x1, x2, x3)  =  DEL_IN_AAG(x3)

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:

DEL_IN_AAG(X, cons(H, T), cons(H, T1)) → DEL_IN_AAG(X, T, T1)

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

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:

DEL_IN_AAG(cons(T1)) → DEL_IN_AAG(T1)

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:

• DEL_IN_AAG(cons(T1)) → DEL_IN_AAG(T1)
  The graph contains the following edges 1 > 1

(13) TRUE

(14) Obligation:

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

DEL_IN_AGA(X, cons(H, T), cons(H, T1)) → DEL_IN_AGA(X, T, T1)

The TRS R consists of the following rules:

conf_in_g(X) → U1_g(X, del2_in_ga(X, Z))
del2_in_ga(X, Y) → U4_ga(X, Y, del_in_aga(U, X, Z))
del_in_aga(X, cons(X, T), T) → del_out_aga(X, cons(X, T), T)
del_in_aga(X, cons(H, T), cons(H, T1)) → U6_aga(X, H, T, T1, del_in_aga(X, T, T1))
U6_aga(X, H, T, T1, del_out_aga(X, T, T1)) → del_out_aga(X, cons(H, T), cons(H, T1))
U4_ga(X, Y, del_out_aga(U, X, Z)) → U5_ga(X, Y, del_in_aga(V, Z, Y))
U5_ga(X, Y, del_out_aga(V, Z, Y)) → del2_out_ga(X, Y)
U1_g(X, del2_out_ga(X, Z)) → U2_g(X, del_in_aag(U, Y, Z))
del_in_aag(X, cons(X, T), T) → del_out_aag(X, cons(X, T), T)
del_in_aag(X, cons(H, T), cons(H, T1)) → U6_aag(X, H, T, T1, del_in_aag(X, T, T1))
U6_aag(X, H, T, T1, del_out_aag(X, T, T1)) → del_out_aag(X, cons(H, T), cons(H, T1))
U2_g(X, del_out_aag(U, Y, Z)) → U3_g(X, conf_in_g(Y))
U3_g(X, conf_out_g(Y)) → conf_out_g(X)

The argument filtering Pi contains the following mapping:
conf_in_g(x1)  =  conf_in_g(x1)
U1_g(x1, x2)  =  U1_g(x1, x2)
del2_in_ga(x1, x2)  =  del2_in_ga(x1)
U4_ga(x1, x2, x3)  =  U4_ga(x1, x3)
del_in_aga(x1, x2, x3)  =  del_in_aga(x2)
cons(x1, x2)  =  cons(x2)
del_out_aga(x1, x2, x3)  =  del_out_aga(x2, x3)
U6_aga(x1, x2, x3, x4, x5)  =  U6_aga(x3, x5)
U5_ga(x1, x2, x3)  =  U5_ga(x1, x3)
del2_out_ga(x1, x2)  =  del2_out_ga(x1, x2)
U2_g(x1, x2)  =  U2_g(x1, x2)
del_in_aag(x1, x2, x3)  =  del_in_aag(x3)
del_out_aag(x1, x2, x3)  =  del_out_aag(x2, x3)
U6_aag(x1, x2, x3, x4, x5)  =  U6_aag(x4, x5)
U3_g(x1, x2)  =  U3_g(x1, x2)
conf_out_g(x1)  =  conf_out_g(x1)
DEL_IN_AGA(x1, x2, x3)  =  DEL_IN_AGA(x2)

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:

DEL_IN_AGA(X, cons(H, T), cons(H, T1)) → DEL_IN_AGA(X, T, T1)

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

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:

DEL_IN_AGA(cons(T)) → DEL_IN_AGA(T)

R is empty.
Q is empty.
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:

• DEL_IN_AGA(cons(T)) → DEL_IN_AGA(T)
  The graph contains the following edges 1 > 1

(20) TRUE

(21) Obligation:

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

U1_G(X, del2_out_ga(X, Z)) → U2_G(X, del_in_aag(U, Y, Z))
U2_G(X, del_out_aag(U, Y, Z)) → CONF_IN_G(Y)
CONF_IN_G(X) → U1_G(X, del2_in_ga(X, Z))

The TRS R consists of the following rules:

conf_in_g(X) → U1_g(X, del2_in_ga(X, Z))
del2_in_ga(X, Y) → U4_ga(X, Y, del_in_aga(U, X, Z))
del_in_aga(X, cons(X, T), T) → del_out_aga(X, cons(X, T), T)
del_in_aga(X, cons(H, T), cons(H, T1)) → U6_aga(X, H, T, T1, del_in_aga(X, T, T1))
U6_aga(X, H, T, T1, del_out_aga(X, T, T1)) → del_out_aga(X, cons(H, T), cons(H, T1))
U4_ga(X, Y, del_out_aga(U, X, Z)) → U5_ga(X, Y, del_in_aga(V, Z, Y))
U5_ga(X, Y, del_out_aga(V, Z, Y)) → del2_out_ga(X, Y)
U1_g(X, del2_out_ga(X, Z)) → U2_g(X, del_in_aag(U, Y, Z))
del_in_aag(X, cons(X, T), T) → del_out_aag(X, cons(X, T), T)
del_in_aag(X, cons(H, T), cons(H, T1)) → U6_aag(X, H, T, T1, del_in_aag(X, T, T1))
U6_aag(X, H, T, T1, del_out_aag(X, T, T1)) → del_out_aag(X, cons(H, T), cons(H, T1))
U2_g(X, del_out_aag(U, Y, Z)) → U3_g(X, conf_in_g(Y))
U3_g(X, conf_out_g(Y)) → conf_out_g(X)

The argument filtering Pi contains the following mapping:
conf_in_g(x1)  =  conf_in_g(x1)
U1_g(x1, x2)  =  U1_g(x1, x2)
del2_in_ga(x1, x2)  =  del2_in_ga(x1)
U4_ga(x1, x2, x3)  =  U4_ga(x1, x3)
del_in_aga(x1, x2, x3)  =  del_in_aga(x2)
cons(x1, x2)  =  cons(x2)
del_out_aga(x1, x2, x3)  =  del_out_aga(x2, x3)
U6_aga(x1, x2, x3, x4, x5)  =  U6_aga(x3, x5)
U5_ga(x1, x2, x3)  =  U5_ga(x1, x3)
del2_out_ga(x1, x2)  =  del2_out_ga(x1, x2)
U2_g(x1, x2)  =  U2_g(x1, x2)
del_in_aag(x1, x2, x3)  =  del_in_aag(x3)
del_out_aag(x1, x2, x3)  =  del_out_aag(x2, x3)
U6_aag(x1, x2, x3, x4, x5)  =  U6_aag(x4, x5)
U3_g(x1, x2)  =  U3_g(x1, x2)
conf_out_g(x1)  =  conf_out_g(x1)
CONF_IN_G(x1)  =  CONF_IN_G(x1)
U1_G(x1, x2)  =  U1_G(x1, x2)
U2_G(x1, x2)  =  U2_G(x1, x2)

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

(22) UsableRulesProof (EQUIVALENT transformation)

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

(23) Obligation:

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

U1_G(X, del2_out_ga(X, Z)) → U2_G(X, del_in_aag(U, Y, Z))
U2_G(X, del_out_aag(U, Y, Z)) → CONF_IN_G(Y)
CONF_IN_G(X) → U1_G(X, del2_in_ga(X, Z))

The TRS R consists of the following rules:

del_in_aag(X, cons(X, T), T) → del_out_aag(X, cons(X, T), T)
del_in_aag(X, cons(H, T), cons(H, T1)) → U6_aag(X, H, T, T1, del_in_aag(X, T, T1))
del2_in_ga(X, Y) → U4_ga(X, Y, del_in_aga(U, X, Z))
U6_aag(X, H, T, T1, del_out_aag(X, T, T1)) → del_out_aag(X, cons(H, T), cons(H, T1))
U4_ga(X, Y, del_out_aga(U, X, Z)) → U5_ga(X, Y, del_in_aga(V, Z, Y))
del_in_aga(X, cons(X, T), T) → del_out_aga(X, cons(X, T), T)
del_in_aga(X, cons(H, T), cons(H, T1)) → U6_aga(X, H, T, T1, del_in_aga(X, T, T1))
U5_ga(X, Y, del_out_aga(V, Z, Y)) → del2_out_ga(X, Y)
U6_aga(X, H, T, T1, del_out_aga(X, T, T1)) → del_out_aga(X, cons(H, T), cons(H, T1))

The argument filtering Pi contains the following mapping:
del2_in_ga(x1, x2)  =  del2_in_ga(x1)
U4_ga(x1, x2, x3)  =  U4_ga(x1, x3)
del_in_aga(x1, x2, x3)  =  del_in_aga(x2)
cons(x1, x2)  =  cons(x2)
del_out_aga(x1, x2, x3)  =  del_out_aga(x2, x3)
U6_aga(x1, x2, x3, x4, x5)  =  U6_aga(x3, x5)
U5_ga(x1, x2, x3)  =  U5_ga(x1, x3)
del2_out_ga(x1, x2)  =  del2_out_ga(x1, x2)
del_in_aag(x1, x2, x3)  =  del_in_aag(x3)
del_out_aag(x1, x2, x3)  =  del_out_aag(x2, x3)
U6_aag(x1, x2, x3, x4, x5)  =  U6_aag(x4, x5)
CONF_IN_G(x1)  =  CONF_IN_G(x1)
U1_G(x1, x2)  =  U1_G(x1, x2)
U2_G(x1, x2)  =  U2_G(x1, x2)

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

(24) PiDPToQDPProof (SOUND transformation)

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

(25) Obligation:

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

U1_G(X, del2_out_ga(X, Z)) → U2_G(X, del_in_aag(Z))
U2_G(X, del_out_aag(Y, Z)) → CONF_IN_G(Y)
CONF_IN_G(X) → U1_G(X, del2_in_ga(X))

The TRS R consists of the following rules:

del_in_aag(T) → del_out_aag(cons(T), T)
del_in_aag(cons(T1)) → U6_aag(T1, del_in_aag(T1))
del2_in_ga(X) → U4_ga(X, del_in_aga(X))
U6_aag(T1, del_out_aag(T, T1)) → del_out_aag(cons(T), cons(T1))
U4_ga(X, del_out_aga(X, Z)) → U5_ga(X, del_in_aga(Z))
del_in_aga(cons(T)) → del_out_aga(cons(T), T)
del_in_aga(cons(T)) → U6_aga(T, del_in_aga(T))
U5_ga(X, del_out_aga(Z, Y)) → del2_out_ga(X, Y)
U6_aga(T, del_out_aga(T, T1)) → del_out_aga(cons(T), cons(T1))

The set Q consists of the following terms:

del_in_aag(x0)
del2_in_ga(x0)
U6_aag(x0, x1)
U4_ga(x0, x1)
del_in_aga(x0)
U5_ga(x0, x1)
U6_aga(x0, x1)

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

(26) PrologToPiTRSProof (SOUND transformation)

We use the technique of [LOPSTR]. With regard to the inferred argument filtering the predicates were used in the following modes:
conf_in: (b)
del2_in: (b,f)
del_in: (f,b,f) (f,f,b)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

conf_in_g(X) → U1_g(X, del2_in_ga(X, Z))
del2_in_ga(X, Y) → U4_ga(X, Y, del_in_aga(U, X, Z))
del_in_aga(X, cons(X, T), T) → del_out_aga(X, cons(X, T), T)
del_in_aga(X, cons(H, T), cons(H, T1)) → U6_aga(X, H, T, T1, del_in_aga(X, T, T1))
U6_aga(X, H, T, T1, del_out_aga(X, T, T1)) → del_out_aga(X, cons(H, T), cons(H, T1))
U4_ga(X, Y, del_out_aga(U, X, Z)) → U5_ga(X, Y, del_in_aga(V, Z, Y))
U5_ga(X, Y, del_out_aga(V, Z, Y)) → del2_out_ga(X, Y)
U1_g(X, del2_out_ga(X, Z)) → U2_g(X, del_in_aag(U, Y, Z))
del_in_aag(X, cons(X, T), T) → del_out_aag(X, cons(X, T), T)
del_in_aag(X, cons(H, T), cons(H, T1)) → U6_aag(X, H, T, T1, del_in_aag(X, T, T1))
U6_aag(X, H, T, T1, del_out_aag(X, T, T1)) → del_out_aag(X, cons(H, T), cons(H, T1))
U2_g(X, del_out_aag(U, Y, Z)) → U3_g(X, conf_in_g(Y))
U3_g(X, conf_out_g(Y)) → conf_out_g(X)

The argument filtering Pi contains the following mapping:
conf_in_g(x1)  =  conf_in_g(x1)
U1_g(x1, x2)  =  U1_g(x2)
del2_in_ga(x1, x2)  =  del2_in_ga(x1)
U4_ga(x1, x2, x3)  =  U4_ga(x3)
del_in_aga(x1, x2, x3)  =  del_in_aga(x2)
cons(x1, x2)  =  cons(x2)
del_out_aga(x1, x2, x3)  =  del_out_aga(x3)
U6_aga(x1, x2, x3, x4, x5)  =  U6_aga(x5)
U5_ga(x1, x2, x3)  =  U5_ga(x3)
del2_out_ga(x1, x2)  =  del2_out_ga(x2)
U2_g(x1, x2)  =  U2_g(x2)
del_in_aag(x1, x2, x3)  =  del_in_aag(x3)
del_out_aag(x1, x2, x3)  =  del_out_aag(x2)
U6_aag(x1, x2, x3, x4, x5)  =  U6_aag(x5)
U3_g(x1, x2)  =  U3_g(x2)
conf_out_g(x1)  =  conf_out_g

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(27) Obligation:

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

conf_in_g(X) → U1_g(X, del2_in_ga(X, Z))
del2_in_ga(X, Y) → U4_ga(X, Y, del_in_aga(U, X, Z))
del_in_aga(X, cons(X, T), T) → del_out_aga(X, cons(X, T), T)
del_in_aga(X, cons(H, T), cons(H, T1)) → U6_aga(X, H, T, T1, del_in_aga(X, T, T1))
U6_aga(X, H, T, T1, del_out_aga(X, T, T1)) → del_out_aga(X, cons(H, T), cons(H, T1))
U4_ga(X, Y, del_out_aga(U, X, Z)) → U5_ga(X, Y, del_in_aga(V, Z, Y))
U5_ga(X, Y, del_out_aga(V, Z, Y)) → del2_out_ga(X, Y)
U1_g(X, del2_out_ga(X, Z)) → U2_g(X, del_in_aag(U, Y, Z))
del_in_aag(X, cons(X, T), T) → del_out_aag(X, cons(X, T), T)
del_in_aag(X, cons(H, T), cons(H, T1)) → U6_aag(X, H, T, T1, del_in_aag(X, T, T1))
U6_aag(X, H, T, T1, del_out_aag(X, T, T1)) → del_out_aag(X, cons(H, T), cons(H, T1))
U2_g(X, del_out_aag(U, Y, Z)) → U3_g(X, conf_in_g(Y))
U3_g(X, conf_out_g(Y)) → conf_out_g(X)

The argument filtering Pi contains the following mapping:
conf_in_g(x1)  =  conf_in_g(x1)
U1_g(x1, x2)  =  U1_g(x2)
del2_in_ga(x1, x2)  =  del2_in_ga(x1)
U4_ga(x1, x2, x3)  =  U4_ga(x3)
del_in_aga(x1, x2, x3)  =  del_in_aga(x2)
cons(x1, x2)  =  cons(x2)
del_out_aga(x1, x2, x3)  =  del_out_aga(x3)
U6_aga(x1, x2, x3, x4, x5)  =  U6_aga(x5)
U5_ga(x1, x2, x3)  =  U5_ga(x3)
del2_out_ga(x1, x2)  =  del2_out_ga(x2)
U2_g(x1, x2)  =  U2_g(x2)
del_in_aag(x1, x2, x3)  =  del_in_aag(x3)
del_out_aag(x1, x2, x3)  =  del_out_aag(x2)
U6_aag(x1, x2, x3, x4, x5)  =  U6_aag(x5)
U3_g(x1, x2)  =  U3_g(x2)
conf_out_g(x1)  =  conf_out_g

(28) 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:

CONF_IN_G(X) → U1_G(X, del2_in_ga(X, Z))
CONF_IN_G(X) → DEL2_IN_GA(X, Z)
DEL2_IN_GA(X, Y) → U4_GA(X, Y, del_in_aga(U, X, Z))
DEL2_IN_GA(X, Y) → DEL_IN_AGA(U, X, Z)
DEL_IN_AGA(X, cons(H, T), cons(H, T1)) → U6_AGA(X, H, T, T1, del_in_aga(X, T, T1))
DEL_IN_AGA(X, cons(H, T), cons(H, T1)) → DEL_IN_AGA(X, T, T1)
U4_GA(X, Y, del_out_aga(U, X, Z)) → U5_GA(X, Y, del_in_aga(V, Z, Y))
U4_GA(X, Y, del_out_aga(U, X, Z)) → DEL_IN_AGA(V, Z, Y)
U1_G(X, del2_out_ga(X, Z)) → U2_G(X, del_in_aag(U, Y, Z))
U1_G(X, del2_out_ga(X, Z)) → DEL_IN_AAG(U, Y, Z)
DEL_IN_AAG(X, cons(H, T), cons(H, T1)) → U6_AAG(X, H, T, T1, del_in_aag(X, T, T1))
DEL_IN_AAG(X, cons(H, T), cons(H, T1)) → DEL_IN_AAG(X, T, T1)
U2_G(X, del_out_aag(U, Y, Z)) → U3_G(X, conf_in_g(Y))
U2_G(X, del_out_aag(U, Y, Z)) → CONF_IN_G(Y)

The TRS R consists of the following rules:

conf_in_g(X) → U1_g(X, del2_in_ga(X, Z))
del2_in_ga(X, Y) → U4_ga(X, Y, del_in_aga(U, X, Z))
del_in_aga(X, cons(X, T), T) → del_out_aga(X, cons(X, T), T)
del_in_aga(X, cons(H, T), cons(H, T1)) → U6_aga(X, H, T, T1, del_in_aga(X, T, T1))
U6_aga(X, H, T, T1, del_out_aga(X, T, T1)) → del_out_aga(X, cons(H, T), cons(H, T1))
U4_ga(X, Y, del_out_aga(U, X, Z)) → U5_ga(X, Y, del_in_aga(V, Z, Y))
U5_ga(X, Y, del_out_aga(V, Z, Y)) → del2_out_ga(X, Y)
U1_g(X, del2_out_ga(X, Z)) → U2_g(X, del_in_aag(U, Y, Z))
del_in_aag(X, cons(X, T), T) → del_out_aag(X, cons(X, T), T)
del_in_aag(X, cons(H, T), cons(H, T1)) → U6_aag(X, H, T, T1, del_in_aag(X, T, T1))
U6_aag(X, H, T, T1, del_out_aag(X, T, T1)) → del_out_aag(X, cons(H, T), cons(H, T1))
U2_g(X, del_out_aag(U, Y, Z)) → U3_g(X, conf_in_g(Y))
U3_g(X, conf_out_g(Y)) → conf_out_g(X)

The argument filtering Pi contains the following mapping:
conf_in_g(x1)  =  conf_in_g(x1)
U1_g(x1, x2)  =  U1_g(x2)
del2_in_ga(x1, x2)  =  del2_in_ga(x1)
U4_ga(x1, x2, x3)  =  U4_ga(x3)
del_in_aga(x1, x2, x3)  =  del_in_aga(x2)
cons(x1, x2)  =  cons(x2)
del_out_aga(x1, x2, x3)  =  del_out_aga(x3)
U6_aga(x1, x2, x3, x4, x5)  =  U6_aga(x5)
U5_ga(x1, x2, x3)  =  U5_ga(x3)
del2_out_ga(x1, x2)  =  del2_out_ga(x2)
U2_g(x1, x2)  =  U2_g(x2)
del_in_aag(x1, x2, x3)  =  del_in_aag(x3)
del_out_aag(x1, x2, x3)  =  del_out_aag(x2)
U6_aag(x1, x2, x3, x4, x5)  =  U6_aag(x5)
U3_g(x1, x2)  =  U3_g(x2)
conf_out_g(x1)  =  conf_out_g
CONF_IN_G(x1)  =  CONF_IN_G(x1)
U1_G(x1, x2)  =  U1_G(x2)
DEL2_IN_GA(x1, x2)  =  DEL2_IN_GA(x1)
U4_GA(x1, x2, x3)  =  U4_GA(x3)
DEL_IN_AGA(x1, x2, x3)  =  DEL_IN_AGA(x2)
U6_AGA(x1, x2, x3, x4, x5)  =  U6_AGA(x5)
U5_GA(x1, x2, x3)  =  U5_GA(x3)
U2_G(x1, x2)  =  U2_G(x2)
DEL_IN_AAG(x1, x2, x3)  =  DEL_IN_AAG(x3)
U6_AAG(x1, x2, x3, x4, x5)  =  U6_AAG(x5)
U3_G(x1, x2)  =  U3_G(x2)

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

(29) Obligation:

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

CONF_IN_G(X) → U1_G(X, del2_in_ga(X, Z))
CONF_IN_G(X) → DEL2_IN_GA(X, Z)
DEL2_IN_GA(X, Y) → U4_GA(X, Y, del_in_aga(U, X, Z))
DEL2_IN_GA(X, Y) → DEL_IN_AGA(U, X, Z)
DEL_IN_AGA(X, cons(H, T), cons(H, T1)) → U6_AGA(X, H, T, T1, del_in_aga(X, T, T1))
DEL_IN_AGA(X, cons(H, T), cons(H, T1)) → DEL_IN_AGA(X, T, T1)
U4_GA(X, Y, del_out_aga(U, X, Z)) → U5_GA(X, Y, del_in_aga(V, Z, Y))
U4_GA(X, Y, del_out_aga(U, X, Z)) → DEL_IN_AGA(V, Z, Y)
U1_G(X, del2_out_ga(X, Z)) → U2_G(X, del_in_aag(U, Y, Z))
U1_G(X, del2_out_ga(X, Z)) → DEL_IN_AAG(U, Y, Z)
DEL_IN_AAG(X, cons(H, T), cons(H, T1)) → U6_AAG(X, H, T, T1, del_in_aag(X, T, T1))
DEL_IN_AAG(X, cons(H, T), cons(H, T1)) → DEL_IN_AAG(X, T, T1)
U2_G(X, del_out_aag(U, Y, Z)) → U3_G(X, conf_in_g(Y))
U2_G(X, del_out_aag(U, Y, Z)) → CONF_IN_G(Y)

The TRS R consists of the following rules:

conf_in_g(X) → U1_g(X, del2_in_ga(X, Z))
del2_in_ga(X, Y) → U4_ga(X, Y, del_in_aga(U, X, Z))
del_in_aga(X, cons(X, T), T) → del_out_aga(X, cons(X, T), T)
del_in_aga(X, cons(H, T), cons(H, T1)) → U6_aga(X, H, T, T1, del_in_aga(X, T, T1))
U6_aga(X, H, T, T1, del_out_aga(X, T, T1)) → del_out_aga(X, cons(H, T), cons(H, T1))
U4_ga(X, Y, del_out_aga(U, X, Z)) → U5_ga(X, Y, del_in_aga(V, Z, Y))
U5_ga(X, Y, del_out_aga(V, Z, Y)) → del2_out_ga(X, Y)
U1_g(X, del2_out_ga(X, Z)) → U2_g(X, del_in_aag(U, Y, Z))
del_in_aag(X, cons(X, T), T) → del_out_aag(X, cons(X, T), T)
del_in_aag(X, cons(H, T), cons(H, T1)) → U6_aag(X, H, T, T1, del_in_aag(X, T, T1))
U6_aag(X, H, T, T1, del_out_aag(X, T, T1)) → del_out_aag(X, cons(H, T), cons(H, T1))
U2_g(X, del_out_aag(U, Y, Z)) → U3_g(X, conf_in_g(Y))
U3_g(X, conf_out_g(Y)) → conf_out_g(X)

The argument filtering Pi contains the following mapping:
conf_in_g(x1)  =  conf_in_g(x1)
U1_g(x1, x2)  =  U1_g(x2)
del2_in_ga(x1, x2)  =  del2_in_ga(x1)
U4_ga(x1, x2, x3)  =  U4_ga(x3)
del_in_aga(x1, x2, x3)  =  del_in_aga(x2)
cons(x1, x2)  =  cons(x2)
del_out_aga(x1, x2, x3)  =  del_out_aga(x3)
U6_aga(x1, x2, x3, x4, x5)  =  U6_aga(x5)
U5_ga(x1, x2, x3)  =  U5_ga(x3)
del2_out_ga(x1, x2)  =  del2_out_ga(x2)
U2_g(x1, x2)  =  U2_g(x2)
del_in_aag(x1, x2, x3)  =  del_in_aag(x3)
del_out_aag(x1, x2, x3)  =  del_out_aag(x2)
U6_aag(x1, x2, x3, x4, x5)  =  U6_aag(x5)
U3_g(x1, x2)  =  U3_g(x2)
conf_out_g(x1)  =  conf_out_g
CONF_IN_G(x1)  =  CONF_IN_G(x1)
U1_G(x1, x2)  =  U1_G(x2)
DEL2_IN_GA(x1, x2)  =  DEL2_IN_GA(x1)
U4_GA(x1, x2, x3)  =  U4_GA(x3)
DEL_IN_AGA(x1, x2, x3)  =  DEL_IN_AGA(x2)
U6_AGA(x1, x2, x3, x4, x5)  =  U6_AGA(x5)
U5_GA(x1, x2, x3)  =  U5_GA(x3)
U2_G(x1, x2)  =  U2_G(x2)
DEL_IN_AAG(x1, x2, x3)  =  DEL_IN_AAG(x3)
U6_AAG(x1, x2, x3, x4, x5)  =  U6_AAG(x5)
U3_G(x1, x2)  =  U3_G(x2)

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

(30) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 3 SCCs with 9 less nodes.

(31) Complex Obligation (AND)

(32) Obligation:

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

DEL_IN_AAG(X, cons(H, T), cons(H, T1)) → DEL_IN_AAG(X, T, T1)

The TRS R consists of the following rules:

conf_in_g(X) → U1_g(X, del2_in_ga(X, Z))
del2_in_ga(X, Y) → U4_ga(X, Y, del_in_aga(U, X, Z))
del_in_aga(X, cons(X, T), T) → del_out_aga(X, cons(X, T), T)
del_in_aga(X, cons(H, T), cons(H, T1)) → U6_aga(X, H, T, T1, del_in_aga(X, T, T1))
U6_aga(X, H, T, T1, del_out_aga(X, T, T1)) → del_out_aga(X, cons(H, T), cons(H, T1))
U4_ga(X, Y, del_out_aga(U, X, Z)) → U5_ga(X, Y, del_in_aga(V, Z, Y))
U5_ga(X, Y, del_out_aga(V, Z, Y)) → del2_out_ga(X, Y)
U1_g(X, del2_out_ga(X, Z)) → U2_g(X, del_in_aag(U, Y, Z))
del_in_aag(X, cons(X, T), T) → del_out_aag(X, cons(X, T), T)
del_in_aag(X, cons(H, T), cons(H, T1)) → U6_aag(X, H, T, T1, del_in_aag(X, T, T1))
U6_aag(X, H, T, T1, del_out_aag(X, T, T1)) → del_out_aag(X, cons(H, T), cons(H, T1))
U2_g(X, del_out_aag(U, Y, Z)) → U3_g(X, conf_in_g(Y))
U3_g(X, conf_out_g(Y)) → conf_out_g(X)

The argument filtering Pi contains the following mapping:
conf_in_g(x1)  =  conf_in_g(x1)
U1_g(x1, x2)  =  U1_g(x2)
del2_in_ga(x1, x2)  =  del2_in_ga(x1)
U4_ga(x1, x2, x3)  =  U4_ga(x3)
del_in_aga(x1, x2, x3)  =  del_in_aga(x2)
cons(x1, x2)  =  cons(x2)
del_out_aga(x1, x2, x3)  =  del_out_aga(x3)
U6_aga(x1, x2, x3, x4, x5)  =  U6_aga(x5)
U5_ga(x1, x2, x3)  =  U5_ga(x3)
del2_out_ga(x1, x2)  =  del2_out_ga(x2)
U2_g(x1, x2)  =  U2_g(x2)
del_in_aag(x1, x2, x3)  =  del_in_aag(x3)
del_out_aag(x1, x2, x3)  =  del_out_aag(x2)
U6_aag(x1, x2, x3, x4, x5)  =  U6_aag(x5)
U3_g(x1, x2)  =  U3_g(x2)
conf_out_g(x1)  =  conf_out_g
DEL_IN_AAG(x1, x2, x3)  =  DEL_IN_AAG(x3)

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:

DEL_IN_AAG(X, cons(H, T), cons(H, T1)) → DEL_IN_AAG(X, T, T1)

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

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:

DEL_IN_AAG(cons(T1)) → DEL_IN_AAG(T1)

R is empty.
Q is empty.
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:

• DEL_IN_AAG(cons(T1)) → DEL_IN_AAG(T1)
  The graph contains the following edges 1 > 1

(38) TRUE

(39) Obligation:

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

DEL_IN_AGA(X, cons(H, T), cons(H, T1)) → DEL_IN_AGA(X, T, T1)

The TRS R consists of the following rules:

conf_in_g(X) → U1_g(X, del2_in_ga(X, Z))
del2_in_ga(X, Y) → U4_ga(X, Y, del_in_aga(U, X, Z))
del_in_aga(X, cons(X, T), T) → del_out_aga(X, cons(X, T), T)
del_in_aga(X, cons(H, T), cons(H, T1)) → U6_aga(X, H, T, T1, del_in_aga(X, T, T1))
U6_aga(X, H, T, T1, del_out_aga(X, T, T1)) → del_out_aga(X, cons(H, T), cons(H, T1))
U4_ga(X, Y, del_out_aga(U, X, Z)) → U5_ga(X, Y, del_in_aga(V, Z, Y))
U5_ga(X, Y, del_out_aga(V, Z, Y)) → del2_out_ga(X, Y)
U1_g(X, del2_out_ga(X, Z)) → U2_g(X, del_in_aag(U, Y, Z))
del_in_aag(X, cons(X, T), T) → del_out_aag(X, cons(X, T), T)
del_in_aag(X, cons(H, T), cons(H, T1)) → U6_aag(X, H, T, T1, del_in_aag(X, T, T1))
U6_aag(X, H, T, T1, del_out_aag(X, T, T1)) → del_out_aag(X, cons(H, T), cons(H, T1))
U2_g(X, del_out_aag(U, Y, Z)) → U3_g(X, conf_in_g(Y))
U3_g(X, conf_out_g(Y)) → conf_out_g(X)

The argument filtering Pi contains the following mapping:
conf_in_g(x1)  =  conf_in_g(x1)
U1_g(x1, x2)  =  U1_g(x2)
del2_in_ga(x1, x2)  =  del2_in_ga(x1)
U4_ga(x1, x2, x3)  =  U4_ga(x3)
del_in_aga(x1, x2, x3)  =  del_in_aga(x2)
cons(x1, x2)  =  cons(x2)
del_out_aga(x1, x2, x3)  =  del_out_aga(x3)
U6_aga(x1, x2, x3, x4, x5)  =  U6_aga(x5)
U5_ga(x1, x2, x3)  =  U5_ga(x3)
del2_out_ga(x1, x2)  =  del2_out_ga(x2)
U2_g(x1, x2)  =  U2_g(x2)
del_in_aag(x1, x2, x3)  =  del_in_aag(x3)
del_out_aag(x1, x2, x3)  =  del_out_aag(x2)
U6_aag(x1, x2, x3, x4, x5)  =  U6_aag(x5)
U3_g(x1, x2)  =  U3_g(x2)
conf_out_g(x1)  =  conf_out_g
DEL_IN_AGA(x1, x2, x3)  =  DEL_IN_AGA(x2)

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

(40) UsableRulesProof (EQUIVALENT transformation)

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

(41) Obligation:

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

DEL_IN_AGA(X, cons(H, T), cons(H, T1)) → DEL_IN_AGA(X, T, T1)

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

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

(42) PiDPToQDPProof (SOUND transformation)

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

(43) Obligation:

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

DEL_IN_AGA(cons(T)) → DEL_IN_AGA(T)

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

(44) 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:

• DEL_IN_AGA(cons(T)) → DEL_IN_AGA(T)
  The graph contains the following edges 1 > 1

(45) TRUE

(46) Obligation:

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

U1_G(X, del2_out_ga(X, Z)) → U2_G(X, del_in_aag(U, Y, Z))
U2_G(X, del_out_aag(U, Y, Z)) → CONF_IN_G(Y)
CONF_IN_G(X) → U1_G(X, del2_in_ga(X, Z))

The TRS R consists of the following rules:

conf_in_g(X) → U1_g(X, del2_in_ga(X, Z))
del2_in_ga(X, Y) → U4_ga(X, Y, del_in_aga(U, X, Z))
del_in_aga(X, cons(X, T), T) → del_out_aga(X, cons(X, T), T)
del_in_aga(X, cons(H, T), cons(H, T1)) → U6_aga(X, H, T, T1, del_in_aga(X, T, T1))
U6_aga(X, H, T, T1, del_out_aga(X, T, T1)) → del_out_aga(X, cons(H, T), cons(H, T1))
U4_ga(X, Y, del_out_aga(U, X, Z)) → U5_ga(X, Y, del_in_aga(V, Z, Y))
U5_ga(X, Y, del_out_aga(V, Z, Y)) → del2_out_ga(X, Y)
U1_g(X, del2_out_ga(X, Z)) → U2_g(X, del_in_aag(U, Y, Z))
del_in_aag(X, cons(X, T), T) → del_out_aag(X, cons(X, T), T)
del_in_aag(X, cons(H, T), cons(H, T1)) → U6_aag(X, H, T, T1, del_in_aag(X, T, T1))
U6_aag(X, H, T, T1, del_out_aag(X, T, T1)) → del_out_aag(X, cons(H, T), cons(H, T1))
U2_g(X, del_out_aag(U, Y, Z)) → U3_g(X, conf_in_g(Y))
U3_g(X, conf_out_g(Y)) → conf_out_g(X)

The argument filtering Pi contains the following mapping:
conf_in_g(x1)  =  conf_in_g(x1)
U1_g(x1, x2)  =  U1_g(x2)
del2_in_ga(x1, x2)  =  del2_in_ga(x1)
U4_ga(x1, x2, x3)  =  U4_ga(x3)
del_in_aga(x1, x2, x3)  =  del_in_aga(x2)
cons(x1, x2)  =  cons(x2)
del_out_aga(x1, x2, x3)  =  del_out_aga(x3)
U6_aga(x1, x2, x3, x4, x5)  =  U6_aga(x5)
U5_ga(x1, x2, x3)  =  U5_ga(x3)
del2_out_ga(x1, x2)  =  del2_out_ga(x2)
U2_g(x1, x2)  =  U2_g(x2)
del_in_aag(x1, x2, x3)  =  del_in_aag(x3)
del_out_aag(x1, x2, x3)  =  del_out_aag(x2)
U6_aag(x1, x2, x3, x4, x5)  =  U6_aag(x5)
U3_g(x1, x2)  =  U3_g(x2)
conf_out_g(x1)  =  conf_out_g
CONF_IN_G(x1)  =  CONF_IN_G(x1)
U1_G(x1, x2)  =  U1_G(x2)
U2_G(x1, x2)  =  U2_G(x2)

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

(47) UsableRulesProof (EQUIVALENT transformation)

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

(48) Obligation:

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

U1_G(X, del2_out_ga(X, Z)) → U2_G(X, del_in_aag(U, Y, Z))
U2_G(X, del_out_aag(U, Y, Z)) → CONF_IN_G(Y)
CONF_IN_G(X) → U1_G(X, del2_in_ga(X, Z))

The TRS R consists of the following rules:

del_in_aag(X, cons(X, T), T) → del_out_aag(X, cons(X, T), T)
del_in_aag(X, cons(H, T), cons(H, T1)) → U6_aag(X, H, T, T1, del_in_aag(X, T, T1))
del2_in_ga(X, Y) → U4_ga(X, Y, del_in_aga(U, X, Z))
U6_aag(X, H, T, T1, del_out_aag(X, T, T1)) → del_out_aag(X, cons(H, T), cons(H, T1))
U4_ga(X, Y, del_out_aga(U, X, Z)) → U5_ga(X, Y, del_in_aga(V, Z, Y))
del_in_aga(X, cons(X, T), T) → del_out_aga(X, cons(X, T), T)
del_in_aga(X, cons(H, T), cons(H, T1)) → U6_aga(X, H, T, T1, del_in_aga(X, T, T1))
U5_ga(X, Y, del_out_aga(V, Z, Y)) → del2_out_ga(X, Y)
U6_aga(X, H, T, T1, del_out_aga(X, T, T1)) → del_out_aga(X, cons(H, T), cons(H, T1))

The argument filtering Pi contains the following mapping:
del2_in_ga(x1, x2)  =  del2_in_ga(x1)
U4_ga(x1, x2, x3)  =  U4_ga(x3)
del_in_aga(x1, x2, x3)  =  del_in_aga(x2)
cons(x1, x2)  =  cons(x2)
del_out_aga(x1, x2, x3)  =  del_out_aga(x3)
U6_aga(x1, x2, x3, x4, x5)  =  U6_aga(x5)
U5_ga(x1, x2, x3)  =  U5_ga(x3)
del2_out_ga(x1, x2)  =  del2_out_ga(x2)
del_in_aag(x1, x2, x3)  =  del_in_aag(x3)
del_out_aag(x1, x2, x3)  =  del_out_aag(x2)
U6_aag(x1, x2, x3, x4, x5)  =  U6_aag(x5)
CONF_IN_G(x1)  =  CONF_IN_G(x1)
U1_G(x1, x2)  =  U1_G(x2)
U2_G(x1, x2)  =  U2_G(x2)

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

(49) PiDPToQDPProof (SOUND transformation)

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

(50) Obligation:

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

U1_G(del2_out_ga(Z)) → U2_G(del_in_aag(Z))
U2_G(del_out_aag(Y)) → CONF_IN_G(Y)
CONF_IN_G(X) → U1_G(del2_in_ga(X))

The TRS R consists of the following rules:

del_in_aag(T) → del_out_aag(cons(T))
del_in_aag(cons(T1)) → U6_aag(del_in_aag(T1))
del2_in_ga(X) → U4_ga(del_in_aga(X))
U6_aag(del_out_aag(T)) → del_out_aag(cons(T))
U4_ga(del_out_aga(Z)) → U5_ga(del_in_aga(Z))
del_in_aga(cons(T)) → del_out_aga(T)
del_in_aga(cons(T)) → U6_aga(del_in_aga(T))
U5_ga(del_out_aga(Y)) → del2_out_ga(Y)
U6_aga(del_out_aga(T1)) → del_out_aga(cons(T1))

The set Q consists of the following terms:

del_in_aag(x0)
del2_in_ga(x0)
U6_aag(x0)
U4_ga(x0)
del_in_aga(x0)
U5_ga(x0)
U6_aga(x0)

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

(51) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

U1_G(del2_out_ga(Z)) → U2_G(del_in_aag(Z))
U2_G(del_out_aag(Y)) → CONF_IN_G(Y)
CONF_IN_G(X) → U1_G(del2_in_ga(X))

Strictly oriented rules of the TRS R:

del_in_aag(T) → del_out_aag(cons(T))
del2_in_ga(X) → U4_ga(del_in_aga(X))
U4_ga(del_out_aga(Z)) → U5_ga(del_in_aga(Z))
del_in_aga(cons(T)) → del_out_aga(T)
U5_ga(del_out_aga(Y)) → del2_out_ga(Y)

Used ordering: Polynomial interpretation [POLO]:

POL(CONF_IN_G(x₁)) = 2 + x₁   
POL(U1_G(x₁)) = x₁   
POL(U2_G(x₁)) = x₁   
POL(U4_ga(x₁)) = x₁   
POL(U5_ga(x₁)) = 7 + x₁   
POL(U6_aag(x₁)) = 9 + x₁   
POL(U6_aga(x₁)) = 9 + x₁   
POL(cons(x₁)) = 9 + x₁   
POL(del2_in_ga(x₁)) = 1 + x₁   
POL(del2_out_ga(x₁)) = 14 + x₁   
POL(del_in_aag(x₁)) = 13 + x₁   
POL(del_in_aga(x₁)) = x₁   
POL(del_out_aag(x₁)) = 3 + x₁   
POL(del_out_aga(x₁)) = 8 + x₁   

(52) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

del_in_aag(cons(T1)) → U6_aag(del_in_aag(T1))
U6_aag(del_out_aag(T)) → del_out_aag(cons(T))
del_in_aga(cons(T)) → U6_aga(del_in_aga(T))
U6_aga(del_out_aga(T1)) → del_out_aga(cons(T1))

The set Q consists of the following terms:

del_in_aag(x0)
del2_in_ga(x0)
U6_aag(x0)
U4_ga(x0)
del_in_aga(x0)
U5_ga(x0)
U6_aga(x0)

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

(53) PisEmptyProof (EQUIVALENT transformation)

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

(54) TRUE