Left Termination of the query pattern conf_in_1(g) w.r.t. the given Prolog program could not be shown:



Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof

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).

We use the technique of [30].Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

conf_in(X) → U1(X, del2_in(X, Z))
del2_in(X, Y) → U4(X, Y, del_in(U, X, Z))
del_in(X, cons(H, T), cons(H, T1)) → U6(X, H, T, T1, del_in(X, T, T1))
del_in(X, cons(X, T), T) → del_out(X, cons(X, T), T)
U6(X, H, T, T1, del_out(X, T, T1)) → del_out(X, cons(H, T), cons(H, T1))
U4(X, Y, del_out(U, X, Z)) → U5(X, Y, del_in(V, Z, Y))
U5(X, Y, del_out(V, Z, Y)) → del2_out(X, Y)
U1(X, del2_out(X, Z)) → U2(X, del_in(U, Y, Z))
U2(X, del_out(U, Y, Z)) → U3(X, conf_in(Y))
U3(X, conf_out(Y)) → conf_out(X)

The argument filtering Pi contains the following mapping:
conf_in(x1)  =  conf_in
U1(x1, x2)  =  U1(x2)
del2_in(x1, x2)  =  del2_in
U4(x1, x2, x3)  =  U4(x3)
del_in(x1, x2, x3)  =  del_in
cons(x1, x2)  =  cons(x2)
U6(x1, x2, x3, x4, x5)  =  U6(x5)
del_out(x1, x2, x3)  =  del_out
U5(x1, x2, x3)  =  U5(x3)
del2_out(x1, x2)  =  del2_out
U2(x1, x2)  =  U2(x2)
U3(x1, x2)  =  U3(x2)
conf_out(x1)  =  conf_out

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog



↳ Prolog
  ↳ PrologToPiTRSProof
PiTRS
      ↳ DependencyPairsProof
  ↳ PrologToPiTRSProof

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

conf_in(X) → U1(X, del2_in(X, Z))
del2_in(X, Y) → U4(X, Y, del_in(U, X, Z))
del_in(X, cons(H, T), cons(H, T1)) → U6(X, H, T, T1, del_in(X, T, T1))
del_in(X, cons(X, T), T) → del_out(X, cons(X, T), T)
U6(X, H, T, T1, del_out(X, T, T1)) → del_out(X, cons(H, T), cons(H, T1))
U4(X, Y, del_out(U, X, Z)) → U5(X, Y, del_in(V, Z, Y))
U5(X, Y, del_out(V, Z, Y)) → del2_out(X, Y)
U1(X, del2_out(X, Z)) → U2(X, del_in(U, Y, Z))
U2(X, del_out(U, Y, Z)) → U3(X, conf_in(Y))
U3(X, conf_out(Y)) → conf_out(X)

The argument filtering Pi contains the following mapping:
conf_in(x1)  =  conf_in
U1(x1, x2)  =  U1(x2)
del2_in(x1, x2)  =  del2_in
U4(x1, x2, x3)  =  U4(x3)
del_in(x1, x2, x3)  =  del_in
cons(x1, x2)  =  cons(x2)
U6(x1, x2, x3, x4, x5)  =  U6(x5)
del_out(x1, x2, x3)  =  del_out
U5(x1, x2, x3)  =  U5(x3)
del2_out(x1, x2)  =  del2_out
U2(x1, x2)  =  U2(x2)
U3(x1, x2)  =  U3(x2)
conf_out(x1)  =  conf_out


Using Dependency Pairs [1,30] we result in the following initial DP problem:
Pi DP problem:
The TRS P consists of the following rules:

CONF_IN(X) → U11(X, del2_in(X, Z))
CONF_IN(X) → DEL2_IN(X, Z)
DEL2_IN(X, Y) → U41(X, Y, del_in(U, X, Z))
DEL2_IN(X, Y) → DEL_IN(U, X, Z)
DEL_IN(X, cons(H, T), cons(H, T1)) → U61(X, H, T, T1, del_in(X, T, T1))
DEL_IN(X, cons(H, T), cons(H, T1)) → DEL_IN(X, T, T1)
U41(X, Y, del_out(U, X, Z)) → U51(X, Y, del_in(V, Z, Y))
U41(X, Y, del_out(U, X, Z)) → DEL_IN(V, Z, Y)
U11(X, del2_out(X, Z)) → U21(X, del_in(U, Y, Z))
U11(X, del2_out(X, Z)) → DEL_IN(U, Y, Z)
U21(X, del_out(U, Y, Z)) → U31(X, conf_in(Y))
U21(X, del_out(U, Y, Z)) → CONF_IN(Y)

The TRS R consists of the following rules:

conf_in(X) → U1(X, del2_in(X, Z))
del2_in(X, Y) → U4(X, Y, del_in(U, X, Z))
del_in(X, cons(H, T), cons(H, T1)) → U6(X, H, T, T1, del_in(X, T, T1))
del_in(X, cons(X, T), T) → del_out(X, cons(X, T), T)
U6(X, H, T, T1, del_out(X, T, T1)) → del_out(X, cons(H, T), cons(H, T1))
U4(X, Y, del_out(U, X, Z)) → U5(X, Y, del_in(V, Z, Y))
U5(X, Y, del_out(V, Z, Y)) → del2_out(X, Y)
U1(X, del2_out(X, Z)) → U2(X, del_in(U, Y, Z))
U2(X, del_out(U, Y, Z)) → U3(X, conf_in(Y))
U3(X, conf_out(Y)) → conf_out(X)

The argument filtering Pi contains the following mapping:
conf_in(x1)  =  conf_in
U1(x1, x2)  =  U1(x2)
del2_in(x1, x2)  =  del2_in
U4(x1, x2, x3)  =  U4(x3)
del_in(x1, x2, x3)  =  del_in
cons(x1, x2)  =  cons(x2)
U6(x1, x2, x3, x4, x5)  =  U6(x5)
del_out(x1, x2, x3)  =  del_out
U5(x1, x2, x3)  =  U5(x3)
del2_out(x1, x2)  =  del2_out
U2(x1, x2)  =  U2(x2)
U3(x1, x2)  =  U3(x2)
conf_out(x1)  =  conf_out
U51(x1, x2, x3)  =  U51(x3)
DEL_IN(x1, x2, x3)  =  DEL_IN
U41(x1, x2, x3)  =  U41(x3)
U11(x1, x2)  =  U11(x2)
CONF_IN(x1)  =  CONF_IN
U31(x1, x2)  =  U31(x2)
U21(x1, x2)  =  U21(x2)
U61(x1, x2, x3, x4, x5)  =  U61(x5)
DEL2_IN(x1, x2)  =  DEL2_IN

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

↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
PiDP
          ↳ DependencyGraphProof
  ↳ PrologToPiTRSProof

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

CONF_IN(X) → U11(X, del2_in(X, Z))
CONF_IN(X) → DEL2_IN(X, Z)
DEL2_IN(X, Y) → U41(X, Y, del_in(U, X, Z))
DEL2_IN(X, Y) → DEL_IN(U, X, Z)
DEL_IN(X, cons(H, T), cons(H, T1)) → U61(X, H, T, T1, del_in(X, T, T1))
DEL_IN(X, cons(H, T), cons(H, T1)) → DEL_IN(X, T, T1)
U41(X, Y, del_out(U, X, Z)) → U51(X, Y, del_in(V, Z, Y))
U41(X, Y, del_out(U, X, Z)) → DEL_IN(V, Z, Y)
U11(X, del2_out(X, Z)) → U21(X, del_in(U, Y, Z))
U11(X, del2_out(X, Z)) → DEL_IN(U, Y, Z)
U21(X, del_out(U, Y, Z)) → U31(X, conf_in(Y))
U21(X, del_out(U, Y, Z)) → CONF_IN(Y)

The TRS R consists of the following rules:

conf_in(X) → U1(X, del2_in(X, Z))
del2_in(X, Y) → U4(X, Y, del_in(U, X, Z))
del_in(X, cons(H, T), cons(H, T1)) → U6(X, H, T, T1, del_in(X, T, T1))
del_in(X, cons(X, T), T) → del_out(X, cons(X, T), T)
U6(X, H, T, T1, del_out(X, T, T1)) → del_out(X, cons(H, T), cons(H, T1))
U4(X, Y, del_out(U, X, Z)) → U5(X, Y, del_in(V, Z, Y))
U5(X, Y, del_out(V, Z, Y)) → del2_out(X, Y)
U1(X, del2_out(X, Z)) → U2(X, del_in(U, Y, Z))
U2(X, del_out(U, Y, Z)) → U3(X, conf_in(Y))
U3(X, conf_out(Y)) → conf_out(X)

The argument filtering Pi contains the following mapping:
conf_in(x1)  =  conf_in
U1(x1, x2)  =  U1(x2)
del2_in(x1, x2)  =  del2_in
U4(x1, x2, x3)  =  U4(x3)
del_in(x1, x2, x3)  =  del_in
cons(x1, x2)  =  cons(x2)
U6(x1, x2, x3, x4, x5)  =  U6(x5)
del_out(x1, x2, x3)  =  del_out
U5(x1, x2, x3)  =  U5(x3)
del2_out(x1, x2)  =  del2_out
U2(x1, x2)  =  U2(x2)
U3(x1, x2)  =  U3(x2)
conf_out(x1)  =  conf_out
U51(x1, x2, x3)  =  U51(x3)
DEL_IN(x1, x2, x3)  =  DEL_IN
U41(x1, x2, x3)  =  U41(x3)
U11(x1, x2)  =  U11(x2)
CONF_IN(x1)  =  CONF_IN
U31(x1, x2)  =  U31(x2)
U21(x1, x2)  =  U21(x2)
U61(x1, x2, x3, x4, x5)  =  U61(x5)
DEL2_IN(x1, x2)  =  DEL2_IN

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 2 SCCs with 8 less nodes.

↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
PiDP
                ↳ UsableRulesProof
              ↳ PiDP
  ↳ PrologToPiTRSProof

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

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

The TRS R consists of the following rules:

conf_in(X) → U1(X, del2_in(X, Z))
del2_in(X, Y) → U4(X, Y, del_in(U, X, Z))
del_in(X, cons(H, T), cons(H, T1)) → U6(X, H, T, T1, del_in(X, T, T1))
del_in(X, cons(X, T), T) → del_out(X, cons(X, T), T)
U6(X, H, T, T1, del_out(X, T, T1)) → del_out(X, cons(H, T), cons(H, T1))
U4(X, Y, del_out(U, X, Z)) → U5(X, Y, del_in(V, Z, Y))
U5(X, Y, del_out(V, Z, Y)) → del2_out(X, Y)
U1(X, del2_out(X, Z)) → U2(X, del_in(U, Y, Z))
U2(X, del_out(U, Y, Z)) → U3(X, conf_in(Y))
U3(X, conf_out(Y)) → conf_out(X)

The argument filtering Pi contains the following mapping:
conf_in(x1)  =  conf_in
U1(x1, x2)  =  U1(x2)
del2_in(x1, x2)  =  del2_in
U4(x1, x2, x3)  =  U4(x3)
del_in(x1, x2, x3)  =  del_in
cons(x1, x2)  =  cons(x2)
U6(x1, x2, x3, x4, x5)  =  U6(x5)
del_out(x1, x2, x3)  =  del_out
U5(x1, x2, x3)  =  U5(x3)
del2_out(x1, x2)  =  del2_out
U2(x1, x2)  =  U2(x2)
U3(x1, x2)  =  U3(x2)
conf_out(x1)  =  conf_out
DEL_IN(x1, x2, x3)  =  DEL_IN

We have to consider all (P,R,Pi)-chains
For (infinitary) constructor rewriting [30] we can delete all non-usable rules from R.

↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
                ↳ UsableRulesProof
PiDP
                    ↳ PiDPToQDPProof
              ↳ PiDP
  ↳ PrologToPiTRSProof

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

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

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

We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem [30] into ordinary QDP problem [15] by application of Pi.

↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
QDP
                        ↳ NonTerminationProof
              ↳ PiDP
  ↳ PrologToPiTRSProof

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

DEL_INDEL_IN

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.

The TRS P consists of the following rules:

DEL_INDEL_IN

The TRS R consists of the following rules:none


s = DEL_IN evaluates to t =DEL_IN

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




Rewriting sequence

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





↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
PiDP
                ↳ UsableRulesProof
  ↳ PrologToPiTRSProof

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

U11(X, del2_out(X, Z)) → U21(X, del_in(U, Y, Z))
U21(X, del_out(U, Y, Z)) → CONF_IN(Y)
CONF_IN(X) → U11(X, del2_in(X, Z))

The TRS R consists of the following rules:

conf_in(X) → U1(X, del2_in(X, Z))
del2_in(X, Y) → U4(X, Y, del_in(U, X, Z))
del_in(X, cons(H, T), cons(H, T1)) → U6(X, H, T, T1, del_in(X, T, T1))
del_in(X, cons(X, T), T) → del_out(X, cons(X, T), T)
U6(X, H, T, T1, del_out(X, T, T1)) → del_out(X, cons(H, T), cons(H, T1))
U4(X, Y, del_out(U, X, Z)) → U5(X, Y, del_in(V, Z, Y))
U5(X, Y, del_out(V, Z, Y)) → del2_out(X, Y)
U1(X, del2_out(X, Z)) → U2(X, del_in(U, Y, Z))
U2(X, del_out(U, Y, Z)) → U3(X, conf_in(Y))
U3(X, conf_out(Y)) → conf_out(X)

The argument filtering Pi contains the following mapping:
conf_in(x1)  =  conf_in
U1(x1, x2)  =  U1(x2)
del2_in(x1, x2)  =  del2_in
U4(x1, x2, x3)  =  U4(x3)
del_in(x1, x2, x3)  =  del_in
cons(x1, x2)  =  cons(x2)
U6(x1, x2, x3, x4, x5)  =  U6(x5)
del_out(x1, x2, x3)  =  del_out
U5(x1, x2, x3)  =  U5(x3)
del2_out(x1, x2)  =  del2_out
U2(x1, x2)  =  U2(x2)
U3(x1, x2)  =  U3(x2)
conf_out(x1)  =  conf_out
U11(x1, x2)  =  U11(x2)
CONF_IN(x1)  =  CONF_IN
U21(x1, x2)  =  U21(x2)

We have to consider all (P,R,Pi)-chains
For (infinitary) constructor rewriting [30] we can delete all non-usable rules from R.

↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
PiDP
                    ↳ PiDPToQDPProof
  ↳ PrologToPiTRSProof

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

U11(X, del2_out(X, Z)) → U21(X, del_in(U, Y, Z))
U21(X, del_out(U, Y, Z)) → CONF_IN(Y)
CONF_IN(X) → U11(X, del2_in(X, Z))

The TRS R consists of the following rules:

del_in(X, cons(H, T), cons(H, T1)) → U6(X, H, T, T1, del_in(X, T, T1))
del_in(X, cons(X, T), T) → del_out(X, cons(X, T), T)
del2_in(X, Y) → U4(X, Y, del_in(U, X, Z))
U6(X, H, T, T1, del_out(X, T, T1)) → del_out(X, cons(H, T), cons(H, T1))
U4(X, Y, del_out(U, X, Z)) → U5(X, Y, del_in(V, Z, Y))
U5(X, Y, del_out(V, Z, Y)) → del2_out(X, Y)

The argument filtering Pi contains the following mapping:
del2_in(x1, x2)  =  del2_in
U4(x1, x2, x3)  =  U4(x3)
del_in(x1, x2, x3)  =  del_in
cons(x1, x2)  =  cons(x2)
U6(x1, x2, x3, x4, x5)  =  U6(x5)
del_out(x1, x2, x3)  =  del_out
U5(x1, x2, x3)  =  U5(x3)
del2_out(x1, x2)  =  del2_out
U11(x1, x2)  =  U11(x2)
CONF_IN(x1)  =  CONF_IN
U21(x1, x2)  =  U21(x2)

We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem [30] into ordinary QDP problem [15] by application of Pi.

↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
QDP
                        ↳ Narrowing
  ↳ PrologToPiTRSProof

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

CONF_INU11(del2_in)
U11(del2_out) → U21(del_in)
U21(del_out) → CONF_IN

The TRS R consists of the following rules:

del_inU6(del_in)
del_indel_out
del2_inU4(del_in)
U6(del_out) → del_out
U4(del_out) → U5(del_in)
U5(del_out) → del2_out

The set Q consists of the following terms:

del_in
del2_in
U6(x0)
U4(x0)
U5(x0)

We have to consider all (P,Q,R)-chains.
By narrowing [15] the rule U11(del2_out) → U21(del_in) at position [0] we obtained the following new rules:

U11(del2_out) → U21(del_out)
U11(del2_out) → U21(U6(del_in))



↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ Narrowing
QDP
                            ↳ Narrowing
  ↳ PrologToPiTRSProof

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

CONF_INU11(del2_in)
U11(del2_out) → U21(del_out)
U21(del_out) → CONF_IN
U11(del2_out) → U21(U6(del_in))

The TRS R consists of the following rules:

del_inU6(del_in)
del_indel_out
del2_inU4(del_in)
U6(del_out) → del_out
U4(del_out) → U5(del_in)
U5(del_out) → del2_out

The set Q consists of the following terms:

del_in
del2_in
U6(x0)
U4(x0)
U5(x0)

We have to consider all (P,Q,R)-chains.
By narrowing [15] the rule CONF_INU11(del2_in) at position [0] we obtained the following new rules:

CONF_INU11(U4(del_in))



↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ Narrowing
QDP
                                ↳ UsableRulesProof
  ↳ PrologToPiTRSProof

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

U11(del2_out) → U21(del_out)
U21(del_out) → CONF_IN
CONF_INU11(U4(del_in))
U11(del2_out) → U21(U6(del_in))

The TRS R consists of the following rules:

del_inU6(del_in)
del_indel_out
del2_inU4(del_in)
U6(del_out) → del_out
U4(del_out) → U5(del_in)
U5(del_out) → del2_out

The set Q consists of the following terms:

del_in
del2_in
U6(x0)
U4(x0)
U5(x0)

We have to consider all (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ Narrowing
                              ↳ QDP
                                ↳ UsableRulesProof
QDP
                                    ↳ QReductionProof
  ↳ PrologToPiTRSProof

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

U11(del2_out) → U21(del_out)
U21(del_out) → CONF_IN
CONF_INU11(U4(del_in))
U11(del2_out) → U21(U6(del_in))

The TRS R consists of the following rules:

del_inU6(del_in)
del_indel_out
U4(del_out) → U5(del_in)
U5(del_out) → del2_out
U6(del_out) → del_out

The set Q consists of the following terms:

del_in
del2_in
U6(x0)
U4(x0)
U5(x0)

We have to consider all (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

del2_in



↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ Narrowing
                              ↳ QDP
                                ↳ UsableRulesProof
                                  ↳ QDP
                                    ↳ QReductionProof
QDP
  ↳ PrologToPiTRSProof

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

U11(del2_out) → U21(del_out)
U21(del_out) → CONF_IN
U11(del2_out) → U21(U6(del_in))
CONF_INU11(U4(del_in))

The TRS R consists of the following rules:

del_inU6(del_in)
del_indel_out
U4(del_out) → U5(del_in)
U5(del_out) → del2_out
U6(del_out) → del_out

The set Q consists of the following terms:

del_in
U6(x0)
U4(x0)
U5(x0)

We have to consider all (P,Q,R)-chains.
We use the technique of [30].Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

conf_in(X) → U1(X, del2_in(X, Z))
del2_in(X, Y) → U4(X, Y, del_in(U, X, Z))
del_in(X, cons(H, T), cons(H, T1)) → U6(X, H, T, T1, del_in(X, T, T1))
del_in(X, cons(X, T), T) → del_out(X, cons(X, T), T)
U6(X, H, T, T1, del_out(X, T, T1)) → del_out(X, cons(H, T), cons(H, T1))
U4(X, Y, del_out(U, X, Z)) → U5(X, Y, del_in(V, Z, Y))
U5(X, Y, del_out(V, Z, Y)) → del2_out(X, Y)
U1(X, del2_out(X, Z)) → U2(X, del_in(U, Y, Z))
U2(X, del_out(U, Y, Z)) → U3(X, conf_in(Y))
U3(X, conf_out(Y)) → conf_out(X)

The argument filtering Pi contains the following mapping:
conf_in(x1)  =  conf_in
U1(x1, x2)  =  U1(x2)
del2_in(x1, x2)  =  del2_in
U4(x1, x2, x3)  =  U4(x3)
del_in(x1, x2, x3)  =  del_in
cons(x1, x2)  =  cons(x2)
U6(x1, x2, x3, x4, x5)  =  U6(x5)
del_out(x1, x2, x3)  =  del_out
U5(x1, x2, x3)  =  U5(x3)
del2_out(x1, x2)  =  del2_out
U2(x1, x2)  =  U2(x2)
U3(x1, x2)  =  U3(x2)
conf_out(x1)  =  conf_out

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog



↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
PiTRS
      ↳ DependencyPairsProof

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

conf_in(X) → U1(X, del2_in(X, Z))
del2_in(X, Y) → U4(X, Y, del_in(U, X, Z))
del_in(X, cons(H, T), cons(H, T1)) → U6(X, H, T, T1, del_in(X, T, T1))
del_in(X, cons(X, T), T) → del_out(X, cons(X, T), T)
U6(X, H, T, T1, del_out(X, T, T1)) → del_out(X, cons(H, T), cons(H, T1))
U4(X, Y, del_out(U, X, Z)) → U5(X, Y, del_in(V, Z, Y))
U5(X, Y, del_out(V, Z, Y)) → del2_out(X, Y)
U1(X, del2_out(X, Z)) → U2(X, del_in(U, Y, Z))
U2(X, del_out(U, Y, Z)) → U3(X, conf_in(Y))
U3(X, conf_out(Y)) → conf_out(X)

The argument filtering Pi contains the following mapping:
conf_in(x1)  =  conf_in
U1(x1, x2)  =  U1(x2)
del2_in(x1, x2)  =  del2_in
U4(x1, x2, x3)  =  U4(x3)
del_in(x1, x2, x3)  =  del_in
cons(x1, x2)  =  cons(x2)
U6(x1, x2, x3, x4, x5)  =  U6(x5)
del_out(x1, x2, x3)  =  del_out
U5(x1, x2, x3)  =  U5(x3)
del2_out(x1, x2)  =  del2_out
U2(x1, x2)  =  U2(x2)
U3(x1, x2)  =  U3(x2)
conf_out(x1)  =  conf_out


Using Dependency Pairs [1,30] we result in the following initial DP problem:
Pi DP problem:
The TRS P consists of the following rules:

CONF_IN(X) → U11(X, del2_in(X, Z))
CONF_IN(X) → DEL2_IN(X, Z)
DEL2_IN(X, Y) → U41(X, Y, del_in(U, X, Z))
DEL2_IN(X, Y) → DEL_IN(U, X, Z)
DEL_IN(X, cons(H, T), cons(H, T1)) → U61(X, H, T, T1, del_in(X, T, T1))
DEL_IN(X, cons(H, T), cons(H, T1)) → DEL_IN(X, T, T1)
U41(X, Y, del_out(U, X, Z)) → U51(X, Y, del_in(V, Z, Y))
U41(X, Y, del_out(U, X, Z)) → DEL_IN(V, Z, Y)
U11(X, del2_out(X, Z)) → U21(X, del_in(U, Y, Z))
U11(X, del2_out(X, Z)) → DEL_IN(U, Y, Z)
U21(X, del_out(U, Y, Z)) → U31(X, conf_in(Y))
U21(X, del_out(U, Y, Z)) → CONF_IN(Y)

The TRS R consists of the following rules:

conf_in(X) → U1(X, del2_in(X, Z))
del2_in(X, Y) → U4(X, Y, del_in(U, X, Z))
del_in(X, cons(H, T), cons(H, T1)) → U6(X, H, T, T1, del_in(X, T, T1))
del_in(X, cons(X, T), T) → del_out(X, cons(X, T), T)
U6(X, H, T, T1, del_out(X, T, T1)) → del_out(X, cons(H, T), cons(H, T1))
U4(X, Y, del_out(U, X, Z)) → U5(X, Y, del_in(V, Z, Y))
U5(X, Y, del_out(V, Z, Y)) → del2_out(X, Y)
U1(X, del2_out(X, Z)) → U2(X, del_in(U, Y, Z))
U2(X, del_out(U, Y, Z)) → U3(X, conf_in(Y))
U3(X, conf_out(Y)) → conf_out(X)

The argument filtering Pi contains the following mapping:
conf_in(x1)  =  conf_in
U1(x1, x2)  =  U1(x2)
del2_in(x1, x2)  =  del2_in
U4(x1, x2, x3)  =  U4(x3)
del_in(x1, x2, x3)  =  del_in
cons(x1, x2)  =  cons(x2)
U6(x1, x2, x3, x4, x5)  =  U6(x5)
del_out(x1, x2, x3)  =  del_out
U5(x1, x2, x3)  =  U5(x3)
del2_out(x1, x2)  =  del2_out
U2(x1, x2)  =  U2(x2)
U3(x1, x2)  =  U3(x2)
conf_out(x1)  =  conf_out
U51(x1, x2, x3)  =  U51(x3)
DEL_IN(x1, x2, x3)  =  DEL_IN
U41(x1, x2, x3)  =  U41(x3)
U11(x1, x2)  =  U11(x2)
CONF_IN(x1)  =  CONF_IN
U31(x1, x2)  =  U31(x2)
U21(x1, x2)  =  U21(x2)
U61(x1, x2, x3, x4, x5)  =  U61(x5)
DEL2_IN(x1, x2)  =  DEL2_IN

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

↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
PiDP
          ↳ DependencyGraphProof

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

CONF_IN(X) → U11(X, del2_in(X, Z))
CONF_IN(X) → DEL2_IN(X, Z)
DEL2_IN(X, Y) → U41(X, Y, del_in(U, X, Z))
DEL2_IN(X, Y) → DEL_IN(U, X, Z)
DEL_IN(X, cons(H, T), cons(H, T1)) → U61(X, H, T, T1, del_in(X, T, T1))
DEL_IN(X, cons(H, T), cons(H, T1)) → DEL_IN(X, T, T1)
U41(X, Y, del_out(U, X, Z)) → U51(X, Y, del_in(V, Z, Y))
U41(X, Y, del_out(U, X, Z)) → DEL_IN(V, Z, Y)
U11(X, del2_out(X, Z)) → U21(X, del_in(U, Y, Z))
U11(X, del2_out(X, Z)) → DEL_IN(U, Y, Z)
U21(X, del_out(U, Y, Z)) → U31(X, conf_in(Y))
U21(X, del_out(U, Y, Z)) → CONF_IN(Y)

The TRS R consists of the following rules:

conf_in(X) → U1(X, del2_in(X, Z))
del2_in(X, Y) → U4(X, Y, del_in(U, X, Z))
del_in(X, cons(H, T), cons(H, T1)) → U6(X, H, T, T1, del_in(X, T, T1))
del_in(X, cons(X, T), T) → del_out(X, cons(X, T), T)
U6(X, H, T, T1, del_out(X, T, T1)) → del_out(X, cons(H, T), cons(H, T1))
U4(X, Y, del_out(U, X, Z)) → U5(X, Y, del_in(V, Z, Y))
U5(X, Y, del_out(V, Z, Y)) → del2_out(X, Y)
U1(X, del2_out(X, Z)) → U2(X, del_in(U, Y, Z))
U2(X, del_out(U, Y, Z)) → U3(X, conf_in(Y))
U3(X, conf_out(Y)) → conf_out(X)

The argument filtering Pi contains the following mapping:
conf_in(x1)  =  conf_in
U1(x1, x2)  =  U1(x2)
del2_in(x1, x2)  =  del2_in
U4(x1, x2, x3)  =  U4(x3)
del_in(x1, x2, x3)  =  del_in
cons(x1, x2)  =  cons(x2)
U6(x1, x2, x3, x4, x5)  =  U6(x5)
del_out(x1, x2, x3)  =  del_out
U5(x1, x2, x3)  =  U5(x3)
del2_out(x1, x2)  =  del2_out
U2(x1, x2)  =  U2(x2)
U3(x1, x2)  =  U3(x2)
conf_out(x1)  =  conf_out
U51(x1, x2, x3)  =  U51(x3)
DEL_IN(x1, x2, x3)  =  DEL_IN
U41(x1, x2, x3)  =  U41(x3)
U11(x1, x2)  =  U11(x2)
CONF_IN(x1)  =  CONF_IN
U31(x1, x2)  =  U31(x2)
U21(x1, x2)  =  U21(x2)
U61(x1, x2, x3, x4, x5)  =  U61(x5)
DEL2_IN(x1, x2)  =  DEL2_IN

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 2 SCCs with 8 less nodes.

↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
PiDP
                ↳ UsableRulesProof
              ↳ PiDP

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

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

The TRS R consists of the following rules:

conf_in(X) → U1(X, del2_in(X, Z))
del2_in(X, Y) → U4(X, Y, del_in(U, X, Z))
del_in(X, cons(H, T), cons(H, T1)) → U6(X, H, T, T1, del_in(X, T, T1))
del_in(X, cons(X, T), T) → del_out(X, cons(X, T), T)
U6(X, H, T, T1, del_out(X, T, T1)) → del_out(X, cons(H, T), cons(H, T1))
U4(X, Y, del_out(U, X, Z)) → U5(X, Y, del_in(V, Z, Y))
U5(X, Y, del_out(V, Z, Y)) → del2_out(X, Y)
U1(X, del2_out(X, Z)) → U2(X, del_in(U, Y, Z))
U2(X, del_out(U, Y, Z)) → U3(X, conf_in(Y))
U3(X, conf_out(Y)) → conf_out(X)

The argument filtering Pi contains the following mapping:
conf_in(x1)  =  conf_in
U1(x1, x2)  =  U1(x2)
del2_in(x1, x2)  =  del2_in
U4(x1, x2, x3)  =  U4(x3)
del_in(x1, x2, x3)  =  del_in
cons(x1, x2)  =  cons(x2)
U6(x1, x2, x3, x4, x5)  =  U6(x5)
del_out(x1, x2, x3)  =  del_out
U5(x1, x2, x3)  =  U5(x3)
del2_out(x1, x2)  =  del2_out
U2(x1, x2)  =  U2(x2)
U3(x1, x2)  =  U3(x2)
conf_out(x1)  =  conf_out
DEL_IN(x1, x2, x3)  =  DEL_IN

We have to consider all (P,R,Pi)-chains
For (infinitary) constructor rewriting [30] we can delete all non-usable rules from R.

↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
                ↳ UsableRulesProof
PiDP
                    ↳ PiDPToQDPProof
              ↳ PiDP

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

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

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

We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem [30] into ordinary QDP problem [15] by application of Pi.

↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
QDP
                        ↳ NonTerminationProof
              ↳ PiDP

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

DEL_INDEL_IN

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.

The TRS P consists of the following rules:

DEL_INDEL_IN

The TRS R consists of the following rules:none


s = DEL_IN evaluates to t =DEL_IN

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




Rewriting sequence

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





↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
PiDP
                ↳ UsableRulesProof

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

U11(X, del2_out(X, Z)) → U21(X, del_in(U, Y, Z))
U21(X, del_out(U, Y, Z)) → CONF_IN(Y)
CONF_IN(X) → U11(X, del2_in(X, Z))

The TRS R consists of the following rules:

conf_in(X) → U1(X, del2_in(X, Z))
del2_in(X, Y) → U4(X, Y, del_in(U, X, Z))
del_in(X, cons(H, T), cons(H, T1)) → U6(X, H, T, T1, del_in(X, T, T1))
del_in(X, cons(X, T), T) → del_out(X, cons(X, T), T)
U6(X, H, T, T1, del_out(X, T, T1)) → del_out(X, cons(H, T), cons(H, T1))
U4(X, Y, del_out(U, X, Z)) → U5(X, Y, del_in(V, Z, Y))
U5(X, Y, del_out(V, Z, Y)) → del2_out(X, Y)
U1(X, del2_out(X, Z)) → U2(X, del_in(U, Y, Z))
U2(X, del_out(U, Y, Z)) → U3(X, conf_in(Y))
U3(X, conf_out(Y)) → conf_out(X)

The argument filtering Pi contains the following mapping:
conf_in(x1)  =  conf_in
U1(x1, x2)  =  U1(x2)
del2_in(x1, x2)  =  del2_in
U4(x1, x2, x3)  =  U4(x3)
del_in(x1, x2, x3)  =  del_in
cons(x1, x2)  =  cons(x2)
U6(x1, x2, x3, x4, x5)  =  U6(x5)
del_out(x1, x2, x3)  =  del_out
U5(x1, x2, x3)  =  U5(x3)
del2_out(x1, x2)  =  del2_out
U2(x1, x2)  =  U2(x2)
U3(x1, x2)  =  U3(x2)
conf_out(x1)  =  conf_out
U11(x1, x2)  =  U11(x2)
CONF_IN(x1)  =  CONF_IN
U21(x1, x2)  =  U21(x2)

We have to consider all (P,R,Pi)-chains
For (infinitary) constructor rewriting [30] we can delete all non-usable rules from R.

↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
PiDP
                    ↳ PiDPToQDPProof

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

U11(X, del2_out(X, Z)) → U21(X, del_in(U, Y, Z))
U21(X, del_out(U, Y, Z)) → CONF_IN(Y)
CONF_IN(X) → U11(X, del2_in(X, Z))

The TRS R consists of the following rules:

del_in(X, cons(H, T), cons(H, T1)) → U6(X, H, T, T1, del_in(X, T, T1))
del_in(X, cons(X, T), T) → del_out(X, cons(X, T), T)
del2_in(X, Y) → U4(X, Y, del_in(U, X, Z))
U6(X, H, T, T1, del_out(X, T, T1)) → del_out(X, cons(H, T), cons(H, T1))
U4(X, Y, del_out(U, X, Z)) → U5(X, Y, del_in(V, Z, Y))
U5(X, Y, del_out(V, Z, Y)) → del2_out(X, Y)

The argument filtering Pi contains the following mapping:
del2_in(x1, x2)  =  del2_in
U4(x1, x2, x3)  =  U4(x3)
del_in(x1, x2, x3)  =  del_in
cons(x1, x2)  =  cons(x2)
U6(x1, x2, x3, x4, x5)  =  U6(x5)
del_out(x1, x2, x3)  =  del_out
U5(x1, x2, x3)  =  U5(x3)
del2_out(x1, x2)  =  del2_out
U11(x1, x2)  =  U11(x2)
CONF_IN(x1)  =  CONF_IN
U21(x1, x2)  =  U21(x2)

We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem [30] into ordinary QDP problem [15] by application of Pi.

↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
QDP
                        ↳ Narrowing

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

CONF_INU11(del2_in)
U11(del2_out) → U21(del_in)
U21(del_out) → CONF_IN

The TRS R consists of the following rules:

del_inU6(del_in)
del_indel_out
del2_inU4(del_in)
U6(del_out) → del_out
U4(del_out) → U5(del_in)
U5(del_out) → del2_out

The set Q consists of the following terms:

del_in
del2_in
U6(x0)
U4(x0)
U5(x0)

We have to consider all (P,Q,R)-chains.
By narrowing [15] the rule U11(del2_out) → U21(del_in) at position [0] we obtained the following new rules:

U11(del2_out) → U21(del_out)
U11(del2_out) → U21(U6(del_in))



↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ Narrowing
QDP
                            ↳ Narrowing

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

CONF_INU11(del2_in)
U11(del2_out) → U21(del_out)
U21(del_out) → CONF_IN
U11(del2_out) → U21(U6(del_in))

The TRS R consists of the following rules:

del_inU6(del_in)
del_indel_out
del2_inU4(del_in)
U6(del_out) → del_out
U4(del_out) → U5(del_in)
U5(del_out) → del2_out

The set Q consists of the following terms:

del_in
del2_in
U6(x0)
U4(x0)
U5(x0)

We have to consider all (P,Q,R)-chains.
By narrowing [15] the rule CONF_INU11(del2_in) at position [0] we obtained the following new rules:

CONF_INU11(U4(del_in))



↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ Narrowing
QDP
                                ↳ UsableRulesProof

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

U11(del2_out) → U21(del_out)
U21(del_out) → CONF_IN
CONF_INU11(U4(del_in))
U11(del2_out) → U21(U6(del_in))

The TRS R consists of the following rules:

del_inU6(del_in)
del_indel_out
del2_inU4(del_in)
U6(del_out) → del_out
U4(del_out) → U5(del_in)
U5(del_out) → del2_out

The set Q consists of the following terms:

del_in
del2_in
U6(x0)
U4(x0)
U5(x0)

We have to consider all (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ Narrowing
                              ↳ QDP
                                ↳ UsableRulesProof
QDP
                                    ↳ QReductionProof

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

U11(del2_out) → U21(del_out)
U21(del_out) → CONF_IN
CONF_INU11(U4(del_in))
U11(del2_out) → U21(U6(del_in))

The TRS R consists of the following rules:

del_inU6(del_in)
del_indel_out
U4(del_out) → U5(del_in)
U5(del_out) → del2_out
U6(del_out) → del_out

The set Q consists of the following terms:

del_in
del2_in
U6(x0)
U4(x0)
U5(x0)

We have to consider all (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

del2_in



↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ Narrowing
                              ↳ QDP
                                ↳ UsableRulesProof
                                  ↳ QDP
                                    ↳ QReductionProof
QDP

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

U11(del2_out) → U21(del_out)
U21(del_out) → CONF_IN
U11(del2_out) → U21(U6(del_in))
CONF_INU11(U4(del_in))

The TRS R consists of the following rules:

del_inU6(del_in)
del_indel_out
U4(del_out) → U5(del_in)
U5(del_out) → del2_out
U6(del_out) → del_out

The set Q consists of the following terms:

del_in
U6(x0)
U4(x0)
U5(x0)

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