Left Termination of the query pattern p_in_1(a) w.r.t. the given Prolog program could not be shown:



↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof

Clauses:

p(X) :- ','(l(X), q(X)).
q(.(A, [])).
r(1).
l([]).
l(.(H, T)) :- ','(r(H), l(T)).

Queries:

p(a).

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:

p_in(X) → U1(X, l_in(X))
l_in(.(H, T)) → U3(H, T, r_in(H))
r_in(1) → r_out(1)
U3(H, T, r_out(H)) → U4(H, T, l_in(T))
l_in([]) → l_out([])
U4(H, T, l_out(T)) → l_out(.(H, T))
U1(X, l_out(X)) → U2(X, q_in(X))
q_in(.(A, [])) → q_out(.(A, []))
U2(X, q_out(X)) → p_out(X)

The argument filtering Pi contains the following mapping:
p_in(x1)  =  p_in
U1(x1, x2)  =  U1(x2)
l_in(x1)  =  l_in
.(x1, x2)  =  .(x1, x2)
U3(x1, x2, x3)  =  U3(x3)
r_in(x1)  =  r_in
1  =  1
r_out(x1)  =  r_out(x1)
U4(x1, x2, x3)  =  U4(x1, x3)
[]  =  []
l_out(x1)  =  l_out(x1)
U2(x1, x2)  =  U2(x1, x2)
q_in(x1)  =  q_in(x1)
q_out(x1)  =  q_out
p_out(x1)  =  p_out(x1)

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:

p_in(X) → U1(X, l_in(X))
l_in(.(H, T)) → U3(H, T, r_in(H))
r_in(1) → r_out(1)
U3(H, T, r_out(H)) → U4(H, T, l_in(T))
l_in([]) → l_out([])
U4(H, T, l_out(T)) → l_out(.(H, T))
U1(X, l_out(X)) → U2(X, q_in(X))
q_in(.(A, [])) → q_out(.(A, []))
U2(X, q_out(X)) → p_out(X)

The argument filtering Pi contains the following mapping:
p_in(x1)  =  p_in
U1(x1, x2)  =  U1(x2)
l_in(x1)  =  l_in
.(x1, x2)  =  .(x1, x2)
U3(x1, x2, x3)  =  U3(x3)
r_in(x1)  =  r_in
1  =  1
r_out(x1)  =  r_out(x1)
U4(x1, x2, x3)  =  U4(x1, x3)
[]  =  []
l_out(x1)  =  l_out(x1)
U2(x1, x2)  =  U2(x1, x2)
q_in(x1)  =  q_in(x1)
q_out(x1)  =  q_out
p_out(x1)  =  p_out(x1)


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

P_IN(X) → U11(X, l_in(X))
P_IN(X) → L_IN(X)
L_IN(.(H, T)) → U31(H, T, r_in(H))
L_IN(.(H, T)) → R_IN(H)
U31(H, T, r_out(H)) → U41(H, T, l_in(T))
U31(H, T, r_out(H)) → L_IN(T)
U11(X, l_out(X)) → U21(X, q_in(X))
U11(X, l_out(X)) → Q_IN(X)

The TRS R consists of the following rules:

p_in(X) → U1(X, l_in(X))
l_in(.(H, T)) → U3(H, T, r_in(H))
r_in(1) → r_out(1)
U3(H, T, r_out(H)) → U4(H, T, l_in(T))
l_in([]) → l_out([])
U4(H, T, l_out(T)) → l_out(.(H, T))
U1(X, l_out(X)) → U2(X, q_in(X))
q_in(.(A, [])) → q_out(.(A, []))
U2(X, q_out(X)) → p_out(X)

The argument filtering Pi contains the following mapping:
p_in(x1)  =  p_in
U1(x1, x2)  =  U1(x2)
l_in(x1)  =  l_in
.(x1, x2)  =  .(x1, x2)
U3(x1, x2, x3)  =  U3(x3)
r_in(x1)  =  r_in
1  =  1
r_out(x1)  =  r_out(x1)
U4(x1, x2, x3)  =  U4(x1, x3)
[]  =  []
l_out(x1)  =  l_out(x1)
U2(x1, x2)  =  U2(x1, x2)
q_in(x1)  =  q_in(x1)
q_out(x1)  =  q_out
p_out(x1)  =  p_out(x1)
P_IN(x1)  =  P_IN
U41(x1, x2, x3)  =  U41(x1, x3)
U31(x1, x2, x3)  =  U31(x3)
Q_IN(x1)  =  Q_IN(x1)
L_IN(x1)  =  L_IN
U11(x1, x2)  =  U11(x2)
U21(x1, x2)  =  U21(x1, x2)
R_IN(x1)  =  R_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:

P_IN(X) → U11(X, l_in(X))
P_IN(X) → L_IN(X)
L_IN(.(H, T)) → U31(H, T, r_in(H))
L_IN(.(H, T)) → R_IN(H)
U31(H, T, r_out(H)) → U41(H, T, l_in(T))
U31(H, T, r_out(H)) → L_IN(T)
U11(X, l_out(X)) → U21(X, q_in(X))
U11(X, l_out(X)) → Q_IN(X)

The TRS R consists of the following rules:

p_in(X) → U1(X, l_in(X))
l_in(.(H, T)) → U3(H, T, r_in(H))
r_in(1) → r_out(1)
U3(H, T, r_out(H)) → U4(H, T, l_in(T))
l_in([]) → l_out([])
U4(H, T, l_out(T)) → l_out(.(H, T))
U1(X, l_out(X)) → U2(X, q_in(X))
q_in(.(A, [])) → q_out(.(A, []))
U2(X, q_out(X)) → p_out(X)

The argument filtering Pi contains the following mapping:
p_in(x1)  =  p_in
U1(x1, x2)  =  U1(x2)
l_in(x1)  =  l_in
.(x1, x2)  =  .(x1, x2)
U3(x1, x2, x3)  =  U3(x3)
r_in(x1)  =  r_in
1  =  1
r_out(x1)  =  r_out(x1)
U4(x1, x2, x3)  =  U4(x1, x3)
[]  =  []
l_out(x1)  =  l_out(x1)
U2(x1, x2)  =  U2(x1, x2)
q_in(x1)  =  q_in(x1)
q_out(x1)  =  q_out
p_out(x1)  =  p_out(x1)
P_IN(x1)  =  P_IN
U41(x1, x2, x3)  =  U41(x1, x3)
U31(x1, x2, x3)  =  U31(x3)
Q_IN(x1)  =  Q_IN(x1)
L_IN(x1)  =  L_IN
U11(x1, x2)  =  U11(x2)
U21(x1, x2)  =  U21(x1, x2)
R_IN(x1)  =  R_IN

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 1 SCC with 6 less nodes.

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

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

L_IN(.(H, T)) → U31(H, T, r_in(H))
U31(H, T, r_out(H)) → L_IN(T)

The TRS R consists of the following rules:

p_in(X) → U1(X, l_in(X))
l_in(.(H, T)) → U3(H, T, r_in(H))
r_in(1) → r_out(1)
U3(H, T, r_out(H)) → U4(H, T, l_in(T))
l_in([]) → l_out([])
U4(H, T, l_out(T)) → l_out(.(H, T))
U1(X, l_out(X)) → U2(X, q_in(X))
q_in(.(A, [])) → q_out(.(A, []))
U2(X, q_out(X)) → p_out(X)

The argument filtering Pi contains the following mapping:
p_in(x1)  =  p_in
U1(x1, x2)  =  U1(x2)
l_in(x1)  =  l_in
.(x1, x2)  =  .(x1, x2)
U3(x1, x2, x3)  =  U3(x3)
r_in(x1)  =  r_in
1  =  1
r_out(x1)  =  r_out(x1)
U4(x1, x2, x3)  =  U4(x1, x3)
[]  =  []
l_out(x1)  =  l_out(x1)
U2(x1, x2)  =  U2(x1, x2)
q_in(x1)  =  q_in(x1)
q_out(x1)  =  q_out
p_out(x1)  =  p_out(x1)
U31(x1, x2, x3)  =  U31(x3)
L_IN(x1)  =  L_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
            ↳ PiDP
              ↳ UsableRulesProof
                ↳ PiDP
                  ↳ PiDPToQDPProof
  ↳ PrologToPiTRSProof

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

L_IN(.(H, T)) → U31(H, T, r_in(H))
U31(H, T, r_out(H)) → L_IN(T)

The TRS R consists of the following rules:

r_in(1) → r_out(1)

The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x1, x2)
r_in(x1)  =  r_in
1  =  1
r_out(x1)  =  r_out(x1)
U31(x1, x2, x3)  =  U31(x3)
L_IN(x1)  =  L_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
            ↳ PiDP
              ↳ UsableRulesProof
                ↳ PiDP
                  ↳ PiDPToQDPProof
                    ↳ QDP
                      ↳ Rewriting
  ↳ PrologToPiTRSProof

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

U31(r_out(H)) → L_IN
L_IN → U31(r_in)

The TRS R consists of the following rules:

r_in → r_out(1)

The set Q consists of the following terms:

r_in

We have to consider all (P,Q,R)-chains.
By rewriting [15] the rule L_IN → U31(r_in) at position [0] we obtained the following new rules:

L_IN → U31(r_out(1))



↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ PiDP
              ↳ UsableRulesProof
                ↳ PiDP
                  ↳ PiDPToQDPProof
                    ↳ QDP
                      ↳ Rewriting
                        ↳ QDP
                          ↳ UsableRulesProof
  ↳ PrologToPiTRSProof

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

U31(r_out(H)) → L_IN
L_IN → U31(r_out(1))

The TRS R consists of the following rules:

r_in → r_out(1)

The set Q consists of the following terms:

r_in

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
            ↳ PiDP
              ↳ UsableRulesProof
                ↳ PiDP
                  ↳ PiDPToQDPProof
                    ↳ QDP
                      ↳ Rewriting
                        ↳ QDP
                          ↳ UsableRulesProof
                            ↳ QDP
                              ↳ QReductionProof
  ↳ PrologToPiTRSProof

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

U31(r_out(H)) → L_IN
L_IN → U31(r_out(1))

R is empty.
The set Q consists of the following terms:

r_in

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.

r_in



↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ PiDP
              ↳ UsableRulesProof
                ↳ PiDP
                  ↳ PiDPToQDPProof
                    ↳ QDP
                      ↳ Rewriting
                        ↳ QDP
                          ↳ UsableRulesProof
                            ↳ QDP
                              ↳ QReductionProof
                                ↳ QDP
                                  ↳ Instantiation
  ↳ PrologToPiTRSProof

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

U31(r_out(H)) → L_IN
L_IN → U31(r_out(1))

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
By instantiating [15] the rule U31(r_out(H)) → L_IN we obtained the following new rules:

U31(r_out(1)) → L_IN



↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ PiDP
              ↳ UsableRulesProof
                ↳ PiDP
                  ↳ PiDPToQDPProof
                    ↳ QDP
                      ↳ Rewriting
                        ↳ QDP
                          ↳ UsableRulesProof
                            ↳ QDP
                              ↳ QReductionProof
                                ↳ QDP
                                  ↳ Instantiation
                                    ↳ QDP
                                      ↳ NonTerminationProof
  ↳ PrologToPiTRSProof

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

L_IN → U31(r_out(1))
U31(r_out(1)) → L_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 narrowing to the left:

The TRS P consists of the following rules:

L_IN → U31(r_out(1))
U31(r_out(1)) → L_IN

The TRS R consists of the following rules:none


s = U31(r_out(1)) evaluates to t =U31(r_out(1))

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



Rewriting sequence

U31(r_out(1)) → L_IN
with rule U31(r_out(1)) → L_IN at position [] and matcher [ ]

L_IN → U31(r_out(1))
with rule L_IN → U31(r_out(1))

Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence


All these steps are and every following step will be a correct step w.r.t to Q.




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:

p_in(X) → U1(X, l_in(X))
l_in(.(H, T)) → U3(H, T, r_in(H))
r_in(1) → r_out(1)
U3(H, T, r_out(H)) → U4(H, T, l_in(T))
l_in([]) → l_out([])
U4(H, T, l_out(T)) → l_out(.(H, T))
U1(X, l_out(X)) → U2(X, q_in(X))
q_in(.(A, [])) → q_out(.(A, []))
U2(X, q_out(X)) → p_out(X)

The argument filtering Pi contains the following mapping:
p_in(x1)  =  p_in
U1(x1, x2)  =  U1(x2)
l_in(x1)  =  l_in
.(x1, x2)  =  .(x1, x2)
U3(x1, x2, x3)  =  U3(x3)
r_in(x1)  =  r_in
1  =  1
r_out(x1)  =  r_out(x1)
U4(x1, x2, x3)  =  U4(x1, x3)
[]  =  []
l_out(x1)  =  l_out(x1)
U2(x1, x2)  =  U2(x1, x2)
q_in(x1)  =  q_in(x1)
q_out(x1)  =  q_out(x1)
p_out(x1)  =  p_out(x1)

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:

p_in(X) → U1(X, l_in(X))
l_in(.(H, T)) → U3(H, T, r_in(H))
r_in(1) → r_out(1)
U3(H, T, r_out(H)) → U4(H, T, l_in(T))
l_in([]) → l_out([])
U4(H, T, l_out(T)) → l_out(.(H, T))
U1(X, l_out(X)) → U2(X, q_in(X))
q_in(.(A, [])) → q_out(.(A, []))
U2(X, q_out(X)) → p_out(X)

The argument filtering Pi contains the following mapping:
p_in(x1)  =  p_in
U1(x1, x2)  =  U1(x2)
l_in(x1)  =  l_in
.(x1, x2)  =  .(x1, x2)
U3(x1, x2, x3)  =  U3(x3)
r_in(x1)  =  r_in
1  =  1
r_out(x1)  =  r_out(x1)
U4(x1, x2, x3)  =  U4(x1, x3)
[]  =  []
l_out(x1)  =  l_out(x1)
U2(x1, x2)  =  U2(x1, x2)
q_in(x1)  =  q_in(x1)
q_out(x1)  =  q_out(x1)
p_out(x1)  =  p_out(x1)


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

P_IN(X) → U11(X, l_in(X))
P_IN(X) → L_IN(X)
L_IN(.(H, T)) → U31(H, T, r_in(H))
L_IN(.(H, T)) → R_IN(H)
U31(H, T, r_out(H)) → U41(H, T, l_in(T))
U31(H, T, r_out(H)) → L_IN(T)
U11(X, l_out(X)) → U21(X, q_in(X))
U11(X, l_out(X)) → Q_IN(X)

The TRS R consists of the following rules:

p_in(X) → U1(X, l_in(X))
l_in(.(H, T)) → U3(H, T, r_in(H))
r_in(1) → r_out(1)
U3(H, T, r_out(H)) → U4(H, T, l_in(T))
l_in([]) → l_out([])
U4(H, T, l_out(T)) → l_out(.(H, T))
U1(X, l_out(X)) → U2(X, q_in(X))
q_in(.(A, [])) → q_out(.(A, []))
U2(X, q_out(X)) → p_out(X)

The argument filtering Pi contains the following mapping:
p_in(x1)  =  p_in
U1(x1, x2)  =  U1(x2)
l_in(x1)  =  l_in
.(x1, x2)  =  .(x1, x2)
U3(x1, x2, x3)  =  U3(x3)
r_in(x1)  =  r_in
1  =  1
r_out(x1)  =  r_out(x1)
U4(x1, x2, x3)  =  U4(x1, x3)
[]  =  []
l_out(x1)  =  l_out(x1)
U2(x1, x2)  =  U2(x1, x2)
q_in(x1)  =  q_in(x1)
q_out(x1)  =  q_out(x1)
p_out(x1)  =  p_out(x1)
P_IN(x1)  =  P_IN
U41(x1, x2, x3)  =  U41(x1, x3)
U31(x1, x2, x3)  =  U31(x3)
Q_IN(x1)  =  Q_IN(x1)
L_IN(x1)  =  L_IN
U11(x1, x2)  =  U11(x2)
U21(x1, x2)  =  U21(x1, x2)
R_IN(x1)  =  R_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:

P_IN(X) → U11(X, l_in(X))
P_IN(X) → L_IN(X)
L_IN(.(H, T)) → U31(H, T, r_in(H))
L_IN(.(H, T)) → R_IN(H)
U31(H, T, r_out(H)) → U41(H, T, l_in(T))
U31(H, T, r_out(H)) → L_IN(T)
U11(X, l_out(X)) → U21(X, q_in(X))
U11(X, l_out(X)) → Q_IN(X)

The TRS R consists of the following rules:

p_in(X) → U1(X, l_in(X))
l_in(.(H, T)) → U3(H, T, r_in(H))
r_in(1) → r_out(1)
U3(H, T, r_out(H)) → U4(H, T, l_in(T))
l_in([]) → l_out([])
U4(H, T, l_out(T)) → l_out(.(H, T))
U1(X, l_out(X)) → U2(X, q_in(X))
q_in(.(A, [])) → q_out(.(A, []))
U2(X, q_out(X)) → p_out(X)

The argument filtering Pi contains the following mapping:
p_in(x1)  =  p_in
U1(x1, x2)  =  U1(x2)
l_in(x1)  =  l_in
.(x1, x2)  =  .(x1, x2)
U3(x1, x2, x3)  =  U3(x3)
r_in(x1)  =  r_in
1  =  1
r_out(x1)  =  r_out(x1)
U4(x1, x2, x3)  =  U4(x1, x3)
[]  =  []
l_out(x1)  =  l_out(x1)
U2(x1, x2)  =  U2(x1, x2)
q_in(x1)  =  q_in(x1)
q_out(x1)  =  q_out(x1)
p_out(x1)  =  p_out(x1)
P_IN(x1)  =  P_IN
U41(x1, x2, x3)  =  U41(x1, x3)
U31(x1, x2, x3)  =  U31(x3)
Q_IN(x1)  =  Q_IN(x1)
L_IN(x1)  =  L_IN
U11(x1, x2)  =  U11(x2)
U21(x1, x2)  =  U21(x1, x2)
R_IN(x1)  =  R_IN

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 1 SCC with 6 less nodes.

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

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

L_IN(.(H, T)) → U31(H, T, r_in(H))
U31(H, T, r_out(H)) → L_IN(T)

The TRS R consists of the following rules:

p_in(X) → U1(X, l_in(X))
l_in(.(H, T)) → U3(H, T, r_in(H))
r_in(1) → r_out(1)
U3(H, T, r_out(H)) → U4(H, T, l_in(T))
l_in([]) → l_out([])
U4(H, T, l_out(T)) → l_out(.(H, T))
U1(X, l_out(X)) → U2(X, q_in(X))
q_in(.(A, [])) → q_out(.(A, []))
U2(X, q_out(X)) → p_out(X)

The argument filtering Pi contains the following mapping:
p_in(x1)  =  p_in
U1(x1, x2)  =  U1(x2)
l_in(x1)  =  l_in
.(x1, x2)  =  .(x1, x2)
U3(x1, x2, x3)  =  U3(x3)
r_in(x1)  =  r_in
1  =  1
r_out(x1)  =  r_out(x1)
U4(x1, x2, x3)  =  U4(x1, x3)
[]  =  []
l_out(x1)  =  l_out(x1)
U2(x1, x2)  =  U2(x1, x2)
q_in(x1)  =  q_in(x1)
q_out(x1)  =  q_out(x1)
p_out(x1)  =  p_out(x1)
U31(x1, x2, x3)  =  U31(x3)
L_IN(x1)  =  L_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
            ↳ PiDP
              ↳ UsableRulesProof
                ↳ PiDP
                  ↳ PiDPToQDPProof

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

L_IN(.(H, T)) → U31(H, T, r_in(H))
U31(H, T, r_out(H)) → L_IN(T)

The TRS R consists of the following rules:

r_in(1) → r_out(1)

The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x1, x2)
r_in(x1)  =  r_in
1  =  1
r_out(x1)  =  r_out(x1)
U31(x1, x2, x3)  =  U31(x3)
L_IN(x1)  =  L_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
            ↳ PiDP
              ↳ UsableRulesProof
                ↳ PiDP
                  ↳ PiDPToQDPProof
                    ↳ QDP
                      ↳ Rewriting

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

U31(r_out(H)) → L_IN
L_IN → U31(r_in)

The TRS R consists of the following rules:

r_in → r_out(1)

The set Q consists of the following terms:

r_in

We have to consider all (P,Q,R)-chains.
By rewriting [15] the rule L_IN → U31(r_in) at position [0] we obtained the following new rules:

L_IN → U31(r_out(1))



↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ PiDP
              ↳ UsableRulesProof
                ↳ PiDP
                  ↳ PiDPToQDPProof
                    ↳ QDP
                      ↳ Rewriting
                        ↳ QDP
                          ↳ UsableRulesProof

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

U31(r_out(H)) → L_IN
L_IN → U31(r_out(1))

The TRS R consists of the following rules:

r_in → r_out(1)

The set Q consists of the following terms:

r_in

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
            ↳ PiDP
              ↳ UsableRulesProof
                ↳ PiDP
                  ↳ PiDPToQDPProof
                    ↳ QDP
                      ↳ Rewriting
                        ↳ QDP
                          ↳ UsableRulesProof
                            ↳ QDP
                              ↳ QReductionProof

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

U31(r_out(H)) → L_IN
L_IN → U31(r_out(1))

R is empty.
The set Q consists of the following terms:

r_in

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.

r_in



↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ PiDP
              ↳ UsableRulesProof
                ↳ PiDP
                  ↳ PiDPToQDPProof
                    ↳ QDP
                      ↳ Rewriting
                        ↳ QDP
                          ↳ UsableRulesProof
                            ↳ QDP
                              ↳ QReductionProof
                                ↳ QDP
                                  ↳ Instantiation

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

U31(r_out(H)) → L_IN
L_IN → U31(r_out(1))

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
By instantiating [15] the rule U31(r_out(H)) → L_IN we obtained the following new rules:

U31(r_out(1)) → L_IN



↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ PiDP
              ↳ UsableRulesProof
                ↳ PiDP
                  ↳ PiDPToQDPProof
                    ↳ QDP
                      ↳ Rewriting
                        ↳ QDP
                          ↳ UsableRulesProof
                            ↳ QDP
                              ↳ QReductionProof
                                ↳ QDP
                                  ↳ Instantiation
                                    ↳ QDP
                                      ↳ NonTerminationProof

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

L_IN → U31(r_out(1))
U31(r_out(1)) → L_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 narrowing to the left:

The TRS P consists of the following rules:

L_IN → U31(r_out(1))
U31(r_out(1)) → L_IN

The TRS R consists of the following rules:none


s = U31(r_out(1)) evaluates to t =U31(r_out(1))

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



Rewriting sequence

U31(r_out(1)) → L_IN
with rule U31(r_out(1)) → L_IN at position [] and matcher [ ]

L_IN → U31(r_out(1))
with rule L_IN → U31(r_out(1))

Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence


All these steps are and every following step will be a correct step w.r.t to Q.