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]. With regard to the inferred argument filtering the predicates were used in the following modes:
p_in: (f)
l_in: (f)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

p_in_a(X) → U1_a(X, l_in_a(X))
l_in_a([]) → l_out_a([])
l_in_a(.(H, T)) → U3_a(H, T, r_in_a(H))
r_in_a(1) → r_out_a(1)
U3_a(H, T, r_out_a(H)) → U4_a(H, T, l_in_a(T))
U4_a(H, T, l_out_a(T)) → l_out_a(.(H, T))
U1_a(X, l_out_a(X)) → U2_a(X, q_in_g(X))
q_in_g(.(A, [])) → q_out_g(.(A, []))
U2_a(X, q_out_g(X)) → p_out_a(X)

The argument filtering Pi contains the following mapping:
p_in_a(x1)  =  p_in_a
U1_a(x1, x2)  =  U1_a(x2)
l_in_a(x1)  =  l_in_a
l_out_a(x1)  =  l_out_a(x1)
U3_a(x1, x2, x3)  =  U3_a(x3)
r_in_a(x1)  =  r_in_a
r_out_a(x1)  =  r_out_a(x1)
U4_a(x1, x2, x3)  =  U4_a(x1, x3)
U2_a(x1, x2)  =  U2_a(x1, x2)
q_in_g(x1)  =  q_in_g(x1)
.(x1, x2)  =  .(x1, x2)
[]  =  []
q_out_g(x1)  =  q_out_g
p_out_a(x1)  =  p_out_a(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_a(X) → U1_a(X, l_in_a(X))
l_in_a([]) → l_out_a([])
l_in_a(.(H, T)) → U3_a(H, T, r_in_a(H))
r_in_a(1) → r_out_a(1)
U3_a(H, T, r_out_a(H)) → U4_a(H, T, l_in_a(T))
U4_a(H, T, l_out_a(T)) → l_out_a(.(H, T))
U1_a(X, l_out_a(X)) → U2_a(X, q_in_g(X))
q_in_g(.(A, [])) → q_out_g(.(A, []))
U2_a(X, q_out_g(X)) → p_out_a(X)

The argument filtering Pi contains the following mapping:
p_in_a(x1)  =  p_in_a
U1_a(x1, x2)  =  U1_a(x2)
l_in_a(x1)  =  l_in_a
l_out_a(x1)  =  l_out_a(x1)
U3_a(x1, x2, x3)  =  U3_a(x3)
r_in_a(x1)  =  r_in_a
r_out_a(x1)  =  r_out_a(x1)
U4_a(x1, x2, x3)  =  U4_a(x1, x3)
U2_a(x1, x2)  =  U2_a(x1, x2)
q_in_g(x1)  =  q_in_g(x1)
.(x1, x2)  =  .(x1, x2)
[]  =  []
q_out_g(x1)  =  q_out_g
p_out_a(x1)  =  p_out_a(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_A(X) → U1_A(X, l_in_a(X))
P_IN_A(X) → L_IN_A(X)
L_IN_A(.(H, T)) → U3_A(H, T, r_in_a(H))
L_IN_A(.(H, T)) → R_IN_A(H)
U3_A(H, T, r_out_a(H)) → U4_A(H, T, l_in_a(T))
U3_A(H, T, r_out_a(H)) → L_IN_A(T)
U1_A(X, l_out_a(X)) → U2_A(X, q_in_g(X))
U1_A(X, l_out_a(X)) → Q_IN_G(X)

The TRS R consists of the following rules:

p_in_a(X) → U1_a(X, l_in_a(X))
l_in_a([]) → l_out_a([])
l_in_a(.(H, T)) → U3_a(H, T, r_in_a(H))
r_in_a(1) → r_out_a(1)
U3_a(H, T, r_out_a(H)) → U4_a(H, T, l_in_a(T))
U4_a(H, T, l_out_a(T)) → l_out_a(.(H, T))
U1_a(X, l_out_a(X)) → U2_a(X, q_in_g(X))
q_in_g(.(A, [])) → q_out_g(.(A, []))
U2_a(X, q_out_g(X)) → p_out_a(X)

The argument filtering Pi contains the following mapping:
p_in_a(x1)  =  p_in_a
U1_a(x1, x2)  =  U1_a(x2)
l_in_a(x1)  =  l_in_a
l_out_a(x1)  =  l_out_a(x1)
U3_a(x1, x2, x3)  =  U3_a(x3)
r_in_a(x1)  =  r_in_a
r_out_a(x1)  =  r_out_a(x1)
U4_a(x1, x2, x3)  =  U4_a(x1, x3)
U2_a(x1, x2)  =  U2_a(x1, x2)
q_in_g(x1)  =  q_in_g(x1)
.(x1, x2)  =  .(x1, x2)
[]  =  []
q_out_g(x1)  =  q_out_g
p_out_a(x1)  =  p_out_a(x1)
R_IN_A(x1)  =  R_IN_A
U3_A(x1, x2, x3)  =  U3_A(x3)
U1_A(x1, x2)  =  U1_A(x2)
U4_A(x1, x2, x3)  =  U4_A(x1, x3)
U2_A(x1, x2)  =  U2_A(x1, x2)
L_IN_A(x1)  =  L_IN_A
P_IN_A(x1)  =  P_IN_A
Q_IN_G(x1)  =  Q_IN_G(x1)

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_A(X) → U1_A(X, l_in_a(X))
P_IN_A(X) → L_IN_A(X)
L_IN_A(.(H, T)) → U3_A(H, T, r_in_a(H))
L_IN_A(.(H, T)) → R_IN_A(H)
U3_A(H, T, r_out_a(H)) → U4_A(H, T, l_in_a(T))
U3_A(H, T, r_out_a(H)) → L_IN_A(T)
U1_A(X, l_out_a(X)) → U2_A(X, q_in_g(X))
U1_A(X, l_out_a(X)) → Q_IN_G(X)

The TRS R consists of the following rules:

p_in_a(X) → U1_a(X, l_in_a(X))
l_in_a([]) → l_out_a([])
l_in_a(.(H, T)) → U3_a(H, T, r_in_a(H))
r_in_a(1) → r_out_a(1)
U3_a(H, T, r_out_a(H)) → U4_a(H, T, l_in_a(T))
U4_a(H, T, l_out_a(T)) → l_out_a(.(H, T))
U1_a(X, l_out_a(X)) → U2_a(X, q_in_g(X))
q_in_g(.(A, [])) → q_out_g(.(A, []))
U2_a(X, q_out_g(X)) → p_out_a(X)

The argument filtering Pi contains the following mapping:
p_in_a(x1)  =  p_in_a
U1_a(x1, x2)  =  U1_a(x2)
l_in_a(x1)  =  l_in_a
l_out_a(x1)  =  l_out_a(x1)
U3_a(x1, x2, x3)  =  U3_a(x3)
r_in_a(x1)  =  r_in_a
r_out_a(x1)  =  r_out_a(x1)
U4_a(x1, x2, x3)  =  U4_a(x1, x3)
U2_a(x1, x2)  =  U2_a(x1, x2)
q_in_g(x1)  =  q_in_g(x1)
.(x1, x2)  =  .(x1, x2)
[]  =  []
q_out_g(x1)  =  q_out_g
p_out_a(x1)  =  p_out_a(x1)
R_IN_A(x1)  =  R_IN_A
U3_A(x1, x2, x3)  =  U3_A(x3)
U1_A(x1, x2)  =  U1_A(x2)
U4_A(x1, x2, x3)  =  U4_A(x1, x3)
U2_A(x1, x2)  =  U2_A(x1, x2)
L_IN_A(x1)  =  L_IN_A
P_IN_A(x1)  =  P_IN_A
Q_IN_G(x1)  =  Q_IN_G(x1)

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:

U3_A(H, T, r_out_a(H)) → L_IN_A(T)
L_IN_A(.(H, T)) → U3_A(H, T, r_in_a(H))

The TRS R consists of the following rules:

p_in_a(X) → U1_a(X, l_in_a(X))
l_in_a([]) → l_out_a([])
l_in_a(.(H, T)) → U3_a(H, T, r_in_a(H))
r_in_a(1) → r_out_a(1)
U3_a(H, T, r_out_a(H)) → U4_a(H, T, l_in_a(T))
U4_a(H, T, l_out_a(T)) → l_out_a(.(H, T))
U1_a(X, l_out_a(X)) → U2_a(X, q_in_g(X))
q_in_g(.(A, [])) → q_out_g(.(A, []))
U2_a(X, q_out_g(X)) → p_out_a(X)

The argument filtering Pi contains the following mapping:
p_in_a(x1)  =  p_in_a
U1_a(x1, x2)  =  U1_a(x2)
l_in_a(x1)  =  l_in_a
l_out_a(x1)  =  l_out_a(x1)
U3_a(x1, x2, x3)  =  U3_a(x3)
r_in_a(x1)  =  r_in_a
r_out_a(x1)  =  r_out_a(x1)
U4_a(x1, x2, x3)  =  U4_a(x1, x3)
U2_a(x1, x2)  =  U2_a(x1, x2)
q_in_g(x1)  =  q_in_g(x1)
.(x1, x2)  =  .(x1, x2)
[]  =  []
q_out_g(x1)  =  q_out_g
p_out_a(x1)  =  p_out_a(x1)
U3_A(x1, x2, x3)  =  U3_A(x3)
L_IN_A(x1)  =  L_IN_A

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:

U3_A(H, T, r_out_a(H)) → L_IN_A(T)
L_IN_A(.(H, T)) → U3_A(H, T, r_in_a(H))

The TRS R consists of the following rules:

r_in_a(1) → r_out_a(1)

The argument filtering Pi contains the following mapping:
r_in_a(x1)  =  r_in_a
r_out_a(x1)  =  r_out_a(x1)
.(x1, x2)  =  .(x1, x2)
U3_A(x1, x2, x3)  =  U3_A(x3)
L_IN_A(x1)  =  L_IN_A

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:

L_IN_A → U3_A(r_in_a)
U3_A(r_out_a(H)) → L_IN_A

The TRS R consists of the following rules:

r_in_a → r_out_a(1)

The set Q consists of the following terms:

r_in_a

We have to consider all (P,Q,R)-chains.
By rewriting [15] the rule L_IN_A → U3_A(r_in_a) at position [0] we obtained the following new rules:

L_IN_A → U3_A(r_out_a(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:

L_IN_A → U3_A(r_out_a(1))
U3_A(r_out_a(H)) → L_IN_A

The TRS R consists of the following rules:

r_in_a → r_out_a(1)

The set Q consists of the following terms:

r_in_a

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:

L_IN_A → U3_A(r_out_a(1))
U3_A(r_out_a(H)) → L_IN_A

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

r_in_a

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_a



↳ 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:

L_IN_A → U3_A(r_out_a(1))
U3_A(r_out_a(H)) → L_IN_A

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
By instantiating [15] the rule U3_A(r_out_a(H)) → L_IN_A we obtained the following new rules:

U3_A(r_out_a(1)) → L_IN_A



↳ 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_A → U3_A(r_out_a(1))
U3_A(r_out_a(1)) → L_IN_A

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_A → U3_A(r_out_a(1))
U3_A(r_out_a(1)) → L_IN_A

The TRS R consists of the following rules:none


s = U3_A(r_out_a(1)) evaluates to t =U3_A(r_out_a(1))

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



Rewriting sequence

U3_A(r_out_a(1)) → L_IN_A
with rule U3_A(r_out_a(1)) → L_IN_A at position [] and matcher [ ]

L_IN_A → U3_A(r_out_a(1))
with rule L_IN_A → U3_A(r_out_a(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]. With regard to the inferred argument filtering the predicates were used in the following modes:
p_in: (f)
l_in: (f)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

p_in_a(X) → U1_a(X, l_in_a(X))
l_in_a([]) → l_out_a([])
l_in_a(.(H, T)) → U3_a(H, T, r_in_a(H))
r_in_a(1) → r_out_a(1)
U3_a(H, T, r_out_a(H)) → U4_a(H, T, l_in_a(T))
U4_a(H, T, l_out_a(T)) → l_out_a(.(H, T))
U1_a(X, l_out_a(X)) → U2_a(X, q_in_g(X))
q_in_g(.(A, [])) → q_out_g(.(A, []))
U2_a(X, q_out_g(X)) → p_out_a(X)

The argument filtering Pi contains the following mapping:
p_in_a(x1)  =  p_in_a
U1_a(x1, x2)  =  U1_a(x2)
l_in_a(x1)  =  l_in_a
l_out_a(x1)  =  l_out_a(x1)
U3_a(x1, x2, x3)  =  U3_a(x3)
r_in_a(x1)  =  r_in_a
r_out_a(x1)  =  r_out_a(x1)
U4_a(x1, x2, x3)  =  U4_a(x1, x3)
U2_a(x1, x2)  =  U2_a(x1, x2)
q_in_g(x1)  =  q_in_g(x1)
.(x1, x2)  =  .(x1, x2)
[]  =  []
q_out_g(x1)  =  q_out_g(x1)
p_out_a(x1)  =  p_out_a(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_a(X) → U1_a(X, l_in_a(X))
l_in_a([]) → l_out_a([])
l_in_a(.(H, T)) → U3_a(H, T, r_in_a(H))
r_in_a(1) → r_out_a(1)
U3_a(H, T, r_out_a(H)) → U4_a(H, T, l_in_a(T))
U4_a(H, T, l_out_a(T)) → l_out_a(.(H, T))
U1_a(X, l_out_a(X)) → U2_a(X, q_in_g(X))
q_in_g(.(A, [])) → q_out_g(.(A, []))
U2_a(X, q_out_g(X)) → p_out_a(X)

The argument filtering Pi contains the following mapping:
p_in_a(x1)  =  p_in_a
U1_a(x1, x2)  =  U1_a(x2)
l_in_a(x1)  =  l_in_a
l_out_a(x1)  =  l_out_a(x1)
U3_a(x1, x2, x3)  =  U3_a(x3)
r_in_a(x1)  =  r_in_a
r_out_a(x1)  =  r_out_a(x1)
U4_a(x1, x2, x3)  =  U4_a(x1, x3)
U2_a(x1, x2)  =  U2_a(x1, x2)
q_in_g(x1)  =  q_in_g(x1)
.(x1, x2)  =  .(x1, x2)
[]  =  []
q_out_g(x1)  =  q_out_g(x1)
p_out_a(x1)  =  p_out_a(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_A(X) → U1_A(X, l_in_a(X))
P_IN_A(X) → L_IN_A(X)
L_IN_A(.(H, T)) → U3_A(H, T, r_in_a(H))
L_IN_A(.(H, T)) → R_IN_A(H)
U3_A(H, T, r_out_a(H)) → U4_A(H, T, l_in_a(T))
U3_A(H, T, r_out_a(H)) → L_IN_A(T)
U1_A(X, l_out_a(X)) → U2_A(X, q_in_g(X))
U1_A(X, l_out_a(X)) → Q_IN_G(X)

The TRS R consists of the following rules:

p_in_a(X) → U1_a(X, l_in_a(X))
l_in_a([]) → l_out_a([])
l_in_a(.(H, T)) → U3_a(H, T, r_in_a(H))
r_in_a(1) → r_out_a(1)
U3_a(H, T, r_out_a(H)) → U4_a(H, T, l_in_a(T))
U4_a(H, T, l_out_a(T)) → l_out_a(.(H, T))
U1_a(X, l_out_a(X)) → U2_a(X, q_in_g(X))
q_in_g(.(A, [])) → q_out_g(.(A, []))
U2_a(X, q_out_g(X)) → p_out_a(X)

The argument filtering Pi contains the following mapping:
p_in_a(x1)  =  p_in_a
U1_a(x1, x2)  =  U1_a(x2)
l_in_a(x1)  =  l_in_a
l_out_a(x1)  =  l_out_a(x1)
U3_a(x1, x2, x3)  =  U3_a(x3)
r_in_a(x1)  =  r_in_a
r_out_a(x1)  =  r_out_a(x1)
U4_a(x1, x2, x3)  =  U4_a(x1, x3)
U2_a(x1, x2)  =  U2_a(x1, x2)
q_in_g(x1)  =  q_in_g(x1)
.(x1, x2)  =  .(x1, x2)
[]  =  []
q_out_g(x1)  =  q_out_g(x1)
p_out_a(x1)  =  p_out_a(x1)
R_IN_A(x1)  =  R_IN_A
U3_A(x1, x2, x3)  =  U3_A(x3)
U1_A(x1, x2)  =  U1_A(x2)
U4_A(x1, x2, x3)  =  U4_A(x1, x3)
U2_A(x1, x2)  =  U2_A(x1, x2)
L_IN_A(x1)  =  L_IN_A
P_IN_A(x1)  =  P_IN_A
Q_IN_G(x1)  =  Q_IN_G(x1)

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_A(X) → U1_A(X, l_in_a(X))
P_IN_A(X) → L_IN_A(X)
L_IN_A(.(H, T)) → U3_A(H, T, r_in_a(H))
L_IN_A(.(H, T)) → R_IN_A(H)
U3_A(H, T, r_out_a(H)) → U4_A(H, T, l_in_a(T))
U3_A(H, T, r_out_a(H)) → L_IN_A(T)
U1_A(X, l_out_a(X)) → U2_A(X, q_in_g(X))
U1_A(X, l_out_a(X)) → Q_IN_G(X)

The TRS R consists of the following rules:

p_in_a(X) → U1_a(X, l_in_a(X))
l_in_a([]) → l_out_a([])
l_in_a(.(H, T)) → U3_a(H, T, r_in_a(H))
r_in_a(1) → r_out_a(1)
U3_a(H, T, r_out_a(H)) → U4_a(H, T, l_in_a(T))
U4_a(H, T, l_out_a(T)) → l_out_a(.(H, T))
U1_a(X, l_out_a(X)) → U2_a(X, q_in_g(X))
q_in_g(.(A, [])) → q_out_g(.(A, []))
U2_a(X, q_out_g(X)) → p_out_a(X)

The argument filtering Pi contains the following mapping:
p_in_a(x1)  =  p_in_a
U1_a(x1, x2)  =  U1_a(x2)
l_in_a(x1)  =  l_in_a
l_out_a(x1)  =  l_out_a(x1)
U3_a(x1, x2, x3)  =  U3_a(x3)
r_in_a(x1)  =  r_in_a
r_out_a(x1)  =  r_out_a(x1)
U4_a(x1, x2, x3)  =  U4_a(x1, x3)
U2_a(x1, x2)  =  U2_a(x1, x2)
q_in_g(x1)  =  q_in_g(x1)
.(x1, x2)  =  .(x1, x2)
[]  =  []
q_out_g(x1)  =  q_out_g(x1)
p_out_a(x1)  =  p_out_a(x1)
R_IN_A(x1)  =  R_IN_A
U3_A(x1, x2, x3)  =  U3_A(x3)
U1_A(x1, x2)  =  U1_A(x2)
U4_A(x1, x2, x3)  =  U4_A(x1, x3)
U2_A(x1, x2)  =  U2_A(x1, x2)
L_IN_A(x1)  =  L_IN_A
P_IN_A(x1)  =  P_IN_A
Q_IN_G(x1)  =  Q_IN_G(x1)

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:

U3_A(H, T, r_out_a(H)) → L_IN_A(T)
L_IN_A(.(H, T)) → U3_A(H, T, r_in_a(H))

The TRS R consists of the following rules:

p_in_a(X) → U1_a(X, l_in_a(X))
l_in_a([]) → l_out_a([])
l_in_a(.(H, T)) → U3_a(H, T, r_in_a(H))
r_in_a(1) → r_out_a(1)
U3_a(H, T, r_out_a(H)) → U4_a(H, T, l_in_a(T))
U4_a(H, T, l_out_a(T)) → l_out_a(.(H, T))
U1_a(X, l_out_a(X)) → U2_a(X, q_in_g(X))
q_in_g(.(A, [])) → q_out_g(.(A, []))
U2_a(X, q_out_g(X)) → p_out_a(X)

The argument filtering Pi contains the following mapping:
p_in_a(x1)  =  p_in_a
U1_a(x1, x2)  =  U1_a(x2)
l_in_a(x1)  =  l_in_a
l_out_a(x1)  =  l_out_a(x1)
U3_a(x1, x2, x3)  =  U3_a(x3)
r_in_a(x1)  =  r_in_a
r_out_a(x1)  =  r_out_a(x1)
U4_a(x1, x2, x3)  =  U4_a(x1, x3)
U2_a(x1, x2)  =  U2_a(x1, x2)
q_in_g(x1)  =  q_in_g(x1)
.(x1, x2)  =  .(x1, x2)
[]  =  []
q_out_g(x1)  =  q_out_g(x1)
p_out_a(x1)  =  p_out_a(x1)
U3_A(x1, x2, x3)  =  U3_A(x3)
L_IN_A(x1)  =  L_IN_A

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:

U3_A(H, T, r_out_a(H)) → L_IN_A(T)
L_IN_A(.(H, T)) → U3_A(H, T, r_in_a(H))

The TRS R consists of the following rules:

r_in_a(1) → r_out_a(1)

The argument filtering Pi contains the following mapping:
r_in_a(x1)  =  r_in_a
r_out_a(x1)  =  r_out_a(x1)
.(x1, x2)  =  .(x1, x2)
U3_A(x1, x2, x3)  =  U3_A(x3)
L_IN_A(x1)  =  L_IN_A

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:

L_IN_A → U3_A(r_in_a)
U3_A(r_out_a(H)) → L_IN_A

The TRS R consists of the following rules:

r_in_a → r_out_a(1)

The set Q consists of the following terms:

r_in_a

We have to consider all (P,Q,R)-chains.
By rewriting [15] the rule L_IN_A → U3_A(r_in_a) at position [0] we obtained the following new rules:

L_IN_A → U3_A(r_out_a(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:

L_IN_A → U3_A(r_out_a(1))
U3_A(r_out_a(H)) → L_IN_A

The TRS R consists of the following rules:

r_in_a → r_out_a(1)

The set Q consists of the following terms:

r_in_a

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:

L_IN_A → U3_A(r_out_a(1))
U3_A(r_out_a(H)) → L_IN_A

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

r_in_a

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_a



↳ 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:

L_IN_A → U3_A(r_out_a(1))
U3_A(r_out_a(H)) → L_IN_A

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
By instantiating [15] the rule U3_A(r_out_a(H)) → L_IN_A we obtained the following new rules:

U3_A(r_out_a(1)) → L_IN_A



↳ 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_A → U3_A(r_out_a(1))
U3_A(r_out_a(1)) → L_IN_A

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_A → U3_A(r_out_a(1))
U3_A(r_out_a(1)) → L_IN_A

The TRS R consists of the following rules:none


s = U3_A(r_out_a(1)) evaluates to t =U3_A(r_out_a(1))

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



Rewriting sequence

U3_A(r_out_a(1)) → L_IN_A
with rule U3_A(r_out_a(1)) → L_IN_A at position [] and matcher [ ]

L_IN_A → U3_A(r_out_a(1))
with rule L_IN_A → U3_A(r_out_a(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.