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



↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof

Clauses:

p(X) :- ','(q(f(Y)), p(Y)).
p(g(X)) :- p(X).
q(g(Y)).

Queries:

p(g).

We use the technique of [30]. With regard to the inferred argument filtering the predicates were used in the following modes:
p_in: (b) (f)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

p_in_g(X) → U1_g(X, q_in_a(f(Y)))
q_in_a(g(Y)) → q_out_a(g(Y))
U1_g(X, q_out_a(f(Y))) → U2_g(X, p_in_a(Y))
p_in_a(X) → U1_a(X, q_in_a(f(Y)))
U1_a(X, q_out_a(f(Y))) → U2_a(X, p_in_a(Y))
p_in_a(g(X)) → U3_a(X, p_in_a(X))
U3_a(X, p_out_a(X)) → p_out_a(g(X))
U2_a(X, p_out_a(Y)) → p_out_a(X)
U2_g(X, p_out_a(Y)) → p_out_g(X)
p_in_g(g(X)) → U3_g(X, p_in_g(X))
U3_g(X, p_out_g(X)) → p_out_g(g(X))

The argument filtering Pi contains the following mapping:
p_in_g(x1)  =  p_in_g(x1)
U1_g(x1, x2)  =  U1_g(x2)
q_in_a(x1)  =  q_in_a
q_out_a(x1)  =  q_out_a
U2_g(x1, x2)  =  U2_g(x2)
p_in_a(x1)  =  p_in_a
U1_a(x1, x2)  =  U1_a(x2)
U2_a(x1, x2)  =  U2_a(x2)
U3_a(x1, x2)  =  U3_a(x2)
p_out_a(x1)  =  p_out_a
p_out_g(x1)  =  p_out_g
g(x1)  =  g(x1)
U3_g(x1, x2)  =  U3_g(x2)

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_g(X) → U1_g(X, q_in_a(f(Y)))
q_in_a(g(Y)) → q_out_a(g(Y))
U1_g(X, q_out_a(f(Y))) → U2_g(X, p_in_a(Y))
p_in_a(X) → U1_a(X, q_in_a(f(Y)))
U1_a(X, q_out_a(f(Y))) → U2_a(X, p_in_a(Y))
p_in_a(g(X)) → U3_a(X, p_in_a(X))
U3_a(X, p_out_a(X)) → p_out_a(g(X))
U2_a(X, p_out_a(Y)) → p_out_a(X)
U2_g(X, p_out_a(Y)) → p_out_g(X)
p_in_g(g(X)) → U3_g(X, p_in_g(X))
U3_g(X, p_out_g(X)) → p_out_g(g(X))

The argument filtering Pi contains the following mapping:
p_in_g(x1)  =  p_in_g(x1)
U1_g(x1, x2)  =  U1_g(x2)
q_in_a(x1)  =  q_in_a
q_out_a(x1)  =  q_out_a
U2_g(x1, x2)  =  U2_g(x2)
p_in_a(x1)  =  p_in_a
U1_a(x1, x2)  =  U1_a(x2)
U2_a(x1, x2)  =  U2_a(x2)
U3_a(x1, x2)  =  U3_a(x2)
p_out_a(x1)  =  p_out_a
p_out_g(x1)  =  p_out_g
g(x1)  =  g(x1)
U3_g(x1, x2)  =  U3_g(x2)


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_G(X) → U1_G(X, q_in_a(f(Y)))
P_IN_G(X) → Q_IN_A(f(Y))
U1_G(X, q_out_a(f(Y))) → U2_G(X, p_in_a(Y))
U1_G(X, q_out_a(f(Y))) → P_IN_A(Y)
P_IN_A(X) → U1_A(X, q_in_a(f(Y)))
P_IN_A(X) → Q_IN_A(f(Y))
U1_A(X, q_out_a(f(Y))) → U2_A(X, p_in_a(Y))
U1_A(X, q_out_a(f(Y))) → P_IN_A(Y)
P_IN_A(g(X)) → U3_A(X, p_in_a(X))
P_IN_A(g(X)) → P_IN_A(X)
P_IN_G(g(X)) → U3_G(X, p_in_g(X))
P_IN_G(g(X)) → P_IN_G(X)

The TRS R consists of the following rules:

p_in_g(X) → U1_g(X, q_in_a(f(Y)))
q_in_a(g(Y)) → q_out_a(g(Y))
U1_g(X, q_out_a(f(Y))) → U2_g(X, p_in_a(Y))
p_in_a(X) → U1_a(X, q_in_a(f(Y)))
U1_a(X, q_out_a(f(Y))) → U2_a(X, p_in_a(Y))
p_in_a(g(X)) → U3_a(X, p_in_a(X))
U3_a(X, p_out_a(X)) → p_out_a(g(X))
U2_a(X, p_out_a(Y)) → p_out_a(X)
U2_g(X, p_out_a(Y)) → p_out_g(X)
p_in_g(g(X)) → U3_g(X, p_in_g(X))
U3_g(X, p_out_g(X)) → p_out_g(g(X))

The argument filtering Pi contains the following mapping:
p_in_g(x1)  =  p_in_g(x1)
U1_g(x1, x2)  =  U1_g(x2)
q_in_a(x1)  =  q_in_a
q_out_a(x1)  =  q_out_a
U2_g(x1, x2)  =  U2_g(x2)
p_in_a(x1)  =  p_in_a
U1_a(x1, x2)  =  U1_a(x2)
U2_a(x1, x2)  =  U2_a(x2)
U3_a(x1, x2)  =  U3_a(x2)
p_out_a(x1)  =  p_out_a
p_out_g(x1)  =  p_out_g
g(x1)  =  g(x1)
U3_g(x1, x2)  =  U3_g(x2)
U2_G(x1, x2)  =  U2_G(x2)
U1_A(x1, x2)  =  U1_A(x2)
P_IN_G(x1)  =  P_IN_G(x1)
U3_A(x1, x2)  =  U3_A(x2)
U2_A(x1, x2)  =  U2_A(x2)
P_IN_A(x1)  =  P_IN_A
U1_G(x1, x2)  =  U1_G(x2)
U3_G(x1, x2)  =  U3_G(x2)
Q_IN_A(x1)  =  Q_IN_A

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_G(X) → U1_G(X, q_in_a(f(Y)))
P_IN_G(X) → Q_IN_A(f(Y))
U1_G(X, q_out_a(f(Y))) → U2_G(X, p_in_a(Y))
U1_G(X, q_out_a(f(Y))) → P_IN_A(Y)
P_IN_A(X) → U1_A(X, q_in_a(f(Y)))
P_IN_A(X) → Q_IN_A(f(Y))
U1_A(X, q_out_a(f(Y))) → U2_A(X, p_in_a(Y))
U1_A(X, q_out_a(f(Y))) → P_IN_A(Y)
P_IN_A(g(X)) → U3_A(X, p_in_a(X))
P_IN_A(g(X)) → P_IN_A(X)
P_IN_G(g(X)) → U3_G(X, p_in_g(X))
P_IN_G(g(X)) → P_IN_G(X)

The TRS R consists of the following rules:

p_in_g(X) → U1_g(X, q_in_a(f(Y)))
q_in_a(g(Y)) → q_out_a(g(Y))
U1_g(X, q_out_a(f(Y))) → U2_g(X, p_in_a(Y))
p_in_a(X) → U1_a(X, q_in_a(f(Y)))
U1_a(X, q_out_a(f(Y))) → U2_a(X, p_in_a(Y))
p_in_a(g(X)) → U3_a(X, p_in_a(X))
U3_a(X, p_out_a(X)) → p_out_a(g(X))
U2_a(X, p_out_a(Y)) → p_out_a(X)
U2_g(X, p_out_a(Y)) → p_out_g(X)
p_in_g(g(X)) → U3_g(X, p_in_g(X))
U3_g(X, p_out_g(X)) → p_out_g(g(X))

The argument filtering Pi contains the following mapping:
p_in_g(x1)  =  p_in_g(x1)
U1_g(x1, x2)  =  U1_g(x2)
q_in_a(x1)  =  q_in_a
q_out_a(x1)  =  q_out_a
U2_g(x1, x2)  =  U2_g(x2)
p_in_a(x1)  =  p_in_a
U1_a(x1, x2)  =  U1_a(x2)
U2_a(x1, x2)  =  U2_a(x2)
U3_a(x1, x2)  =  U3_a(x2)
p_out_a(x1)  =  p_out_a
p_out_g(x1)  =  p_out_g
g(x1)  =  g(x1)
U3_g(x1, x2)  =  U3_g(x2)
U2_G(x1, x2)  =  U2_G(x2)
U1_A(x1, x2)  =  U1_A(x2)
P_IN_G(x1)  =  P_IN_G(x1)
U3_A(x1, x2)  =  U3_A(x2)
U2_A(x1, x2)  =  U2_A(x2)
P_IN_A(x1)  =  P_IN_A
U1_G(x1, x2)  =  U1_G(x2)
U3_G(x1, x2)  =  U3_G(x2)
Q_IN_A(x1)  =  Q_IN_A

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 2 SCCs with 10 less nodes.

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

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

P_IN_A(g(X)) → P_IN_A(X)

The TRS R consists of the following rules:

p_in_g(X) → U1_g(X, q_in_a(f(Y)))
q_in_a(g(Y)) → q_out_a(g(Y))
U1_g(X, q_out_a(f(Y))) → U2_g(X, p_in_a(Y))
p_in_a(X) → U1_a(X, q_in_a(f(Y)))
U1_a(X, q_out_a(f(Y))) → U2_a(X, p_in_a(Y))
p_in_a(g(X)) → U3_a(X, p_in_a(X))
U3_a(X, p_out_a(X)) → p_out_a(g(X))
U2_a(X, p_out_a(Y)) → p_out_a(X)
U2_g(X, p_out_a(Y)) → p_out_g(X)
p_in_g(g(X)) → U3_g(X, p_in_g(X))
U3_g(X, p_out_g(X)) → p_out_g(g(X))

The argument filtering Pi contains the following mapping:
p_in_g(x1)  =  p_in_g(x1)
U1_g(x1, x2)  =  U1_g(x2)
q_in_a(x1)  =  q_in_a
q_out_a(x1)  =  q_out_a
U2_g(x1, x2)  =  U2_g(x2)
p_in_a(x1)  =  p_in_a
U1_a(x1, x2)  =  U1_a(x2)
U2_a(x1, x2)  =  U2_a(x2)
U3_a(x1, x2)  =  U3_a(x2)
p_out_a(x1)  =  p_out_a
p_out_g(x1)  =  p_out_g
g(x1)  =  g(x1)
U3_g(x1, x2)  =  U3_g(x2)
P_IN_A(x1)  =  P_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
            ↳ AND
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
              ↳ PiDP
  ↳ PrologToPiTRSProof

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

P_IN_A(g(X)) → P_IN_A(X)

R is empty.
The argument filtering Pi contains the following mapping:
g(x1)  =  g(x1)
P_IN_A(x1)  =  P_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
            ↳ AND
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ NonTerminationProof
              ↳ PiDP
  ↳ PrologToPiTRSProof

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

P_IN_A → P_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 semiunifying a rule from P directly.

The TRS P consists of the following rules:

P_IN_A → P_IN_A

The TRS R consists of the following rules:none


s = P_IN_A evaluates to t =P_IN_A

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 P_IN_A to P_IN_A.





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

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

P_IN_G(g(X)) → P_IN_G(X)

The TRS R consists of the following rules:

p_in_g(X) → U1_g(X, q_in_a(f(Y)))
q_in_a(g(Y)) → q_out_a(g(Y))
U1_g(X, q_out_a(f(Y))) → U2_g(X, p_in_a(Y))
p_in_a(X) → U1_a(X, q_in_a(f(Y)))
U1_a(X, q_out_a(f(Y))) → U2_a(X, p_in_a(Y))
p_in_a(g(X)) → U3_a(X, p_in_a(X))
U3_a(X, p_out_a(X)) → p_out_a(g(X))
U2_a(X, p_out_a(Y)) → p_out_a(X)
U2_g(X, p_out_a(Y)) → p_out_g(X)
p_in_g(g(X)) → U3_g(X, p_in_g(X))
U3_g(X, p_out_g(X)) → p_out_g(g(X))

The argument filtering Pi contains the following mapping:
p_in_g(x1)  =  p_in_g(x1)
U1_g(x1, x2)  =  U1_g(x2)
q_in_a(x1)  =  q_in_a
q_out_a(x1)  =  q_out_a
U2_g(x1, x2)  =  U2_g(x2)
p_in_a(x1)  =  p_in_a
U1_a(x1, x2)  =  U1_a(x2)
U2_a(x1, x2)  =  U2_a(x2)
U3_a(x1, x2)  =  U3_a(x2)
p_out_a(x1)  =  p_out_a
p_out_g(x1)  =  p_out_g
g(x1)  =  g(x1)
U3_g(x1, x2)  =  U3_g(x2)
P_IN_G(x1)  =  P_IN_G(x1)

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:

P_IN_G(g(X)) → P_IN_G(X)

R is empty.
Pi is empty.
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
                        ↳ QDPSizeChangeProof
  ↳ PrologToPiTRSProof

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

P_IN_G(g(X)) → P_IN_G(X)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] 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:


We use the technique of [30]. With regard to the inferred argument filtering the predicates were used in the following modes:
p_in: (b) (f)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

p_in_g(X) → U1_g(X, q_in_a(f(Y)))
q_in_a(g(Y)) → q_out_a(g(Y))
U1_g(X, q_out_a(f(Y))) → U2_g(X, p_in_a(Y))
p_in_a(X) → U1_a(X, q_in_a(f(Y)))
U1_a(X, q_out_a(f(Y))) → U2_a(X, p_in_a(Y))
p_in_a(g(X)) → U3_a(X, p_in_a(X))
U3_a(X, p_out_a(X)) → p_out_a(g(X))
U2_a(X, p_out_a(Y)) → p_out_a(X)
U2_g(X, p_out_a(Y)) → p_out_g(X)
p_in_g(g(X)) → U3_g(X, p_in_g(X))
U3_g(X, p_out_g(X)) → p_out_g(g(X))

The argument filtering Pi contains the following mapping:
p_in_g(x1)  =  p_in_g(x1)
U1_g(x1, x2)  =  U1_g(x1, x2)
q_in_a(x1)  =  q_in_a
q_out_a(x1)  =  q_out_a
U2_g(x1, x2)  =  U2_g(x1, x2)
p_in_a(x1)  =  p_in_a
U1_a(x1, x2)  =  U1_a(x2)
U2_a(x1, x2)  =  U2_a(x2)
U3_a(x1, x2)  =  U3_a(x2)
p_out_a(x1)  =  p_out_a
p_out_g(x1)  =  p_out_g(x1)
g(x1)  =  g(x1)
U3_g(x1, x2)  =  U3_g(x1, x2)

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_g(X) → U1_g(X, q_in_a(f(Y)))
q_in_a(g(Y)) → q_out_a(g(Y))
U1_g(X, q_out_a(f(Y))) → U2_g(X, p_in_a(Y))
p_in_a(X) → U1_a(X, q_in_a(f(Y)))
U1_a(X, q_out_a(f(Y))) → U2_a(X, p_in_a(Y))
p_in_a(g(X)) → U3_a(X, p_in_a(X))
U3_a(X, p_out_a(X)) → p_out_a(g(X))
U2_a(X, p_out_a(Y)) → p_out_a(X)
U2_g(X, p_out_a(Y)) → p_out_g(X)
p_in_g(g(X)) → U3_g(X, p_in_g(X))
U3_g(X, p_out_g(X)) → p_out_g(g(X))

The argument filtering Pi contains the following mapping:
p_in_g(x1)  =  p_in_g(x1)
U1_g(x1, x2)  =  U1_g(x1, x2)
q_in_a(x1)  =  q_in_a
q_out_a(x1)  =  q_out_a
U2_g(x1, x2)  =  U2_g(x1, x2)
p_in_a(x1)  =  p_in_a
U1_a(x1, x2)  =  U1_a(x2)
U2_a(x1, x2)  =  U2_a(x2)
U3_a(x1, x2)  =  U3_a(x2)
p_out_a(x1)  =  p_out_a
p_out_g(x1)  =  p_out_g(x1)
g(x1)  =  g(x1)
U3_g(x1, x2)  =  U3_g(x1, x2)


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_G(X) → U1_G(X, q_in_a(f(Y)))
P_IN_G(X) → Q_IN_A(f(Y))
U1_G(X, q_out_a(f(Y))) → U2_G(X, p_in_a(Y))
U1_G(X, q_out_a(f(Y))) → P_IN_A(Y)
P_IN_A(X) → U1_A(X, q_in_a(f(Y)))
P_IN_A(X) → Q_IN_A(f(Y))
U1_A(X, q_out_a(f(Y))) → U2_A(X, p_in_a(Y))
U1_A(X, q_out_a(f(Y))) → P_IN_A(Y)
P_IN_A(g(X)) → U3_A(X, p_in_a(X))
P_IN_A(g(X)) → P_IN_A(X)
P_IN_G(g(X)) → U3_G(X, p_in_g(X))
P_IN_G(g(X)) → P_IN_G(X)

The TRS R consists of the following rules:

p_in_g(X) → U1_g(X, q_in_a(f(Y)))
q_in_a(g(Y)) → q_out_a(g(Y))
U1_g(X, q_out_a(f(Y))) → U2_g(X, p_in_a(Y))
p_in_a(X) → U1_a(X, q_in_a(f(Y)))
U1_a(X, q_out_a(f(Y))) → U2_a(X, p_in_a(Y))
p_in_a(g(X)) → U3_a(X, p_in_a(X))
U3_a(X, p_out_a(X)) → p_out_a(g(X))
U2_a(X, p_out_a(Y)) → p_out_a(X)
U2_g(X, p_out_a(Y)) → p_out_g(X)
p_in_g(g(X)) → U3_g(X, p_in_g(X))
U3_g(X, p_out_g(X)) → p_out_g(g(X))

The argument filtering Pi contains the following mapping:
p_in_g(x1)  =  p_in_g(x1)
U1_g(x1, x2)  =  U1_g(x1, x2)
q_in_a(x1)  =  q_in_a
q_out_a(x1)  =  q_out_a
U2_g(x1, x2)  =  U2_g(x1, x2)
p_in_a(x1)  =  p_in_a
U1_a(x1, x2)  =  U1_a(x2)
U2_a(x1, x2)  =  U2_a(x2)
U3_a(x1, x2)  =  U3_a(x2)
p_out_a(x1)  =  p_out_a
p_out_g(x1)  =  p_out_g(x1)
g(x1)  =  g(x1)
U3_g(x1, x2)  =  U3_g(x1, x2)
U2_G(x1, x2)  =  U2_G(x1, x2)
U1_A(x1, x2)  =  U1_A(x2)
P_IN_G(x1)  =  P_IN_G(x1)
U3_A(x1, x2)  =  U3_A(x2)
U2_A(x1, x2)  =  U2_A(x2)
P_IN_A(x1)  =  P_IN_A
U1_G(x1, x2)  =  U1_G(x1, x2)
U3_G(x1, x2)  =  U3_G(x1, x2)
Q_IN_A(x1)  =  Q_IN_A

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_G(X) → U1_G(X, q_in_a(f(Y)))
P_IN_G(X) → Q_IN_A(f(Y))
U1_G(X, q_out_a(f(Y))) → U2_G(X, p_in_a(Y))
U1_G(X, q_out_a(f(Y))) → P_IN_A(Y)
P_IN_A(X) → U1_A(X, q_in_a(f(Y)))
P_IN_A(X) → Q_IN_A(f(Y))
U1_A(X, q_out_a(f(Y))) → U2_A(X, p_in_a(Y))
U1_A(X, q_out_a(f(Y))) → P_IN_A(Y)
P_IN_A(g(X)) → U3_A(X, p_in_a(X))
P_IN_A(g(X)) → P_IN_A(X)
P_IN_G(g(X)) → U3_G(X, p_in_g(X))
P_IN_G(g(X)) → P_IN_G(X)

The TRS R consists of the following rules:

p_in_g(X) → U1_g(X, q_in_a(f(Y)))
q_in_a(g(Y)) → q_out_a(g(Y))
U1_g(X, q_out_a(f(Y))) → U2_g(X, p_in_a(Y))
p_in_a(X) → U1_a(X, q_in_a(f(Y)))
U1_a(X, q_out_a(f(Y))) → U2_a(X, p_in_a(Y))
p_in_a(g(X)) → U3_a(X, p_in_a(X))
U3_a(X, p_out_a(X)) → p_out_a(g(X))
U2_a(X, p_out_a(Y)) → p_out_a(X)
U2_g(X, p_out_a(Y)) → p_out_g(X)
p_in_g(g(X)) → U3_g(X, p_in_g(X))
U3_g(X, p_out_g(X)) → p_out_g(g(X))

The argument filtering Pi contains the following mapping:
p_in_g(x1)  =  p_in_g(x1)
U1_g(x1, x2)  =  U1_g(x1, x2)
q_in_a(x1)  =  q_in_a
q_out_a(x1)  =  q_out_a
U2_g(x1, x2)  =  U2_g(x1, x2)
p_in_a(x1)  =  p_in_a
U1_a(x1, x2)  =  U1_a(x2)
U2_a(x1, x2)  =  U2_a(x2)
U3_a(x1, x2)  =  U3_a(x2)
p_out_a(x1)  =  p_out_a
p_out_g(x1)  =  p_out_g(x1)
g(x1)  =  g(x1)
U3_g(x1, x2)  =  U3_g(x1, x2)
U2_G(x1, x2)  =  U2_G(x1, x2)
U1_A(x1, x2)  =  U1_A(x2)
P_IN_G(x1)  =  P_IN_G(x1)
U3_A(x1, x2)  =  U3_A(x2)
U2_A(x1, x2)  =  U2_A(x2)
P_IN_A(x1)  =  P_IN_A
U1_G(x1, x2)  =  U1_G(x1, x2)
U3_G(x1, x2)  =  U3_G(x1, x2)
Q_IN_A(x1)  =  Q_IN_A

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 2 SCCs with 10 less nodes.

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

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

P_IN_A(g(X)) → P_IN_A(X)

The TRS R consists of the following rules:

p_in_g(X) → U1_g(X, q_in_a(f(Y)))
q_in_a(g(Y)) → q_out_a(g(Y))
U1_g(X, q_out_a(f(Y))) → U2_g(X, p_in_a(Y))
p_in_a(X) → U1_a(X, q_in_a(f(Y)))
U1_a(X, q_out_a(f(Y))) → U2_a(X, p_in_a(Y))
p_in_a(g(X)) → U3_a(X, p_in_a(X))
U3_a(X, p_out_a(X)) → p_out_a(g(X))
U2_a(X, p_out_a(Y)) → p_out_a(X)
U2_g(X, p_out_a(Y)) → p_out_g(X)
p_in_g(g(X)) → U3_g(X, p_in_g(X))
U3_g(X, p_out_g(X)) → p_out_g(g(X))

The argument filtering Pi contains the following mapping:
p_in_g(x1)  =  p_in_g(x1)
U1_g(x1, x2)  =  U1_g(x1, x2)
q_in_a(x1)  =  q_in_a
q_out_a(x1)  =  q_out_a
U2_g(x1, x2)  =  U2_g(x1, x2)
p_in_a(x1)  =  p_in_a
U1_a(x1, x2)  =  U1_a(x2)
U2_a(x1, x2)  =  U2_a(x2)
U3_a(x1, x2)  =  U3_a(x2)
p_out_a(x1)  =  p_out_a
p_out_g(x1)  =  p_out_g(x1)
g(x1)  =  g(x1)
U3_g(x1, x2)  =  U3_g(x1, x2)
P_IN_A(x1)  =  P_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
            ↳ AND
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
              ↳ PiDP

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

P_IN_A(g(X)) → P_IN_A(X)

R is empty.
The argument filtering Pi contains the following mapping:
g(x1)  =  g(x1)
P_IN_A(x1)  =  P_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
            ↳ AND
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ NonTerminationProof
              ↳ PiDP

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

P_IN_A → P_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 semiunifying a rule from P directly.

The TRS P consists of the following rules:

P_IN_A → P_IN_A

The TRS R consists of the following rules:none


s = P_IN_A evaluates to t =P_IN_A

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 P_IN_A to P_IN_A.





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

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

P_IN_G(g(X)) → P_IN_G(X)

The TRS R consists of the following rules:

p_in_g(X) → U1_g(X, q_in_a(f(Y)))
q_in_a(g(Y)) → q_out_a(g(Y))
U1_g(X, q_out_a(f(Y))) → U2_g(X, p_in_a(Y))
p_in_a(X) → U1_a(X, q_in_a(f(Y)))
U1_a(X, q_out_a(f(Y))) → U2_a(X, p_in_a(Y))
p_in_a(g(X)) → U3_a(X, p_in_a(X))
U3_a(X, p_out_a(X)) → p_out_a(g(X))
U2_a(X, p_out_a(Y)) → p_out_a(X)
U2_g(X, p_out_a(Y)) → p_out_g(X)
p_in_g(g(X)) → U3_g(X, p_in_g(X))
U3_g(X, p_out_g(X)) → p_out_g(g(X))

The argument filtering Pi contains the following mapping:
p_in_g(x1)  =  p_in_g(x1)
U1_g(x1, x2)  =  U1_g(x1, x2)
q_in_a(x1)  =  q_in_a
q_out_a(x1)  =  q_out_a
U2_g(x1, x2)  =  U2_g(x1, x2)
p_in_a(x1)  =  p_in_a
U1_a(x1, x2)  =  U1_a(x2)
U2_a(x1, x2)  =  U2_a(x2)
U3_a(x1, x2)  =  U3_a(x2)
p_out_a(x1)  =  p_out_a
p_out_g(x1)  =  p_out_g(x1)
g(x1)  =  g(x1)
U3_g(x1, x2)  =  U3_g(x1, x2)
P_IN_G(x1)  =  P_IN_G(x1)

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:

P_IN_G(g(X)) → P_IN_G(X)

R is empty.
Pi is empty.
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
                        ↳ QDPSizeChangeProof

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

P_IN_G(g(X)) → P_IN_G(X)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] 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: