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

p_in(g(X)) → U3(X, p_in(X))
p_in(X) → U1(X, q_in(f(Y)))
q_in(g(Y)) → q_out(g(Y))
U1(X, q_out(f(Y))) → U2(X, p_in(Y))
U2(X, p_out(Y)) → p_out(X)
U3(X, p_out(X)) → p_out(g(X))

The argument filtering Pi contains the following mapping:
p_in(x1)  =  p_in
g(x1)  =  g(x1)
U3(x1, x2)  =  U3(x2)
U1(x1, x2)  =  U1(x2)
q_in(x1)  =  q_in
f(x1)  =  f(x1)
q_out(x1)  =  q_out
U2(x1, x2)  =  U2(x2)
p_out(x1)  =  p_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:

p_in(g(X)) → U3(X, p_in(X))
p_in(X) → U1(X, q_in(f(Y)))
q_in(g(Y)) → q_out(g(Y))
U1(X, q_out(f(Y))) → U2(X, p_in(Y))
U2(X, p_out(Y)) → p_out(X)
U3(X, p_out(X)) → p_out(g(X))

The argument filtering Pi contains the following mapping:
p_in(x1)  =  p_in
g(x1)  =  g(x1)
U3(x1, x2)  =  U3(x2)
U1(x1, x2)  =  U1(x2)
q_in(x1)  =  q_in
f(x1)  =  f(x1)
q_out(x1)  =  q_out
U2(x1, x2)  =  U2(x2)
p_out(x1)  =  p_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:

P_IN(g(X)) → U31(X, p_in(X))
P_IN(g(X)) → P_IN(X)
P_IN(X) → U11(X, q_in(f(Y)))
P_IN(X) → Q_IN(f(Y))
U11(X, q_out(f(Y))) → U21(X, p_in(Y))
U11(X, q_out(f(Y))) → P_IN(Y)

The TRS R consists of the following rules:

p_in(g(X)) → U3(X, p_in(X))
p_in(X) → U1(X, q_in(f(Y)))
q_in(g(Y)) → q_out(g(Y))
U1(X, q_out(f(Y))) → U2(X, p_in(Y))
U2(X, p_out(Y)) → p_out(X)
U3(X, p_out(X)) → p_out(g(X))

The argument filtering Pi contains the following mapping:
p_in(x1)  =  p_in
g(x1)  =  g(x1)
U3(x1, x2)  =  U3(x2)
U1(x1, x2)  =  U1(x2)
q_in(x1)  =  q_in
f(x1)  =  f(x1)
q_out(x1)  =  q_out
U2(x1, x2)  =  U2(x2)
p_out(x1)  =  p_out
P_IN(x1)  =  P_IN
Q_IN(x1)  =  Q_IN
U11(x1, x2)  =  U11(x2)
U31(x1, x2)  =  U31(x2)
U21(x1, x2)  =  U21(x2)

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)) → U31(X, p_in(X))
P_IN(g(X)) → P_IN(X)
P_IN(X) → U11(X, q_in(f(Y)))
P_IN(X) → Q_IN(f(Y))
U11(X, q_out(f(Y))) → U21(X, p_in(Y))
U11(X, q_out(f(Y))) → P_IN(Y)

The TRS R consists of the following rules:

p_in(g(X)) → U3(X, p_in(X))
p_in(X) → U1(X, q_in(f(Y)))
q_in(g(Y)) → q_out(g(Y))
U1(X, q_out(f(Y))) → U2(X, p_in(Y))
U2(X, p_out(Y)) → p_out(X)
U3(X, p_out(X)) → p_out(g(X))

The argument filtering Pi contains the following mapping:
p_in(x1)  =  p_in
g(x1)  =  g(x1)
U3(x1, x2)  =  U3(x2)
U1(x1, x2)  =  U1(x2)
q_in(x1)  =  q_in
f(x1)  =  f(x1)
q_out(x1)  =  q_out
U2(x1, x2)  =  U2(x2)
p_out(x1)  =  p_out
P_IN(x1)  =  P_IN
Q_IN(x1)  =  Q_IN
U11(x1, x2)  =  U11(x2)
U31(x1, x2)  =  U31(x2)
U21(x1, x2)  =  U21(x2)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 1 SCC with 5 less nodes.

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

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

P_IN(g(X)) → P_IN(X)

The TRS R consists of the following rules:

p_in(g(X)) → U3(X, p_in(X))
p_in(X) → U1(X, q_in(f(Y)))
q_in(g(Y)) → q_out(g(Y))
U1(X, q_out(f(Y))) → U2(X, p_in(Y))
U2(X, p_out(Y)) → p_out(X)
U3(X, p_out(X)) → p_out(g(X))

The argument filtering Pi contains the following mapping:
p_in(x1)  =  p_in
g(x1)  =  g(x1)
U3(x1, x2)  =  U3(x2)
U1(x1, x2)  =  U1(x2)
q_in(x1)  =  q_in
f(x1)  =  f(x1)
q_out(x1)  =  q_out
U2(x1, x2)  =  U2(x2)
p_out(x1)  =  p_out
P_IN(x1)  =  P_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:

P_IN(g(X)) → P_IN(X)

R is empty.
The argument filtering Pi contains the following mapping:
g(x1)  =  g(x1)
P_IN(x1)  =  P_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
                      ↳ NonTerminationProof
  ↳ PrologToPiTRSProof

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

P_IN → P_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:

P_IN → P_IN

The TRS R consists of the following rules:none


s = P_IN evaluates to t =P_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 P_IN to P_IN.




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(g(X)) → U3(X, p_in(X))
p_in(X) → U1(X, q_in(f(Y)))
q_in(g(Y)) → q_out(g(Y))
U1(X, q_out(f(Y))) → U2(X, p_in(Y))
U2(X, p_out(Y)) → p_out(X)
U3(X, p_out(X)) → p_out(g(X))

The argument filtering Pi contains the following mapping:
p_in(x1)  =  p_in
g(x1)  =  g(x1)
U3(x1, x2)  =  U3(x2)
U1(x1, x2)  =  U1(x2)
q_in(x1)  =  q_in
f(x1)  =  f(x1)
q_out(x1)  =  q_out
U2(x1, x2)  =  U2(x2)
p_out(x1)  =  p_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:

p_in(g(X)) → U3(X, p_in(X))
p_in(X) → U1(X, q_in(f(Y)))
q_in(g(Y)) → q_out(g(Y))
U1(X, q_out(f(Y))) → U2(X, p_in(Y))
U2(X, p_out(Y)) → p_out(X)
U3(X, p_out(X)) → p_out(g(X))

The argument filtering Pi contains the following mapping:
p_in(x1)  =  p_in
g(x1)  =  g(x1)
U3(x1, x2)  =  U3(x2)
U1(x1, x2)  =  U1(x2)
q_in(x1)  =  q_in
f(x1)  =  f(x1)
q_out(x1)  =  q_out
U2(x1, x2)  =  U2(x2)
p_out(x1)  =  p_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:

P_IN(g(X)) → U31(X, p_in(X))
P_IN(g(X)) → P_IN(X)
P_IN(X) → U11(X, q_in(f(Y)))
P_IN(X) → Q_IN(f(Y))
U11(X, q_out(f(Y))) → U21(X, p_in(Y))
U11(X, q_out(f(Y))) → P_IN(Y)

The TRS R consists of the following rules:

p_in(g(X)) → U3(X, p_in(X))
p_in(X) → U1(X, q_in(f(Y)))
q_in(g(Y)) → q_out(g(Y))
U1(X, q_out(f(Y))) → U2(X, p_in(Y))
U2(X, p_out(Y)) → p_out(X)
U3(X, p_out(X)) → p_out(g(X))

The argument filtering Pi contains the following mapping:
p_in(x1)  =  p_in
g(x1)  =  g(x1)
U3(x1, x2)  =  U3(x2)
U1(x1, x2)  =  U1(x2)
q_in(x1)  =  q_in
f(x1)  =  f(x1)
q_out(x1)  =  q_out
U2(x1, x2)  =  U2(x2)
p_out(x1)  =  p_out
P_IN(x1)  =  P_IN
Q_IN(x1)  =  Q_IN
U11(x1, x2)  =  U11(x2)
U31(x1, x2)  =  U31(x2)
U21(x1, x2)  =  U21(x2)

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)) → U31(X, p_in(X))
P_IN(g(X)) → P_IN(X)
P_IN(X) → U11(X, q_in(f(Y)))
P_IN(X) → Q_IN(f(Y))
U11(X, q_out(f(Y))) → U21(X, p_in(Y))
U11(X, q_out(f(Y))) → P_IN(Y)

The TRS R consists of the following rules:

p_in(g(X)) → U3(X, p_in(X))
p_in(X) → U1(X, q_in(f(Y)))
q_in(g(Y)) → q_out(g(Y))
U1(X, q_out(f(Y))) → U2(X, p_in(Y))
U2(X, p_out(Y)) → p_out(X)
U3(X, p_out(X)) → p_out(g(X))

The argument filtering Pi contains the following mapping:
p_in(x1)  =  p_in
g(x1)  =  g(x1)
U3(x1, x2)  =  U3(x2)
U1(x1, x2)  =  U1(x2)
q_in(x1)  =  q_in
f(x1)  =  f(x1)
q_out(x1)  =  q_out
U2(x1, x2)  =  U2(x2)
p_out(x1)  =  p_out
P_IN(x1)  =  P_IN
Q_IN(x1)  =  Q_IN
U11(x1, x2)  =  U11(x2)
U31(x1, x2)  =  U31(x2)
U21(x1, x2)  =  U21(x2)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 1 SCC with 5 less nodes.

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

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

P_IN(g(X)) → P_IN(X)

The TRS R consists of the following rules:

p_in(g(X)) → U3(X, p_in(X))
p_in(X) → U1(X, q_in(f(Y)))
q_in(g(Y)) → q_out(g(Y))
U1(X, q_out(f(Y))) → U2(X, p_in(Y))
U2(X, p_out(Y)) → p_out(X)
U3(X, p_out(X)) → p_out(g(X))

The argument filtering Pi contains the following mapping:
p_in(x1)  =  p_in
g(x1)  =  g(x1)
U3(x1, x2)  =  U3(x2)
U1(x1, x2)  =  U1(x2)
q_in(x1)  =  q_in
f(x1)  =  f(x1)
q_out(x1)  =  q_out
U2(x1, x2)  =  U2(x2)
p_out(x1)  =  p_out
P_IN(x1)  =  P_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:

P_IN(g(X)) → P_IN(X)

R is empty.
The argument filtering Pi contains the following mapping:
g(x1)  =  g(x1)
P_IN(x1)  =  P_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
                      ↳ NonTerminationProof

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

P_IN → P_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:

P_IN → P_IN

The TRS R consists of the following rules:none


s = P_IN evaluates to t =P_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 P_IN to P_IN.