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(a).
p(X) :- p(a).
q(b).

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(X) → U1(X, p_in(a))
p_in(a) → p_out(a)
U1(X, p_out(a)) → p_out(X)

The argument filtering Pi contains the following mapping:
p_in(x1)  =  p_in(x1)
U1(x1, x2)  =  U1(x2)
a  =  a
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(X) → U1(X, p_in(a))
p_in(a) → p_out(a)
U1(X, p_out(a)) → p_out(X)

The argument filtering Pi contains the following mapping:
p_in(x1)  =  p_in(x1)
U1(x1, x2)  =  U1(x2)
a  =  a
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(X) → U11(X, p_in(a))
P_IN(X) → P_IN(a)

The TRS R consists of the following rules:

p_in(X) → U1(X, p_in(a))
p_in(a) → p_out(a)
U1(X, p_out(a)) → p_out(X)

The argument filtering Pi contains the following mapping:
p_in(x1)  =  p_in(x1)
U1(x1, x2)  =  U1(x2)
a  =  a
p_out(x1)  =  p_out
P_IN(x1)  =  P_IN(x1)
U11(x1, x2)  =  U11(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(X) → U11(X, p_in(a))
P_IN(X) → P_IN(a)

The TRS R consists of the following rules:

p_in(X) → U1(X, p_in(a))
p_in(a) → p_out(a)
U1(X, p_out(a)) → p_out(X)

The argument filtering Pi contains the following mapping:
p_in(x1)  =  p_in(x1)
U1(x1, x2)  =  U1(x2)
a  =  a
p_out(x1)  =  p_out
P_IN(x1)  =  P_IN(x1)
U11(x1, x2)  =  U11(x2)

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

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

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

P_IN(X) → P_IN(a)

The TRS R consists of the following rules:

p_in(X) → U1(X, p_in(a))
p_in(a) → p_out(a)
U1(X, p_out(a)) → p_out(X)

The argument filtering Pi contains the following mapping:
p_in(x1)  =  p_in(x1)
U1(x1, x2)  =  U1(x2)
a  =  a
p_out(x1)  =  p_out
P_IN(x1)  =  P_IN(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
            ↳ PiDP
              ↳ UsableRulesProof
                ↳ PiDP
                  ↳ PiDPToQDPProof
  ↳ PrologToPiTRSProof

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

P_IN(X) → P_IN(a)

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
            ↳ PiDP
              ↳ UsableRulesProof
                ↳ PiDP
                  ↳ PiDPToQDPProof
                    ↳ QDP
                      ↳ Instantiation
  ↳ PrologToPiTRSProof

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

P_IN(X) → P_IN(a)

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

P_IN(a) → P_IN(a)



↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ PiDP
              ↳ UsableRulesProof
                ↳ PiDP
                  ↳ PiDPToQDPProof
                    ↳ QDP
                      ↳ Instantiation
                        ↳ QDP
                          ↳ NonTerminationProof
  ↳ 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).




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, p_in(a))
p_in(a) → p_out(a)
U1(X, p_out(a)) → p_out(X)

Pi is empty.

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, p_in(a))
p_in(a) → p_out(a)
U1(X, p_out(a)) → p_out(X)

Pi is empty.

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, p_in(a))
P_IN(X) → P_IN(a)

The TRS R consists of the following rules:

p_in(X) → U1(X, p_in(a))
p_in(a) → p_out(a)
U1(X, p_out(a)) → p_out(X)

Pi is empty.
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, p_in(a))
P_IN(X) → P_IN(a)

The TRS R consists of the following rules:

p_in(X) → U1(X, p_in(a))
p_in(a) → p_out(a)
U1(X, p_out(a)) → p_out(X)

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

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

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

P_IN(X) → P_IN(a)

The TRS R consists of the following rules:

p_in(X) → U1(X, p_in(a))
p_in(a) → p_out(a)
U1(X, p_out(a)) → p_out(X)

Pi is empty.
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(X) → P_IN(a)

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
            ↳ PiDP
              ↳ UsableRulesProof
                ↳ PiDP
                  ↳ PiDPToQDPProof
                    ↳ QDP
                      ↳ Instantiation

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

P_IN(X) → P_IN(a)

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

P_IN(a) → P_IN(a)



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

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).