Left Termination of the query pattern p2_in_1(a) w.r.t. the given Prolog program could not be shown:



↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof

Clauses:

p1(f(X)) :- p1(X).
p2(f(X)) :- p2(X).

Queries:

p2(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:

p2_in(f(X)) → U2(X, p2_in(X))
U2(X, p2_out(X)) → p2_out(f(X))

The argument filtering Pi contains the following mapping:
p2_in(x1)  =  p2_in
f(x1)  =  f(x1)
U2(x1, x2)  =  U2(x2)
p2_out(x1)  =  p2_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:

p2_in(f(X)) → U2(X, p2_in(X))
U2(X, p2_out(X)) → p2_out(f(X))

The argument filtering Pi contains the following mapping:
p2_in(x1)  =  p2_in
f(x1)  =  f(x1)
U2(x1, x2)  =  U2(x2)
p2_out(x1)  =  p2_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:

P2_IN(f(X)) → U21(X, p2_in(X))
P2_IN(f(X)) → P2_IN(X)

The TRS R consists of the following rules:

p2_in(f(X)) → U2(X, p2_in(X))
U2(X, p2_out(X)) → p2_out(f(X))

The argument filtering Pi contains the following mapping:
p2_in(x1)  =  p2_in
f(x1)  =  f(x1)
U2(x1, x2)  =  U2(x2)
p2_out(x1)  =  p2_out(x1)
P2_IN(x1)  =  P2_IN
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:

P2_IN(f(X)) → U21(X, p2_in(X))
P2_IN(f(X)) → P2_IN(X)

The TRS R consists of the following rules:

p2_in(f(X)) → U2(X, p2_in(X))
U2(X, p2_out(X)) → p2_out(f(X))

The argument filtering Pi contains the following mapping:
p2_in(x1)  =  p2_in
f(x1)  =  f(x1)
U2(x1, x2)  =  U2(x2)
p2_out(x1)  =  p2_out(x1)
P2_IN(x1)  =  P2_IN
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 1 less node.

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

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

P2_IN(f(X)) → P2_IN(X)

The TRS R consists of the following rules:

p2_in(f(X)) → U2(X, p2_in(X))
U2(X, p2_out(X)) → p2_out(f(X))

The argument filtering Pi contains the following mapping:
p2_in(x1)  =  p2_in
f(x1)  =  f(x1)
U2(x1, x2)  =  U2(x2)
p2_out(x1)  =  p2_out(x1)
P2_IN(x1)  =  P2_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:

P2_IN(f(X)) → P2_IN(X)

R is empty.
The argument filtering Pi contains the following mapping:
f(x1)  =  f(x1)
P2_IN(x1)  =  P2_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:

P2_IN → P2_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:

P2_IN → P2_IN

The TRS R consists of the following rules:none


s = P2_IN evaluates to t =P2_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 P2_IN to P2_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:

p2_in(f(X)) → U2(X, p2_in(X))
U2(X, p2_out(X)) → p2_out(f(X))

The argument filtering Pi contains the following mapping:
p2_in(x1)  =  p2_in
f(x1)  =  f(x1)
U2(x1, x2)  =  U2(x2)
p2_out(x1)  =  p2_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:

p2_in(f(X)) → U2(X, p2_in(X))
U2(X, p2_out(X)) → p2_out(f(X))

The argument filtering Pi contains the following mapping:
p2_in(x1)  =  p2_in
f(x1)  =  f(x1)
U2(x1, x2)  =  U2(x2)
p2_out(x1)  =  p2_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:

P2_IN(f(X)) → U21(X, p2_in(X))
P2_IN(f(X)) → P2_IN(X)

The TRS R consists of the following rules:

p2_in(f(X)) → U2(X, p2_in(X))
U2(X, p2_out(X)) → p2_out(f(X))

The argument filtering Pi contains the following mapping:
p2_in(x1)  =  p2_in
f(x1)  =  f(x1)
U2(x1, x2)  =  U2(x2)
p2_out(x1)  =  p2_out(x1)
P2_IN(x1)  =  P2_IN
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:

P2_IN(f(X)) → U21(X, p2_in(X))
P2_IN(f(X)) → P2_IN(X)

The TRS R consists of the following rules:

p2_in(f(X)) → U2(X, p2_in(X))
U2(X, p2_out(X)) → p2_out(f(X))

The argument filtering Pi contains the following mapping:
p2_in(x1)  =  p2_in
f(x1)  =  f(x1)
U2(x1, x2)  =  U2(x2)
p2_out(x1)  =  p2_out(x1)
P2_IN(x1)  =  P2_IN
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 1 less node.

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

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

P2_IN(f(X)) → P2_IN(X)

The TRS R consists of the following rules:

p2_in(f(X)) → U2(X, p2_in(X))
U2(X, p2_out(X)) → p2_out(f(X))

The argument filtering Pi contains the following mapping:
p2_in(x1)  =  p2_in
f(x1)  =  f(x1)
U2(x1, x2)  =  U2(x2)
p2_out(x1)  =  p2_out(x1)
P2_IN(x1)  =  P2_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:

P2_IN(f(X)) → P2_IN(X)

R is empty.
The argument filtering Pi contains the following mapping:
f(x1)  =  f(x1)
P2_IN(x1)  =  P2_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:

P2_IN → P2_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:

P2_IN → P2_IN

The TRS R consists of the following rules:none


s = P2_IN evaluates to t =P2_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 P2_IN to P2_IN.