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(b).
p(a) :- p1(X).
p1(b).
p1(a) :- p1(X).

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(a) → U1(p1_in(X))
p1_in(a) → U2(p1_in(X))
p1_in(b) → p1_out(b)
U2(p1_out(X)) → p1_out(a)
U1(p1_out(X)) → p_out(a)
p_in(b) → p_out(b)

The argument filtering Pi contains the following mapping:
p_in(x1)  =  p_in(x1)
a  =  a
U1(x1)  =  U1(x1)
p1_in(x1)  =  p1_in
U2(x1)  =  U2(x1)
b  =  b
p1_out(x1)  =  p1_out(x1)
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(a) → U1(p1_in(X))
p1_in(a) → U2(p1_in(X))
p1_in(b) → p1_out(b)
U2(p1_out(X)) → p1_out(a)
U1(p1_out(X)) → p_out(a)
p_in(b) → p_out(b)

The argument filtering Pi contains the following mapping:
p_in(x1)  =  p_in(x1)
a  =  a
U1(x1)  =  U1(x1)
p1_in(x1)  =  p1_in
U2(x1)  =  U2(x1)
b  =  b
p1_out(x1)  =  p1_out(x1)
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(a) → U11(p1_in(X))
P_IN(a) → P1_IN(X)
P1_IN(a) → U21(p1_in(X))
P1_IN(a) → P1_IN(X)

The TRS R consists of the following rules:

p_in(a) → U1(p1_in(X))
p1_in(a) → U2(p1_in(X))
p1_in(b) → p1_out(b)
U2(p1_out(X)) → p1_out(a)
U1(p1_out(X)) → p_out(a)
p_in(b) → p_out(b)

The argument filtering Pi contains the following mapping:
p_in(x1)  =  p_in(x1)
a  =  a
U1(x1)  =  U1(x1)
p1_in(x1)  =  p1_in
U2(x1)  =  U2(x1)
b  =  b
p1_out(x1)  =  p1_out(x1)
p_out(x1)  =  p_out
P_IN(x1)  =  P_IN(x1)
P1_IN(x1)  =  P1_IN
U11(x1)  =  U11(x1)
U21(x1)  =  U21(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) → U11(p1_in(X))
P_IN(a) → P1_IN(X)
P1_IN(a) → U21(p1_in(X))
P1_IN(a) → P1_IN(X)

The TRS R consists of the following rules:

p_in(a) → U1(p1_in(X))
p1_in(a) → U2(p1_in(X))
p1_in(b) → p1_out(b)
U2(p1_out(X)) → p1_out(a)
U1(p1_out(X)) → p_out(a)
p_in(b) → p_out(b)

The argument filtering Pi contains the following mapping:
p_in(x1)  =  p_in(x1)
a  =  a
U1(x1)  =  U1(x1)
p1_in(x1)  =  p1_in
U2(x1)  =  U2(x1)
b  =  b
p1_out(x1)  =  p1_out(x1)
p_out(x1)  =  p_out
P_IN(x1)  =  P_IN(x1)
P1_IN(x1)  =  P1_IN
U11(x1)  =  U11(x1)
U21(x1)  =  U21(x1)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 1 SCC with 3 less nodes.

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

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

P1_IN(a) → P1_IN(X)

The TRS R consists of the following rules:

p_in(a) → U1(p1_in(X))
p1_in(a) → U2(p1_in(X))
p1_in(b) → p1_out(b)
U2(p1_out(X)) → p1_out(a)
U1(p1_out(X)) → p_out(a)
p_in(b) → p_out(b)

The argument filtering Pi contains the following mapping:
p_in(x1)  =  p_in(x1)
a  =  a
U1(x1)  =  U1(x1)
p1_in(x1)  =  p1_in
U2(x1)  =  U2(x1)
b  =  b
p1_out(x1)  =  p1_out(x1)
p_out(x1)  =  p_out
P1_IN(x1)  =  P1_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:

P1_IN(a) → P1_IN(X)

R is empty.
The argument filtering Pi contains the following mapping:
a  =  a
P1_IN(x1)  =  P1_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:

P1_IN → P1_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:

P1_IN → P1_IN

The TRS R consists of the following rules:none


s = P1_IN evaluates to t =P1_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 P1_IN to P1_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(a) → U1(p1_in(X))
p1_in(a) → U2(p1_in(X))
p1_in(b) → p1_out(b)
U2(p1_out(X)) → p1_out(a)
U1(p1_out(X)) → p_out(a)
p_in(b) → p_out(b)

The argument filtering Pi contains the following mapping:
p_in(x1)  =  p_in(x1)
a  =  a
U1(x1)  =  U1(x1)
p1_in(x1)  =  p1_in
U2(x1)  =  U2(x1)
b  =  b
p1_out(x1)  =  p1_out(x1)
p_out(x1)  =  p_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:

p_in(a) → U1(p1_in(X))
p1_in(a) → U2(p1_in(X))
p1_in(b) → p1_out(b)
U2(p1_out(X)) → p1_out(a)
U1(p1_out(X)) → p_out(a)
p_in(b) → p_out(b)

The argument filtering Pi contains the following mapping:
p_in(x1)  =  p_in(x1)
a  =  a
U1(x1)  =  U1(x1)
p1_in(x1)  =  p1_in
U2(x1)  =  U2(x1)
b  =  b
p1_out(x1)  =  p1_out(x1)
p_out(x1)  =  p_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:

P_IN(a) → U11(p1_in(X))
P_IN(a) → P1_IN(X)
P1_IN(a) → U21(p1_in(X))
P1_IN(a) → P1_IN(X)

The TRS R consists of the following rules:

p_in(a) → U1(p1_in(X))
p1_in(a) → U2(p1_in(X))
p1_in(b) → p1_out(b)
U2(p1_out(X)) → p1_out(a)
U1(p1_out(X)) → p_out(a)
p_in(b) → p_out(b)

The argument filtering Pi contains the following mapping:
p_in(x1)  =  p_in(x1)
a  =  a
U1(x1)  =  U1(x1)
p1_in(x1)  =  p1_in
U2(x1)  =  U2(x1)
b  =  b
p1_out(x1)  =  p1_out(x1)
p_out(x1)  =  p_out(x1)
P_IN(x1)  =  P_IN(x1)
P1_IN(x1)  =  P1_IN
U11(x1)  =  U11(x1)
U21(x1)  =  U21(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) → U11(p1_in(X))
P_IN(a) → P1_IN(X)
P1_IN(a) → U21(p1_in(X))
P1_IN(a) → P1_IN(X)

The TRS R consists of the following rules:

p_in(a) → U1(p1_in(X))
p1_in(a) → U2(p1_in(X))
p1_in(b) → p1_out(b)
U2(p1_out(X)) → p1_out(a)
U1(p1_out(X)) → p_out(a)
p_in(b) → p_out(b)

The argument filtering Pi contains the following mapping:
p_in(x1)  =  p_in(x1)
a  =  a
U1(x1)  =  U1(x1)
p1_in(x1)  =  p1_in
U2(x1)  =  U2(x1)
b  =  b
p1_out(x1)  =  p1_out(x1)
p_out(x1)  =  p_out(x1)
P_IN(x1)  =  P_IN(x1)
P1_IN(x1)  =  P1_IN
U11(x1)  =  U11(x1)
U21(x1)  =  U21(x1)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 1 SCC with 3 less nodes.

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

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

P1_IN(a) → P1_IN(X)

The TRS R consists of the following rules:

p_in(a) → U1(p1_in(X))
p1_in(a) → U2(p1_in(X))
p1_in(b) → p1_out(b)
U2(p1_out(X)) → p1_out(a)
U1(p1_out(X)) → p_out(a)
p_in(b) → p_out(b)

The argument filtering Pi contains the following mapping:
p_in(x1)  =  p_in(x1)
a  =  a
U1(x1)  =  U1(x1)
p1_in(x1)  =  p1_in
U2(x1)  =  U2(x1)
b  =  b
p1_out(x1)  =  p1_out(x1)
p_out(x1)  =  p_out(x1)
P1_IN(x1)  =  P1_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:

P1_IN(a) → P1_IN(X)

R is empty.
The argument filtering Pi contains the following mapping:
a  =  a
P1_IN(x1)  =  P1_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:

P1_IN → P1_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:

P1_IN → P1_IN

The TRS R consists of the following rules:none


s = P1_IN evaluates to t =P1_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 P1_IN to P1_IN.