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



PROLOG
  ↳ UnrequestedClauseRemoverProof

p1(a0).
p1(X) :- p1(a0).
q1(b0).


The clause

q1(b0).

can be ignored, as it is not needed by any of the given querys.

Deleting this clauses results in the following prolog program:

p1(a0).
p1(X) :- p1(a0).



↳ PROLOG
  ↳ UnrequestedClauseRemoverProof
PROLOG
      ↳ PrologToPiTRSProof
      ↳ PrologToPiTRSProof

p1(a0).
p1(X) :- p1(a0).


With regard to the inferred argument filtering the predicates were used in the following modes:
p1: (b)
Transforming PROLOG into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:


p_1_in_g1(a_0) -> p_1_out_g1(a_0)
p_1_in_g1(X) -> if_p_1_in_1_g2(X, p_1_in_g1(a_0))
if_p_1_in_1_g2(X, p_1_out_g1(a_0)) -> p_1_out_g1(X)

The argument filtering Pi contains the following mapping:
p_1_in_g1(x1)  =  p_1_in_g1(x1)
a_0  =  a_0
p_1_out_g1(x1)  =  p_1_out_g
if_p_1_in_1_g2(x1, x2)  =  if_p_1_in_1_g1(x2)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of PROLOG



↳ PROLOG
  ↳ UnrequestedClauseRemoverProof
    ↳ PROLOG
      ↳ PrologToPiTRSProof
PiTRS
          ↳ DependencyPairsProof
      ↳ PrologToPiTRSProof

Pi-finite rewrite system:
The TRS R consists of the following rules:

p_1_in_g1(a_0) -> p_1_out_g1(a_0)
p_1_in_g1(X) -> if_p_1_in_1_g2(X, p_1_in_g1(a_0))
if_p_1_in_1_g2(X, p_1_out_g1(a_0)) -> p_1_out_g1(X)

The argument filtering Pi contains the following mapping:
p_1_in_g1(x1)  =  p_1_in_g1(x1)
a_0  =  a_0
p_1_out_g1(x1)  =  p_1_out_g
if_p_1_in_1_g2(x1, x2)  =  if_p_1_in_1_g1(x2)


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

P_1_IN_G1(X) -> IF_P_1_IN_1_G2(X, p_1_in_g1(a_0))
P_1_IN_G1(X) -> P_1_IN_G1(a_0)

The TRS R consists of the following rules:

p_1_in_g1(a_0) -> p_1_out_g1(a_0)
p_1_in_g1(X) -> if_p_1_in_1_g2(X, p_1_in_g1(a_0))
if_p_1_in_1_g2(X, p_1_out_g1(a_0)) -> p_1_out_g1(X)

The argument filtering Pi contains the following mapping:
p_1_in_g1(x1)  =  p_1_in_g1(x1)
a_0  =  a_0
p_1_out_g1(x1)  =  p_1_out_g
if_p_1_in_1_g2(x1, x2)  =  if_p_1_in_1_g1(x2)
IF_P_1_IN_1_G2(x1, x2)  =  IF_P_1_IN_1_G1(x2)
P_1_IN_G1(x1)  =  P_1_IN_G1(x1)

We have to consider all (P,R,Pi)-chains

↳ PROLOG
  ↳ UnrequestedClauseRemoverProof
    ↳ PROLOG
      ↳ PrologToPiTRSProof
        ↳ PiTRS
          ↳ DependencyPairsProof
PiDP
              ↳ DependencyGraphProof
      ↳ PrologToPiTRSProof

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

P_1_IN_G1(X) -> IF_P_1_IN_1_G2(X, p_1_in_g1(a_0))
P_1_IN_G1(X) -> P_1_IN_G1(a_0)

The TRS R consists of the following rules:

p_1_in_g1(a_0) -> p_1_out_g1(a_0)
p_1_in_g1(X) -> if_p_1_in_1_g2(X, p_1_in_g1(a_0))
if_p_1_in_1_g2(X, p_1_out_g1(a_0)) -> p_1_out_g1(X)

The argument filtering Pi contains the following mapping:
p_1_in_g1(x1)  =  p_1_in_g1(x1)
a_0  =  a_0
p_1_out_g1(x1)  =  p_1_out_g
if_p_1_in_1_g2(x1, x2)  =  if_p_1_in_1_g1(x2)
IF_P_1_IN_1_G2(x1, x2)  =  IF_P_1_IN_1_G1(x2)
P_1_IN_G1(x1)  =  P_1_IN_G1(x1)

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

↳ PROLOG
  ↳ UnrequestedClauseRemoverProof
    ↳ PROLOG
      ↳ PrologToPiTRSProof
        ↳ PiTRS
          ↳ DependencyPairsProof
            ↳ PiDP
              ↳ DependencyGraphProof
PiDP
                  ↳ UsableRulesProof
      ↳ PrologToPiTRSProof

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

P_1_IN_G1(X) -> P_1_IN_G1(a_0)

The TRS R consists of the following rules:

p_1_in_g1(a_0) -> p_1_out_g1(a_0)
p_1_in_g1(X) -> if_p_1_in_1_g2(X, p_1_in_g1(a_0))
if_p_1_in_1_g2(X, p_1_out_g1(a_0)) -> p_1_out_g1(X)

The argument filtering Pi contains the following mapping:
p_1_in_g1(x1)  =  p_1_in_g1(x1)
a_0  =  a_0
p_1_out_g1(x1)  =  p_1_out_g
if_p_1_in_1_g2(x1, x2)  =  if_p_1_in_1_g1(x2)
P_1_IN_G1(x1)  =  P_1_IN_G1(x1)

We have to consider all (P,R,Pi)-chains
For (infinitary) constructor rewriting we can delete all non-usable rules from R.

↳ PROLOG
  ↳ UnrequestedClauseRemoverProof
    ↳ PROLOG
      ↳ PrologToPiTRSProof
        ↳ PiTRS
          ↳ DependencyPairsProof
            ↳ PiDP
              ↳ DependencyGraphProof
                ↳ PiDP
                  ↳ UsableRulesProof
PiDP
                      ↳ PiDPToQDPProof
      ↳ PrologToPiTRSProof

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

P_1_IN_G1(X) -> P_1_IN_G1(a_0)

R is empty.
Pi is empty.
We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem into ordinary QDP problem by application of Pi.

↳ PROLOG
  ↳ UnrequestedClauseRemoverProof
    ↳ PROLOG
      ↳ PrologToPiTRSProof
        ↳ PiTRS
          ↳ DependencyPairsProof
            ↳ PiDP
              ↳ DependencyGraphProof
                ↳ PiDP
                  ↳ UsableRulesProof
                    ↳ PiDP
                      ↳ PiDPToQDPProof
QDP
      ↳ PrologToPiTRSProof

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

P_1_IN_G1(X) -> P_1_IN_G1(a_0)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {P_1_IN_G1}.
With regard to the inferred argument filtering the predicates were used in the following modes:
p1: (b)
Transforming PROLOG into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

p_1_in_g1(a_0) -> p_1_out_g1(a_0)
p_1_in_g1(X) -> if_p_1_in_1_g2(X, p_1_in_g1(a_0))
if_p_1_in_1_g2(X, p_1_out_g1(a_0)) -> p_1_out_g1(X)

Pi is empty.

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of PROLOG



↳ PROLOG
  ↳ UnrequestedClauseRemoverProof
    ↳ PROLOG
      ↳ PrologToPiTRSProof
      ↳ PrologToPiTRSProof
PiTRS
          ↳ DependencyPairsProof

Pi-finite rewrite system:
The TRS R consists of the following rules:

p_1_in_g1(a_0) -> p_1_out_g1(a_0)
p_1_in_g1(X) -> if_p_1_in_1_g2(X, p_1_in_g1(a_0))
if_p_1_in_1_g2(X, p_1_out_g1(a_0)) -> p_1_out_g1(X)

Pi is empty.

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

P_1_IN_G1(X) -> IF_P_1_IN_1_G2(X, p_1_in_g1(a_0))
P_1_IN_G1(X) -> P_1_IN_G1(a_0)

The TRS R consists of the following rules:

p_1_in_g1(a_0) -> p_1_out_g1(a_0)
p_1_in_g1(X) -> if_p_1_in_1_g2(X, p_1_in_g1(a_0))
if_p_1_in_1_g2(X, p_1_out_g1(a_0)) -> p_1_out_g1(X)

Pi is empty.
We have to consider all (P,R,Pi)-chains

↳ PROLOG
  ↳ UnrequestedClauseRemoverProof
    ↳ PROLOG
      ↳ PrologToPiTRSProof
      ↳ PrologToPiTRSProof
        ↳ PiTRS
          ↳ DependencyPairsProof
PiDP
              ↳ DependencyGraphProof

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

P_1_IN_G1(X) -> IF_P_1_IN_1_G2(X, p_1_in_g1(a_0))
P_1_IN_G1(X) -> P_1_IN_G1(a_0)

The TRS R consists of the following rules:

p_1_in_g1(a_0) -> p_1_out_g1(a_0)
p_1_in_g1(X) -> if_p_1_in_1_g2(X, p_1_in_g1(a_0))
if_p_1_in_1_g2(X, p_1_out_g1(a_0)) -> p_1_out_g1(X)

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

↳ PROLOG
  ↳ UnrequestedClauseRemoverProof
    ↳ PROLOG
      ↳ PrologToPiTRSProof
      ↳ PrologToPiTRSProof
        ↳ PiTRS
          ↳ DependencyPairsProof
            ↳ PiDP
              ↳ DependencyGraphProof
PiDP
                  ↳ UsableRulesProof

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

P_1_IN_G1(X) -> P_1_IN_G1(a_0)

The TRS R consists of the following rules:

p_1_in_g1(a_0) -> p_1_out_g1(a_0)
p_1_in_g1(X) -> if_p_1_in_1_g2(X, p_1_in_g1(a_0))
if_p_1_in_1_g2(X, p_1_out_g1(a_0)) -> p_1_out_g1(X)

Pi is empty.
We have to consider all (P,R,Pi)-chains
For (infinitary) constructor rewriting we can delete all non-usable rules from R.

↳ PROLOG
  ↳ UnrequestedClauseRemoverProof
    ↳ PROLOG
      ↳ PrologToPiTRSProof
      ↳ PrologToPiTRSProof
        ↳ PiTRS
          ↳ DependencyPairsProof
            ↳ PiDP
              ↳ DependencyGraphProof
                ↳ PiDP
                  ↳ UsableRulesProof
PiDP
                      ↳ PiDPToQDPProof

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

P_1_IN_G1(X) -> P_1_IN_G1(a_0)

R is empty.
Pi is empty.
We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem into ordinary QDP problem by application of Pi.

↳ PROLOG
  ↳ UnrequestedClauseRemoverProof
    ↳ PROLOG
      ↳ PrologToPiTRSProof
      ↳ PrologToPiTRSProof
        ↳ PiTRS
          ↳ DependencyPairsProof
            ↳ PiDP
              ↳ DependencyGraphProof
                ↳ PiDP
                  ↳ UsableRulesProof
                    ↳ PiDP
                      ↳ PiDPToQDPProof
QDP

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

P_1_IN_G1(X) -> P_1_IN_G1(a_0)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {P_1_IN_G1}.