Left Termination proof of /tmp/tmpYLGFVr/pl4.5.3b.pl

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)

↳ 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}.