Left Termination of the query pattern p_in_2(g, a) w.r.t. the given Prolog program could successfully be proven:



↳ Prolog
  ↳ PrologToPiTRSProof

Clauses:

p(X, Z) :- ','(q(X, Y), p(Y, Z)).
p(X, X).
q(a, b).

Queries:

p(g,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, X) → p_out(X, X)
p_in(X, Z) → U1(X, Z, q_in(X, Y))
q_in(a, b) → q_out(a, b)
U1(X, Z, q_out(X, Y)) → U2(X, Z, p_in(Y, Z))
U2(X, Z, p_out(Y, Z)) → p_out(X, Z)

The argument filtering Pi contains the following mapping:
p_in(x1, x2)  =  p_in(x1)
p_out(x1, x2)  =  p_out(x2)
U1(x1, x2, x3)  =  U1(x3)
q_in(x1, x2)  =  q_in(x1)
a  =  a
b  =  b
q_out(x1, x2)  =  q_out(x2)
U2(x1, x2, x3)  =  U2(x3)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog



↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof

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

p_in(X, X) → p_out(X, X)
p_in(X, Z) → U1(X, Z, q_in(X, Y))
q_in(a, b) → q_out(a, b)
U1(X, Z, q_out(X, Y)) → U2(X, Z, p_in(Y, Z))
U2(X, Z, p_out(Y, Z)) → p_out(X, Z)

The argument filtering Pi contains the following mapping:
p_in(x1, x2)  =  p_in(x1)
p_out(x1, x2)  =  p_out(x2)
U1(x1, x2, x3)  =  U1(x3)
q_in(x1, x2)  =  q_in(x1)
a  =  a
b  =  b
q_out(x1, x2)  =  q_out(x2)
U2(x1, x2, x3)  =  U2(x3)


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, Z) → U11(X, Z, q_in(X, Y))
P_IN(X, Z) → Q_IN(X, Y)
U11(X, Z, q_out(X, Y)) → U21(X, Z, p_in(Y, Z))
U11(X, Z, q_out(X, Y)) → P_IN(Y, Z)

The TRS R consists of the following rules:

p_in(X, X) → p_out(X, X)
p_in(X, Z) → U1(X, Z, q_in(X, Y))
q_in(a, b) → q_out(a, b)
U1(X, Z, q_out(X, Y)) → U2(X, Z, p_in(Y, Z))
U2(X, Z, p_out(Y, Z)) → p_out(X, Z)

The argument filtering Pi contains the following mapping:
p_in(x1, x2)  =  p_in(x1)
p_out(x1, x2)  =  p_out(x2)
U1(x1, x2, x3)  =  U1(x3)
q_in(x1, x2)  =  q_in(x1)
a  =  a
b  =  b
q_out(x1, x2)  =  q_out(x2)
U2(x1, x2, x3)  =  U2(x3)
P_IN(x1, x2)  =  P_IN(x1)
Q_IN(x1, x2)  =  Q_IN(x1)
U21(x1, x2, x3)  =  U21(x3)
U11(x1, x2, x3)  =  U11(x3)

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

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

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

P_IN(X, Z) → U11(X, Z, q_in(X, Y))
P_IN(X, Z) → Q_IN(X, Y)
U11(X, Z, q_out(X, Y)) → U21(X, Z, p_in(Y, Z))
U11(X, Z, q_out(X, Y)) → P_IN(Y, Z)

The TRS R consists of the following rules:

p_in(X, X) → p_out(X, X)
p_in(X, Z) → U1(X, Z, q_in(X, Y))
q_in(a, b) → q_out(a, b)
U1(X, Z, q_out(X, Y)) → U2(X, Z, p_in(Y, Z))
U2(X, Z, p_out(Y, Z)) → p_out(X, Z)

The argument filtering Pi contains the following mapping:
p_in(x1, x2)  =  p_in(x1)
p_out(x1, x2)  =  p_out(x2)
U1(x1, x2, x3)  =  U1(x3)
q_in(x1, x2)  =  q_in(x1)
a  =  a
b  =  b
q_out(x1, x2)  =  q_out(x2)
U2(x1, x2, x3)  =  U2(x3)
P_IN(x1, x2)  =  P_IN(x1)
Q_IN(x1, x2)  =  Q_IN(x1)
U21(x1, x2, x3)  =  U21(x3)
U11(x1, x2, x3)  =  U11(x3)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 1 SCC with 2 less nodes.

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

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

U11(X, Z, q_out(X, Y)) → P_IN(Y, Z)
P_IN(X, Z) → U11(X, Z, q_in(X, Y))

The TRS R consists of the following rules:

p_in(X, X) → p_out(X, X)
p_in(X, Z) → U1(X, Z, q_in(X, Y))
q_in(a, b) → q_out(a, b)
U1(X, Z, q_out(X, Y)) → U2(X, Z, p_in(Y, Z))
U2(X, Z, p_out(Y, Z)) → p_out(X, Z)

The argument filtering Pi contains the following mapping:
p_in(x1, x2)  =  p_in(x1)
p_out(x1, x2)  =  p_out(x2)
U1(x1, x2, x3)  =  U1(x3)
q_in(x1, x2)  =  q_in(x1)
a  =  a
b  =  b
q_out(x1, x2)  =  q_out(x2)
U2(x1, x2, x3)  =  U2(x3)
P_IN(x1, x2)  =  P_IN(x1)
U11(x1, x2, x3)  =  U11(x3)

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

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

U11(X, Z, q_out(X, Y)) → P_IN(Y, Z)
P_IN(X, Z) → U11(X, Z, q_in(X, Y))

The TRS R consists of the following rules:

q_in(a, b) → q_out(a, b)

The argument filtering Pi contains the following mapping:
q_in(x1, x2)  =  q_in(x1)
a  =  a
b  =  b
q_out(x1, x2)  =  q_out(x2)
P_IN(x1, x2)  =  P_IN(x1)
U11(x1, x2, x3)  =  U11(x3)

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
                      ↳ UsableRulesReductionPairsProof

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

U11(q_out(Y)) → P_IN(Y)
P_IN(X) → U11(q_in(X))

The TRS R consists of the following rules:

q_in(a) → q_out(b)

The set Q consists of the following terms:

q_in(x0)

We have to consider all (P,Q,R)-chains.
By using the usable rules with reduction pair processor [15] with a polynomial ordering [25], all dependency pairs and the corresponding usable rules [17] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

The following dependency pairs can be deleted:

U11(q_out(Y)) → P_IN(Y)
The following rules are removed from R:

q_in(a) → q_out(b)
Used ordering: POLO with Polynomial interpretation [25]:

POL(P_IN(x1)) = 2 + 2·x1   
POL(U11(x1)) = 2 + x1   
POL(a) = 2   
POL(b) = 0   
POL(q_in(x1)) = x1   
POL(q_out(x1)) = 1 + 2·x1   



↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ PiDP
              ↳ UsableRulesProof
                ↳ PiDP
                  ↳ PiDPToQDPProof
                    ↳ QDP
                      ↳ UsableRulesReductionPairsProof
                        ↳ QDP
                          ↳ DependencyGraphProof

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

P_IN(X) → U11(q_in(X))

R is empty.
The set Q consists of the following terms:

q_in(x0)

We have to consider all (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.