Left Termination of the query pattern goal_in_1(g) w.r.t. the given Prolog program could not be shown:



↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof

Clauses:

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

Queries:

goal(g).

We use the technique of [30]. With regard to the inferred argument filtering the predicates were used in the following modes:
goal_in: (b)
p_in: (b) (f)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

goal_in_g(X) → U2_g(X, p_in_g(X))
p_in_g(a) → p_out_g(a)
p_in_g(X) → U1_g(X, p_in_a(Y))
p_in_a(a) → p_out_a(a)
p_in_a(X) → U1_a(X, p_in_a(Y))
U1_a(X, p_out_a(Y)) → p_out_a(X)
U1_g(X, p_out_a(Y)) → p_out_g(X)
U2_g(X, p_out_g(X)) → U3_g(X, q_in_g(X))
q_in_g(b) → q_out_g(b)
U3_g(X, q_out_g(X)) → goal_out_g(X)

The argument filtering Pi contains the following mapping:
goal_in_g(x1)  =  goal_in_g(x1)
U2_g(x1, x2)  =  U2_g(x1, x2)
p_in_g(x1)  =  p_in_g(x1)
a  =  a
p_out_g(x1)  =  p_out_g
U1_g(x1, x2)  =  U1_g(x2)
p_in_a(x1)  =  p_in_a
p_out_a(x1)  =  p_out_a
U1_a(x1, x2)  =  U1_a(x2)
U3_g(x1, x2)  =  U3_g(x2)
q_in_g(x1)  =  q_in_g(x1)
b  =  b
q_out_g(x1)  =  q_out_g
goal_out_g(x1)  =  goal_out_g

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:

goal_in_g(X) → U2_g(X, p_in_g(X))
p_in_g(a) → p_out_g(a)
p_in_g(X) → U1_g(X, p_in_a(Y))
p_in_a(a) → p_out_a(a)
p_in_a(X) → U1_a(X, p_in_a(Y))
U1_a(X, p_out_a(Y)) → p_out_a(X)
U1_g(X, p_out_a(Y)) → p_out_g(X)
U2_g(X, p_out_g(X)) → U3_g(X, q_in_g(X))
q_in_g(b) → q_out_g(b)
U3_g(X, q_out_g(X)) → goal_out_g(X)

The argument filtering Pi contains the following mapping:
goal_in_g(x1)  =  goal_in_g(x1)
U2_g(x1, x2)  =  U2_g(x1, x2)
p_in_g(x1)  =  p_in_g(x1)
a  =  a
p_out_g(x1)  =  p_out_g
U1_g(x1, x2)  =  U1_g(x2)
p_in_a(x1)  =  p_in_a
p_out_a(x1)  =  p_out_a
U1_a(x1, x2)  =  U1_a(x2)
U3_g(x1, x2)  =  U3_g(x2)
q_in_g(x1)  =  q_in_g(x1)
b  =  b
q_out_g(x1)  =  q_out_g
goal_out_g(x1)  =  goal_out_g


Using Dependency Pairs [1,30] we result in the following initial DP problem:
Pi DP problem:
The TRS P consists of the following rules:

GOAL_IN_G(X) → U2_G(X, p_in_g(X))
GOAL_IN_G(X) → P_IN_G(X)
P_IN_G(X) → U1_G(X, p_in_a(Y))
P_IN_G(X) → P_IN_A(Y)
P_IN_A(X) → U1_A(X, p_in_a(Y))
P_IN_A(X) → P_IN_A(Y)
U2_G(X, p_out_g(X)) → U3_G(X, q_in_g(X))
U2_G(X, p_out_g(X)) → Q_IN_G(X)

The TRS R consists of the following rules:

goal_in_g(X) → U2_g(X, p_in_g(X))
p_in_g(a) → p_out_g(a)
p_in_g(X) → U1_g(X, p_in_a(Y))
p_in_a(a) → p_out_a(a)
p_in_a(X) → U1_a(X, p_in_a(Y))
U1_a(X, p_out_a(Y)) → p_out_a(X)
U1_g(X, p_out_a(Y)) → p_out_g(X)
U2_g(X, p_out_g(X)) → U3_g(X, q_in_g(X))
q_in_g(b) → q_out_g(b)
U3_g(X, q_out_g(X)) → goal_out_g(X)

The argument filtering Pi contains the following mapping:
goal_in_g(x1)  =  goal_in_g(x1)
U2_g(x1, x2)  =  U2_g(x1, x2)
p_in_g(x1)  =  p_in_g(x1)
a  =  a
p_out_g(x1)  =  p_out_g
U1_g(x1, x2)  =  U1_g(x2)
p_in_a(x1)  =  p_in_a
p_out_a(x1)  =  p_out_a
U1_a(x1, x2)  =  U1_a(x2)
U3_g(x1, x2)  =  U3_g(x2)
q_in_g(x1)  =  q_in_g(x1)
b  =  b
q_out_g(x1)  =  q_out_g
goal_out_g(x1)  =  goal_out_g
U2_G(x1, x2)  =  U2_G(x1, x2)
U1_A(x1, x2)  =  U1_A(x2)
P_IN_G(x1)  =  P_IN_G(x1)
GOAL_IN_G(x1)  =  GOAL_IN_G(x1)
P_IN_A(x1)  =  P_IN_A
Q_IN_G(x1)  =  Q_IN_G(x1)
U1_G(x1, x2)  =  U1_G(x2)
U3_G(x1, x2)  =  U3_G(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:

GOAL_IN_G(X) → U2_G(X, p_in_g(X))
GOAL_IN_G(X) → P_IN_G(X)
P_IN_G(X) → U1_G(X, p_in_a(Y))
P_IN_G(X) → P_IN_A(Y)
P_IN_A(X) → U1_A(X, p_in_a(Y))
P_IN_A(X) → P_IN_A(Y)
U2_G(X, p_out_g(X)) → U3_G(X, q_in_g(X))
U2_G(X, p_out_g(X)) → Q_IN_G(X)

The TRS R consists of the following rules:

goal_in_g(X) → U2_g(X, p_in_g(X))
p_in_g(a) → p_out_g(a)
p_in_g(X) → U1_g(X, p_in_a(Y))
p_in_a(a) → p_out_a(a)
p_in_a(X) → U1_a(X, p_in_a(Y))
U1_a(X, p_out_a(Y)) → p_out_a(X)
U1_g(X, p_out_a(Y)) → p_out_g(X)
U2_g(X, p_out_g(X)) → U3_g(X, q_in_g(X))
q_in_g(b) → q_out_g(b)
U3_g(X, q_out_g(X)) → goal_out_g(X)

The argument filtering Pi contains the following mapping:
goal_in_g(x1)  =  goal_in_g(x1)
U2_g(x1, x2)  =  U2_g(x1, x2)
p_in_g(x1)  =  p_in_g(x1)
a  =  a
p_out_g(x1)  =  p_out_g
U1_g(x1, x2)  =  U1_g(x2)
p_in_a(x1)  =  p_in_a
p_out_a(x1)  =  p_out_a
U1_a(x1, x2)  =  U1_a(x2)
U3_g(x1, x2)  =  U3_g(x2)
q_in_g(x1)  =  q_in_g(x1)
b  =  b
q_out_g(x1)  =  q_out_g
goal_out_g(x1)  =  goal_out_g
U2_G(x1, x2)  =  U2_G(x1, x2)
U1_A(x1, x2)  =  U1_A(x2)
P_IN_G(x1)  =  P_IN_G(x1)
GOAL_IN_G(x1)  =  GOAL_IN_G(x1)
P_IN_A(x1)  =  P_IN_A
Q_IN_G(x1)  =  Q_IN_G(x1)
U1_G(x1, x2)  =  U1_G(x2)
U3_G(x1, x2)  =  U3_G(x2)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 1 SCC with 7 less nodes.

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

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

P_IN_A(X) → P_IN_A(Y)

The TRS R consists of the following rules:

goal_in_g(X) → U2_g(X, p_in_g(X))
p_in_g(a) → p_out_g(a)
p_in_g(X) → U1_g(X, p_in_a(Y))
p_in_a(a) → p_out_a(a)
p_in_a(X) → U1_a(X, p_in_a(Y))
U1_a(X, p_out_a(Y)) → p_out_a(X)
U1_g(X, p_out_a(Y)) → p_out_g(X)
U2_g(X, p_out_g(X)) → U3_g(X, q_in_g(X))
q_in_g(b) → q_out_g(b)
U3_g(X, q_out_g(X)) → goal_out_g(X)

The argument filtering Pi contains the following mapping:
goal_in_g(x1)  =  goal_in_g(x1)
U2_g(x1, x2)  =  U2_g(x1, x2)
p_in_g(x1)  =  p_in_g(x1)
a  =  a
p_out_g(x1)  =  p_out_g
U1_g(x1, x2)  =  U1_g(x2)
p_in_a(x1)  =  p_in_a
p_out_a(x1)  =  p_out_a
U1_a(x1, x2)  =  U1_a(x2)
U3_g(x1, x2)  =  U3_g(x2)
q_in_g(x1)  =  q_in_g(x1)
b  =  b
q_out_g(x1)  =  q_out_g
goal_out_g(x1)  =  goal_out_g
P_IN_A(x1)  =  P_IN_A

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:

P_IN_A(X) → P_IN_A(Y)

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

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:

P_IN_A → P_IN_A

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:

P_IN_A → P_IN_A

The TRS R consists of the following rules:none


s = P_IN_A evaluates to t =P_IN_A

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 P_IN_A to P_IN_A.




We use the technique of [30]. With regard to the inferred argument filtering the predicates were used in the following modes:
goal_in: (b)
p_in: (b) (f)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

goal_in_g(X) → U2_g(X, p_in_g(X))
p_in_g(a) → p_out_g(a)
p_in_g(X) → U1_g(X, p_in_a(Y))
p_in_a(a) → p_out_a(a)
p_in_a(X) → U1_a(X, p_in_a(Y))
U1_a(X, p_out_a(Y)) → p_out_a(X)
U1_g(X, p_out_a(Y)) → p_out_g(X)
U2_g(X, p_out_g(X)) → U3_g(X, q_in_g(X))
q_in_g(b) → q_out_g(b)
U3_g(X, q_out_g(X)) → goal_out_g(X)

The argument filtering Pi contains the following mapping:
goal_in_g(x1)  =  goal_in_g(x1)
U2_g(x1, x2)  =  U2_g(x1, x2)
p_in_g(x1)  =  p_in_g(x1)
a  =  a
p_out_g(x1)  =  p_out_g(x1)
U1_g(x1, x2)  =  U1_g(x1, x2)
p_in_a(x1)  =  p_in_a
p_out_a(x1)  =  p_out_a
U1_a(x1, x2)  =  U1_a(x2)
U3_g(x1, x2)  =  U3_g(x1, x2)
q_in_g(x1)  =  q_in_g(x1)
b  =  b
q_out_g(x1)  =  q_out_g(x1)
goal_out_g(x1)  =  goal_out_g(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:

goal_in_g(X) → U2_g(X, p_in_g(X))
p_in_g(a) → p_out_g(a)
p_in_g(X) → U1_g(X, p_in_a(Y))
p_in_a(a) → p_out_a(a)
p_in_a(X) → U1_a(X, p_in_a(Y))
U1_a(X, p_out_a(Y)) → p_out_a(X)
U1_g(X, p_out_a(Y)) → p_out_g(X)
U2_g(X, p_out_g(X)) → U3_g(X, q_in_g(X))
q_in_g(b) → q_out_g(b)
U3_g(X, q_out_g(X)) → goal_out_g(X)

The argument filtering Pi contains the following mapping:
goal_in_g(x1)  =  goal_in_g(x1)
U2_g(x1, x2)  =  U2_g(x1, x2)
p_in_g(x1)  =  p_in_g(x1)
a  =  a
p_out_g(x1)  =  p_out_g(x1)
U1_g(x1, x2)  =  U1_g(x1, x2)
p_in_a(x1)  =  p_in_a
p_out_a(x1)  =  p_out_a
U1_a(x1, x2)  =  U1_a(x2)
U3_g(x1, x2)  =  U3_g(x1, x2)
q_in_g(x1)  =  q_in_g(x1)
b  =  b
q_out_g(x1)  =  q_out_g(x1)
goal_out_g(x1)  =  goal_out_g(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:

GOAL_IN_G(X) → U2_G(X, p_in_g(X))
GOAL_IN_G(X) → P_IN_G(X)
P_IN_G(X) → U1_G(X, p_in_a(Y))
P_IN_G(X) → P_IN_A(Y)
P_IN_A(X) → U1_A(X, p_in_a(Y))
P_IN_A(X) → P_IN_A(Y)
U2_G(X, p_out_g(X)) → U3_G(X, q_in_g(X))
U2_G(X, p_out_g(X)) → Q_IN_G(X)

The TRS R consists of the following rules:

goal_in_g(X) → U2_g(X, p_in_g(X))
p_in_g(a) → p_out_g(a)
p_in_g(X) → U1_g(X, p_in_a(Y))
p_in_a(a) → p_out_a(a)
p_in_a(X) → U1_a(X, p_in_a(Y))
U1_a(X, p_out_a(Y)) → p_out_a(X)
U1_g(X, p_out_a(Y)) → p_out_g(X)
U2_g(X, p_out_g(X)) → U3_g(X, q_in_g(X))
q_in_g(b) → q_out_g(b)
U3_g(X, q_out_g(X)) → goal_out_g(X)

The argument filtering Pi contains the following mapping:
goal_in_g(x1)  =  goal_in_g(x1)
U2_g(x1, x2)  =  U2_g(x1, x2)
p_in_g(x1)  =  p_in_g(x1)
a  =  a
p_out_g(x1)  =  p_out_g(x1)
U1_g(x1, x2)  =  U1_g(x1, x2)
p_in_a(x1)  =  p_in_a
p_out_a(x1)  =  p_out_a
U1_a(x1, x2)  =  U1_a(x2)
U3_g(x1, x2)  =  U3_g(x1, x2)
q_in_g(x1)  =  q_in_g(x1)
b  =  b
q_out_g(x1)  =  q_out_g(x1)
goal_out_g(x1)  =  goal_out_g(x1)
U2_G(x1, x2)  =  U2_G(x1, x2)
U1_A(x1, x2)  =  U1_A(x2)
P_IN_G(x1)  =  P_IN_G(x1)
GOAL_IN_G(x1)  =  GOAL_IN_G(x1)
P_IN_A(x1)  =  P_IN_A
Q_IN_G(x1)  =  Q_IN_G(x1)
U1_G(x1, x2)  =  U1_G(x1, x2)
U3_G(x1, x2)  =  U3_G(x1, 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:

GOAL_IN_G(X) → U2_G(X, p_in_g(X))
GOAL_IN_G(X) → P_IN_G(X)
P_IN_G(X) → U1_G(X, p_in_a(Y))
P_IN_G(X) → P_IN_A(Y)
P_IN_A(X) → U1_A(X, p_in_a(Y))
P_IN_A(X) → P_IN_A(Y)
U2_G(X, p_out_g(X)) → U3_G(X, q_in_g(X))
U2_G(X, p_out_g(X)) → Q_IN_G(X)

The TRS R consists of the following rules:

goal_in_g(X) → U2_g(X, p_in_g(X))
p_in_g(a) → p_out_g(a)
p_in_g(X) → U1_g(X, p_in_a(Y))
p_in_a(a) → p_out_a(a)
p_in_a(X) → U1_a(X, p_in_a(Y))
U1_a(X, p_out_a(Y)) → p_out_a(X)
U1_g(X, p_out_a(Y)) → p_out_g(X)
U2_g(X, p_out_g(X)) → U3_g(X, q_in_g(X))
q_in_g(b) → q_out_g(b)
U3_g(X, q_out_g(X)) → goal_out_g(X)

The argument filtering Pi contains the following mapping:
goal_in_g(x1)  =  goal_in_g(x1)
U2_g(x1, x2)  =  U2_g(x1, x2)
p_in_g(x1)  =  p_in_g(x1)
a  =  a
p_out_g(x1)  =  p_out_g(x1)
U1_g(x1, x2)  =  U1_g(x1, x2)
p_in_a(x1)  =  p_in_a
p_out_a(x1)  =  p_out_a
U1_a(x1, x2)  =  U1_a(x2)
U3_g(x1, x2)  =  U3_g(x1, x2)
q_in_g(x1)  =  q_in_g(x1)
b  =  b
q_out_g(x1)  =  q_out_g(x1)
goal_out_g(x1)  =  goal_out_g(x1)
U2_G(x1, x2)  =  U2_G(x1, x2)
U1_A(x1, x2)  =  U1_A(x2)
P_IN_G(x1)  =  P_IN_G(x1)
GOAL_IN_G(x1)  =  GOAL_IN_G(x1)
P_IN_A(x1)  =  P_IN_A
Q_IN_G(x1)  =  Q_IN_G(x1)
U1_G(x1, x2)  =  U1_G(x1, x2)
U3_G(x1, x2)  =  U3_G(x1, x2)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 1 SCC with 7 less nodes.

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

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

P_IN_A(X) → P_IN_A(Y)

The TRS R consists of the following rules:

goal_in_g(X) → U2_g(X, p_in_g(X))
p_in_g(a) → p_out_g(a)
p_in_g(X) → U1_g(X, p_in_a(Y))
p_in_a(a) → p_out_a(a)
p_in_a(X) → U1_a(X, p_in_a(Y))
U1_a(X, p_out_a(Y)) → p_out_a(X)
U1_g(X, p_out_a(Y)) → p_out_g(X)
U2_g(X, p_out_g(X)) → U3_g(X, q_in_g(X))
q_in_g(b) → q_out_g(b)
U3_g(X, q_out_g(X)) → goal_out_g(X)

The argument filtering Pi contains the following mapping:
goal_in_g(x1)  =  goal_in_g(x1)
U2_g(x1, x2)  =  U2_g(x1, x2)
p_in_g(x1)  =  p_in_g(x1)
a  =  a
p_out_g(x1)  =  p_out_g(x1)
U1_g(x1, x2)  =  U1_g(x1, x2)
p_in_a(x1)  =  p_in_a
p_out_a(x1)  =  p_out_a
U1_a(x1, x2)  =  U1_a(x2)
U3_g(x1, x2)  =  U3_g(x1, x2)
q_in_g(x1)  =  q_in_g(x1)
b  =  b
q_out_g(x1)  =  q_out_g(x1)
goal_out_g(x1)  =  goal_out_g(x1)
P_IN_A(x1)  =  P_IN_A

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:

P_IN_A(X) → P_IN_A(Y)

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

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:

P_IN_A → P_IN_A

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:

P_IN_A → P_IN_A

The TRS R consists of the following rules:none


s = P_IN_A evaluates to t =P_IN_A

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 P_IN_A to P_IN_A.