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].Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

goal_in(X) → U2(X, p_in(X))
p_in(X) → U1(X, p_in(Y))
p_in(a) → p_out(a)
U1(X, p_out(Y)) → p_out(X)
U2(X, p_out(X)) → U3(X, q_in(X))
q_in(b) → q_out(b)
U3(X, q_out(X)) → goal_out(X)

The argument filtering Pi contains the following mapping:
goal_in(x1)  =  goal_in(x1)
U2(x1, x2)  =  U2(x1, x2)
p_in(x1)  =  p_in
U1(x1, x2)  =  U1(x2)
a  =  a
p_out(x1)  =  p_out
U3(x1, x2)  =  U3(x2)
q_in(x1)  =  q_in(x1)
b  =  b
q_out(x1)  =  q_out
goal_out(x1)  =  goal_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:

goal_in(X) → U2(X, p_in(X))
p_in(X) → U1(X, p_in(Y))
p_in(a) → p_out(a)
U1(X, p_out(Y)) → p_out(X)
U2(X, p_out(X)) → U3(X, q_in(X))
q_in(b) → q_out(b)
U3(X, q_out(X)) → goal_out(X)

The argument filtering Pi contains the following mapping:
goal_in(x1)  =  goal_in(x1)
U2(x1, x2)  =  U2(x1, x2)
p_in(x1)  =  p_in
U1(x1, x2)  =  U1(x2)
a  =  a
p_out(x1)  =  p_out
U3(x1, x2)  =  U3(x2)
q_in(x1)  =  q_in(x1)
b  =  b
q_out(x1)  =  q_out
goal_out(x1)  =  goal_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:

GOAL_IN(X) → U21(X, p_in(X))
GOAL_IN(X) → P_IN(X)
P_IN(X) → U11(X, p_in(Y))
P_IN(X) → P_IN(Y)
U21(X, p_out(X)) → U31(X, q_in(X))
U21(X, p_out(X)) → Q_IN(X)

The TRS R consists of the following rules:

goal_in(X) → U2(X, p_in(X))
p_in(X) → U1(X, p_in(Y))
p_in(a) → p_out(a)
U1(X, p_out(Y)) → p_out(X)
U2(X, p_out(X)) → U3(X, q_in(X))
q_in(b) → q_out(b)
U3(X, q_out(X)) → goal_out(X)

The argument filtering Pi contains the following mapping:
goal_in(x1)  =  goal_in(x1)
U2(x1, x2)  =  U2(x1, x2)
p_in(x1)  =  p_in
U1(x1, x2)  =  U1(x2)
a  =  a
p_out(x1)  =  p_out
U3(x1, x2)  =  U3(x2)
q_in(x1)  =  q_in(x1)
b  =  b
q_out(x1)  =  q_out
goal_out(x1)  =  goal_out
P_IN(x1)  =  P_IN
Q_IN(x1)  =  Q_IN(x1)
U11(x1, x2)  =  U11(x2)
U31(x1, x2)  =  U31(x2)
U21(x1, x2)  =  U21(x1, x2)
GOAL_IN(x1)  =  GOAL_IN(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:

GOAL_IN(X) → U21(X, p_in(X))
GOAL_IN(X) → P_IN(X)
P_IN(X) → U11(X, p_in(Y))
P_IN(X) → P_IN(Y)
U21(X, p_out(X)) → U31(X, q_in(X))
U21(X, p_out(X)) → Q_IN(X)

The TRS R consists of the following rules:

goal_in(X) → U2(X, p_in(X))
p_in(X) → U1(X, p_in(Y))
p_in(a) → p_out(a)
U1(X, p_out(Y)) → p_out(X)
U2(X, p_out(X)) → U3(X, q_in(X))
q_in(b) → q_out(b)
U3(X, q_out(X)) → goal_out(X)

The argument filtering Pi contains the following mapping:
goal_in(x1)  =  goal_in(x1)
U2(x1, x2)  =  U2(x1, x2)
p_in(x1)  =  p_in
U1(x1, x2)  =  U1(x2)
a  =  a
p_out(x1)  =  p_out
U3(x1, x2)  =  U3(x2)
q_in(x1)  =  q_in(x1)
b  =  b
q_out(x1)  =  q_out
goal_out(x1)  =  goal_out
P_IN(x1)  =  P_IN
Q_IN(x1)  =  Q_IN(x1)
U11(x1, x2)  =  U11(x2)
U31(x1, x2)  =  U31(x2)
U21(x1, x2)  =  U21(x1, x2)
GOAL_IN(x1)  =  GOAL_IN(x1)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 1 SCC with 5 less nodes.

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

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

P_IN(X) → P_IN(Y)

The TRS R consists of the following rules:

goal_in(X) → U2(X, p_in(X))
p_in(X) → U1(X, p_in(Y))
p_in(a) → p_out(a)
U1(X, p_out(Y)) → p_out(X)
U2(X, p_out(X)) → U3(X, q_in(X))
q_in(b) → q_out(b)
U3(X, q_out(X)) → goal_out(X)

The argument filtering Pi contains the following mapping:
goal_in(x1)  =  goal_in(x1)
U2(x1, x2)  =  U2(x1, x2)
p_in(x1)  =  p_in
U1(x1, x2)  =  U1(x2)
a  =  a
p_out(x1)  =  p_out
U3(x1, x2)  =  U3(x2)
q_in(x1)  =  q_in(x1)
b  =  b
q_out(x1)  =  q_out
goal_out(x1)  =  goal_out
P_IN(x1)  =  P_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:

P_IN(X) → P_IN(Y)

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

P_IN → P_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:

P_IN → P_IN

The TRS R consists of the following rules:none


s = P_IN evaluates to t =P_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 P_IN to P_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:

goal_in(X) → U2(X, p_in(X))
p_in(X) → U1(X, p_in(Y))
p_in(a) → p_out(a)
U1(X, p_out(Y)) → p_out(X)
U2(X, p_out(X)) → U3(X, q_in(X))
q_in(b) → q_out(b)
U3(X, q_out(X)) → goal_out(X)

The argument filtering Pi contains the following mapping:
goal_in(x1)  =  goal_in(x1)
U2(x1, x2)  =  U2(x1, x2)
p_in(x1)  =  p_in
U1(x1, x2)  =  U1(x2)
a  =  a
p_out(x1)  =  p_out
U3(x1, x2)  =  U3(x1, x2)
q_in(x1)  =  q_in(x1)
b  =  b
q_out(x1)  =  q_out(x1)
goal_out(x1)  =  goal_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:

goal_in(X) → U2(X, p_in(X))
p_in(X) → U1(X, p_in(Y))
p_in(a) → p_out(a)
U1(X, p_out(Y)) → p_out(X)
U2(X, p_out(X)) → U3(X, q_in(X))
q_in(b) → q_out(b)
U3(X, q_out(X)) → goal_out(X)

The argument filtering Pi contains the following mapping:
goal_in(x1)  =  goal_in(x1)
U2(x1, x2)  =  U2(x1, x2)
p_in(x1)  =  p_in
U1(x1, x2)  =  U1(x2)
a  =  a
p_out(x1)  =  p_out
U3(x1, x2)  =  U3(x1, x2)
q_in(x1)  =  q_in(x1)
b  =  b
q_out(x1)  =  q_out(x1)
goal_out(x1)  =  goal_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:

GOAL_IN(X) → U21(X, p_in(X))
GOAL_IN(X) → P_IN(X)
P_IN(X) → U11(X, p_in(Y))
P_IN(X) → P_IN(Y)
U21(X, p_out(X)) → U31(X, q_in(X))
U21(X, p_out(X)) → Q_IN(X)

The TRS R consists of the following rules:

goal_in(X) → U2(X, p_in(X))
p_in(X) → U1(X, p_in(Y))
p_in(a) → p_out(a)
U1(X, p_out(Y)) → p_out(X)
U2(X, p_out(X)) → U3(X, q_in(X))
q_in(b) → q_out(b)
U3(X, q_out(X)) → goal_out(X)

The argument filtering Pi contains the following mapping:
goal_in(x1)  =  goal_in(x1)
U2(x1, x2)  =  U2(x1, x2)
p_in(x1)  =  p_in
U1(x1, x2)  =  U1(x2)
a  =  a
p_out(x1)  =  p_out
U3(x1, x2)  =  U3(x1, x2)
q_in(x1)  =  q_in(x1)
b  =  b
q_out(x1)  =  q_out(x1)
goal_out(x1)  =  goal_out(x1)
P_IN(x1)  =  P_IN
Q_IN(x1)  =  Q_IN(x1)
U11(x1, x2)  =  U11(x2)
U31(x1, x2)  =  U31(x1, x2)
U21(x1, x2)  =  U21(x1, x2)
GOAL_IN(x1)  =  GOAL_IN(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:

GOAL_IN(X) → U21(X, p_in(X))
GOAL_IN(X) → P_IN(X)
P_IN(X) → U11(X, p_in(Y))
P_IN(X) → P_IN(Y)
U21(X, p_out(X)) → U31(X, q_in(X))
U21(X, p_out(X)) → Q_IN(X)

The TRS R consists of the following rules:

goal_in(X) → U2(X, p_in(X))
p_in(X) → U1(X, p_in(Y))
p_in(a) → p_out(a)
U1(X, p_out(Y)) → p_out(X)
U2(X, p_out(X)) → U3(X, q_in(X))
q_in(b) → q_out(b)
U3(X, q_out(X)) → goal_out(X)

The argument filtering Pi contains the following mapping:
goal_in(x1)  =  goal_in(x1)
U2(x1, x2)  =  U2(x1, x2)
p_in(x1)  =  p_in
U1(x1, x2)  =  U1(x2)
a  =  a
p_out(x1)  =  p_out
U3(x1, x2)  =  U3(x1, x2)
q_in(x1)  =  q_in(x1)
b  =  b
q_out(x1)  =  q_out(x1)
goal_out(x1)  =  goal_out(x1)
P_IN(x1)  =  P_IN
Q_IN(x1)  =  Q_IN(x1)
U11(x1, x2)  =  U11(x2)
U31(x1, x2)  =  U31(x1, x2)
U21(x1, x2)  =  U21(x1, x2)
GOAL_IN(x1)  =  GOAL_IN(x1)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 1 SCC with 5 less nodes.

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

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

P_IN(X) → P_IN(Y)

The TRS R consists of the following rules:

goal_in(X) → U2(X, p_in(X))
p_in(X) → U1(X, p_in(Y))
p_in(a) → p_out(a)
U1(X, p_out(Y)) → p_out(X)
U2(X, p_out(X)) → U3(X, q_in(X))
q_in(b) → q_out(b)
U3(X, q_out(X)) → goal_out(X)

The argument filtering Pi contains the following mapping:
goal_in(x1)  =  goal_in(x1)
U2(x1, x2)  =  U2(x1, x2)
p_in(x1)  =  p_in
U1(x1, x2)  =  U1(x2)
a  =  a
p_out(x1)  =  p_out
U3(x1, x2)  =  U3(x1, x2)
q_in(x1)  =  q_in(x1)
b  =  b
q_out(x1)  =  q_out(x1)
goal_out(x1)  =  goal_out(x1)
P_IN(x1)  =  P_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:

P_IN(X) → P_IN(Y)

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

P_IN → P_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:

P_IN → P_IN

The TRS R consists of the following rules:none


s = P_IN evaluates to t =P_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 P_IN to P_IN.