Left Termination of the query pattern p_in_1(a) w.r.t. the given Prolog program could successfully be proven:



↳ Prolog
  ↳ PrologToPiTRSProof

Clauses:

p(X) :- ','(q(f(Y)), p(Y)).
q(g(Y)).

Queries:

p(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) → U1(X, q_in(f(Y)))
q_in(g(Y)) → q_out(g(Y))
U1(X, q_out(f(Y))) → U2(X, p_in(Y))
U2(X, p_out(Y)) → p_out(X)

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

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) → U1(X, q_in(f(Y)))
q_in(g(Y)) → q_out(g(Y))
U1(X, q_out(f(Y))) → U2(X, p_in(Y))
U2(X, p_out(Y)) → p_out(X)

The argument filtering Pi contains the following mapping:
p_in(x1)  =  p_in
U1(x1, x2)  =  U1(x2)
q_in(x1)  =  q_in(x1)
f(x1)  =  f
g(x1)  =  g(x1)
q_out(x1)  =  q_out
U2(x1, x2)  =  U2(x2)
p_out(x1)  =  p_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:

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

The TRS R consists of the following rules:

p_in(X) → U1(X, q_in(f(Y)))
q_in(g(Y)) → q_out(g(Y))
U1(X, q_out(f(Y))) → U2(X, p_in(Y))
U2(X, p_out(Y)) → p_out(X)

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

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

The TRS R consists of the following rules:

p_in(X) → U1(X, q_in(f(Y)))
q_in(g(Y)) → q_out(g(Y))
U1(X, q_out(f(Y))) → U2(X, p_in(Y))
U2(X, p_out(Y)) → p_out(X)

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

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 0 SCCs with 4 less nodes.