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



↳ PROLOG
  ↳ PrologToPiTRSProof

p1(X) :- q1(f1(Y)), p1(Y).
q1(g1(Y)).


With regard to the inferred argument filtering the predicates were used in the following modes:
p1: (f) (b)
Transforming PROLOG into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:


p_1_in_a1(X) -> if_p_1_in_1_a2(X, q_1_in_a1(f_11(Y)))
q_1_in_a1(g_11(Y)) -> q_1_out_a1(g_11(Y))
if_p_1_in_1_a2(X, q_1_out_a1(f_11(Y))) -> if_p_1_in_2_a3(X, Y, p_1_in_g1(Y))
p_1_in_g1(X) -> if_p_1_in_1_g2(X, q_1_in_a1(f_11(Y)))
if_p_1_in_1_g2(X, q_1_out_a1(f_11(Y))) -> if_p_1_in_2_g3(X, Y, p_1_in_g1(Y))
if_p_1_in_2_g3(X, Y, p_1_out_g1(Y)) -> p_1_out_g1(X)
if_p_1_in_2_a3(X, Y, p_1_out_g1(Y)) -> p_1_out_a1(X)

The argument filtering Pi contains the following mapping:
p_1_in_a1(x1)  =  p_1_in_a
f_11(x1)  =  f_11(x1)
g_11(x1)  =  g_1
if_p_1_in_1_a2(x1, x2)  =  if_p_1_in_1_a1(x2)
q_1_in_a1(x1)  =  q_1_in_a
q_1_out_a1(x1)  =  q_1_out_a1(x1)
if_p_1_in_2_a3(x1, x2, x3)  =  if_p_1_in_2_a1(x3)
p_1_in_g1(x1)  =  p_1_in_g1(x1)
if_p_1_in_1_g2(x1, x2)  =  if_p_1_in_1_g1(x2)
if_p_1_in_2_g3(x1, x2, x3)  =  if_p_1_in_2_g1(x3)
p_1_out_g1(x1)  =  p_1_out_g
p_1_out_a1(x1)  =  p_1_out_a

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_1_in_a1(X) -> if_p_1_in_1_a2(X, q_1_in_a1(f_11(Y)))
q_1_in_a1(g_11(Y)) -> q_1_out_a1(g_11(Y))
if_p_1_in_1_a2(X, q_1_out_a1(f_11(Y))) -> if_p_1_in_2_a3(X, Y, p_1_in_g1(Y))
p_1_in_g1(X) -> if_p_1_in_1_g2(X, q_1_in_a1(f_11(Y)))
if_p_1_in_1_g2(X, q_1_out_a1(f_11(Y))) -> if_p_1_in_2_g3(X, Y, p_1_in_g1(Y))
if_p_1_in_2_g3(X, Y, p_1_out_g1(Y)) -> p_1_out_g1(X)
if_p_1_in_2_a3(X, Y, p_1_out_g1(Y)) -> p_1_out_a1(X)

The argument filtering Pi contains the following mapping:
p_1_in_a1(x1)  =  p_1_in_a
f_11(x1)  =  f_11(x1)
g_11(x1)  =  g_1
if_p_1_in_1_a2(x1, x2)  =  if_p_1_in_1_a1(x2)
q_1_in_a1(x1)  =  q_1_in_a
q_1_out_a1(x1)  =  q_1_out_a1(x1)
if_p_1_in_2_a3(x1, x2, x3)  =  if_p_1_in_2_a1(x3)
p_1_in_g1(x1)  =  p_1_in_g1(x1)
if_p_1_in_1_g2(x1, x2)  =  if_p_1_in_1_g1(x2)
if_p_1_in_2_g3(x1, x2, x3)  =  if_p_1_in_2_g1(x3)
p_1_out_g1(x1)  =  p_1_out_g
p_1_out_a1(x1)  =  p_1_out_a


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

P_1_IN_A1(X) -> IF_P_1_IN_1_A2(X, q_1_in_a1(f_11(Y)))
P_1_IN_A1(X) -> Q_1_IN_A1(f_11(Y))
IF_P_1_IN_1_A2(X, q_1_out_a1(f_11(Y))) -> IF_P_1_IN_2_A3(X, Y, p_1_in_g1(Y))
IF_P_1_IN_1_A2(X, q_1_out_a1(f_11(Y))) -> P_1_IN_G1(Y)
P_1_IN_G1(X) -> IF_P_1_IN_1_G2(X, q_1_in_a1(f_11(Y)))
P_1_IN_G1(X) -> Q_1_IN_A1(f_11(Y))
IF_P_1_IN_1_G2(X, q_1_out_a1(f_11(Y))) -> IF_P_1_IN_2_G3(X, Y, p_1_in_g1(Y))
IF_P_1_IN_1_G2(X, q_1_out_a1(f_11(Y))) -> P_1_IN_G1(Y)

The TRS R consists of the following rules:

p_1_in_a1(X) -> if_p_1_in_1_a2(X, q_1_in_a1(f_11(Y)))
q_1_in_a1(g_11(Y)) -> q_1_out_a1(g_11(Y))
if_p_1_in_1_a2(X, q_1_out_a1(f_11(Y))) -> if_p_1_in_2_a3(X, Y, p_1_in_g1(Y))
p_1_in_g1(X) -> if_p_1_in_1_g2(X, q_1_in_a1(f_11(Y)))
if_p_1_in_1_g2(X, q_1_out_a1(f_11(Y))) -> if_p_1_in_2_g3(X, Y, p_1_in_g1(Y))
if_p_1_in_2_g3(X, Y, p_1_out_g1(Y)) -> p_1_out_g1(X)
if_p_1_in_2_a3(X, Y, p_1_out_g1(Y)) -> p_1_out_a1(X)

The argument filtering Pi contains the following mapping:
p_1_in_a1(x1)  =  p_1_in_a
f_11(x1)  =  f_11(x1)
g_11(x1)  =  g_1
if_p_1_in_1_a2(x1, x2)  =  if_p_1_in_1_a1(x2)
q_1_in_a1(x1)  =  q_1_in_a
q_1_out_a1(x1)  =  q_1_out_a1(x1)
if_p_1_in_2_a3(x1, x2, x3)  =  if_p_1_in_2_a1(x3)
p_1_in_g1(x1)  =  p_1_in_g1(x1)
if_p_1_in_1_g2(x1, x2)  =  if_p_1_in_1_g1(x2)
if_p_1_in_2_g3(x1, x2, x3)  =  if_p_1_in_2_g1(x3)
p_1_out_g1(x1)  =  p_1_out_g
p_1_out_a1(x1)  =  p_1_out_a
IF_P_1_IN_2_G3(x1, x2, x3)  =  IF_P_1_IN_2_G1(x3)
IF_P_1_IN_2_A3(x1, x2, x3)  =  IF_P_1_IN_2_A1(x3)
P_1_IN_A1(x1)  =  P_1_IN_A
Q_1_IN_A1(x1)  =  Q_1_IN_A
IF_P_1_IN_1_G2(x1, x2)  =  IF_P_1_IN_1_G1(x2)
IF_P_1_IN_1_A2(x1, x2)  =  IF_P_1_IN_1_A1(x2)
P_1_IN_G1(x1)  =  P_1_IN_G1(x1)

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_1_IN_A1(X) -> IF_P_1_IN_1_A2(X, q_1_in_a1(f_11(Y)))
P_1_IN_A1(X) -> Q_1_IN_A1(f_11(Y))
IF_P_1_IN_1_A2(X, q_1_out_a1(f_11(Y))) -> IF_P_1_IN_2_A3(X, Y, p_1_in_g1(Y))
IF_P_1_IN_1_A2(X, q_1_out_a1(f_11(Y))) -> P_1_IN_G1(Y)
P_1_IN_G1(X) -> IF_P_1_IN_1_G2(X, q_1_in_a1(f_11(Y)))
P_1_IN_G1(X) -> Q_1_IN_A1(f_11(Y))
IF_P_1_IN_1_G2(X, q_1_out_a1(f_11(Y))) -> IF_P_1_IN_2_G3(X, Y, p_1_in_g1(Y))
IF_P_1_IN_1_G2(X, q_1_out_a1(f_11(Y))) -> P_1_IN_G1(Y)

The TRS R consists of the following rules:

p_1_in_a1(X) -> if_p_1_in_1_a2(X, q_1_in_a1(f_11(Y)))
q_1_in_a1(g_11(Y)) -> q_1_out_a1(g_11(Y))
if_p_1_in_1_a2(X, q_1_out_a1(f_11(Y))) -> if_p_1_in_2_a3(X, Y, p_1_in_g1(Y))
p_1_in_g1(X) -> if_p_1_in_1_g2(X, q_1_in_a1(f_11(Y)))
if_p_1_in_1_g2(X, q_1_out_a1(f_11(Y))) -> if_p_1_in_2_g3(X, Y, p_1_in_g1(Y))
if_p_1_in_2_g3(X, Y, p_1_out_g1(Y)) -> p_1_out_g1(X)
if_p_1_in_2_a3(X, Y, p_1_out_g1(Y)) -> p_1_out_a1(X)

The argument filtering Pi contains the following mapping:
p_1_in_a1(x1)  =  p_1_in_a
f_11(x1)  =  f_11(x1)
g_11(x1)  =  g_1
if_p_1_in_1_a2(x1, x2)  =  if_p_1_in_1_a1(x2)
q_1_in_a1(x1)  =  q_1_in_a
q_1_out_a1(x1)  =  q_1_out_a1(x1)
if_p_1_in_2_a3(x1, x2, x3)  =  if_p_1_in_2_a1(x3)
p_1_in_g1(x1)  =  p_1_in_g1(x1)
if_p_1_in_1_g2(x1, x2)  =  if_p_1_in_1_g1(x2)
if_p_1_in_2_g3(x1, x2, x3)  =  if_p_1_in_2_g1(x3)
p_1_out_g1(x1)  =  p_1_out_g
p_1_out_a1(x1)  =  p_1_out_a
IF_P_1_IN_2_G3(x1, x2, x3)  =  IF_P_1_IN_2_G1(x3)
IF_P_1_IN_2_A3(x1, x2, x3)  =  IF_P_1_IN_2_A1(x3)
P_1_IN_A1(x1)  =  P_1_IN_A
Q_1_IN_A1(x1)  =  Q_1_IN_A
IF_P_1_IN_1_G2(x1, x2)  =  IF_P_1_IN_1_G1(x2)
IF_P_1_IN_1_A2(x1, x2)  =  IF_P_1_IN_1_A1(x2)
P_1_IN_G1(x1)  =  P_1_IN_G1(x1)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph contains 0 SCCs with 8 less nodes.