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



↳ PROLOG
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof

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


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


p_1_in_g1(X) -> if_p_1_in_1_g2(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_g2(X, q_1_out_a1(f_11(Y))) -> if_p_1_in_2_g3(X, Y, p_1_in_g1(Y))
p_1_in_g1(g_11(X)) -> if_p_1_in_3_g2(X, p_1_in_a1(X))
p_1_in_a1(X) -> if_p_1_in_1_a2(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_2_a3(X, Y, p_1_out_g1(Y)) -> p_1_out_a1(X)
p_1_in_a1(g_11(X)) -> if_p_1_in_3_a2(X, p_1_in_a1(X))
if_p_1_in_3_a2(X, p_1_out_a1(X)) -> p_1_out_a1(g_11(X))
if_p_1_in_3_g2(X, p_1_out_a1(X)) -> p_1_out_g1(g_11(X))
if_p_1_in_2_g3(X, Y, p_1_out_g1(Y)) -> p_1_out_g1(X)

The argument filtering Pi contains the following mapping:
p_1_in_g1(x1)  =  p_1_in_g1(x1)
f_11(x1)  =  f_11(x1)
g_11(x1)  =  g_1
if_p_1_in_1_g2(x1, x2)  =  if_p_1_in_1_g1(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_g3(x1, x2, x3)  =  if_p_1_in_2_g1(x3)
if_p_1_in_3_g2(x1, x2)  =  if_p_1_in_3_g1(x2)
p_1_in_a1(x1)  =  p_1_in_a
if_p_1_in_1_a2(x1, x2)  =  if_p_1_in_1_a1(x2)
if_p_1_in_2_a3(x1, x2, x3)  =  if_p_1_in_2_a1(x3)
p_1_out_g1(x1)  =  p_1_out_g
p_1_out_a1(x1)  =  p_1_out_a
if_p_1_in_3_a2(x1, x2)  =  if_p_1_in_3_a1(x2)

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:

p_1_in_g1(X) -> if_p_1_in_1_g2(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_g2(X, q_1_out_a1(f_11(Y))) -> if_p_1_in_2_g3(X, Y, p_1_in_g1(Y))
p_1_in_g1(g_11(X)) -> if_p_1_in_3_g2(X, p_1_in_a1(X))
p_1_in_a1(X) -> if_p_1_in_1_a2(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_2_a3(X, Y, p_1_out_g1(Y)) -> p_1_out_a1(X)
p_1_in_a1(g_11(X)) -> if_p_1_in_3_a2(X, p_1_in_a1(X))
if_p_1_in_3_a2(X, p_1_out_a1(X)) -> p_1_out_a1(g_11(X))
if_p_1_in_3_g2(X, p_1_out_a1(X)) -> p_1_out_g1(g_11(X))
if_p_1_in_2_g3(X, Y, p_1_out_g1(Y)) -> p_1_out_g1(X)

The argument filtering Pi contains the following mapping:
p_1_in_g1(x1)  =  p_1_in_g1(x1)
f_11(x1)  =  f_11(x1)
g_11(x1)  =  g_1
if_p_1_in_1_g2(x1, x2)  =  if_p_1_in_1_g1(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_g3(x1, x2, x3)  =  if_p_1_in_2_g1(x3)
if_p_1_in_3_g2(x1, x2)  =  if_p_1_in_3_g1(x2)
p_1_in_a1(x1)  =  p_1_in_a
if_p_1_in_1_a2(x1, x2)  =  if_p_1_in_1_a1(x2)
if_p_1_in_2_a3(x1, x2, x3)  =  if_p_1_in_2_a1(x3)
p_1_out_g1(x1)  =  p_1_out_g
p_1_out_a1(x1)  =  p_1_out_a
if_p_1_in_3_a2(x1, x2)  =  if_p_1_in_3_a1(x2)


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

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)
P_1_IN_G1(g_11(X)) -> IF_P_1_IN_3_G2(X, p_1_in_a1(X))
P_1_IN_G1(g_11(X)) -> P_1_IN_A1(X)
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_A1(g_11(X)) -> IF_P_1_IN_3_A2(X, p_1_in_a1(X))
P_1_IN_A1(g_11(X)) -> P_1_IN_A1(X)

The TRS R consists of the following rules:

p_1_in_g1(X) -> if_p_1_in_1_g2(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_g2(X, q_1_out_a1(f_11(Y))) -> if_p_1_in_2_g3(X, Y, p_1_in_g1(Y))
p_1_in_g1(g_11(X)) -> if_p_1_in_3_g2(X, p_1_in_a1(X))
p_1_in_a1(X) -> if_p_1_in_1_a2(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_2_a3(X, Y, p_1_out_g1(Y)) -> p_1_out_a1(X)
p_1_in_a1(g_11(X)) -> if_p_1_in_3_a2(X, p_1_in_a1(X))
if_p_1_in_3_a2(X, p_1_out_a1(X)) -> p_1_out_a1(g_11(X))
if_p_1_in_3_g2(X, p_1_out_a1(X)) -> p_1_out_g1(g_11(X))
if_p_1_in_2_g3(X, Y, p_1_out_g1(Y)) -> p_1_out_g1(X)

The argument filtering Pi contains the following mapping:
p_1_in_g1(x1)  =  p_1_in_g1(x1)
f_11(x1)  =  f_11(x1)
g_11(x1)  =  g_1
if_p_1_in_1_g2(x1, x2)  =  if_p_1_in_1_g1(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_g3(x1, x2, x3)  =  if_p_1_in_2_g1(x3)
if_p_1_in_3_g2(x1, x2)  =  if_p_1_in_3_g1(x2)
p_1_in_a1(x1)  =  p_1_in_a
if_p_1_in_1_a2(x1, x2)  =  if_p_1_in_1_a1(x2)
if_p_1_in_2_a3(x1, x2, x3)  =  if_p_1_in_2_a1(x3)
p_1_out_g1(x1)  =  p_1_out_g
p_1_out_a1(x1)  =  p_1_out_a
if_p_1_in_3_a2(x1, x2)  =  if_p_1_in_3_a1(x2)
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_3_A2(x1, x2)  =  IF_P_1_IN_3_A1(x2)
IF_P_1_IN_1_A2(x1, x2)  =  IF_P_1_IN_1_A1(x2)
IF_P_1_IN_1_G2(x1, x2)  =  IF_P_1_IN_1_G1(x2)
IF_P_1_IN_3_G2(x1, x2)  =  IF_P_1_IN_3_G1(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
  ↳ PrologToPiTRSProof

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

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)
P_1_IN_G1(g_11(X)) -> IF_P_1_IN_3_G2(X, p_1_in_a1(X))
P_1_IN_G1(g_11(X)) -> P_1_IN_A1(X)
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_A1(g_11(X)) -> IF_P_1_IN_3_A2(X, p_1_in_a1(X))
P_1_IN_A1(g_11(X)) -> P_1_IN_A1(X)

The TRS R consists of the following rules:

p_1_in_g1(X) -> if_p_1_in_1_g2(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_g2(X, q_1_out_a1(f_11(Y))) -> if_p_1_in_2_g3(X, Y, p_1_in_g1(Y))
p_1_in_g1(g_11(X)) -> if_p_1_in_3_g2(X, p_1_in_a1(X))
p_1_in_a1(X) -> if_p_1_in_1_a2(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_2_a3(X, Y, p_1_out_g1(Y)) -> p_1_out_a1(X)
p_1_in_a1(g_11(X)) -> if_p_1_in_3_a2(X, p_1_in_a1(X))
if_p_1_in_3_a2(X, p_1_out_a1(X)) -> p_1_out_a1(g_11(X))
if_p_1_in_3_g2(X, p_1_out_a1(X)) -> p_1_out_g1(g_11(X))
if_p_1_in_2_g3(X, Y, p_1_out_g1(Y)) -> p_1_out_g1(X)

The argument filtering Pi contains the following mapping:
p_1_in_g1(x1)  =  p_1_in_g1(x1)
f_11(x1)  =  f_11(x1)
g_11(x1)  =  g_1
if_p_1_in_1_g2(x1, x2)  =  if_p_1_in_1_g1(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_g3(x1, x2, x3)  =  if_p_1_in_2_g1(x3)
if_p_1_in_3_g2(x1, x2)  =  if_p_1_in_3_g1(x2)
p_1_in_a1(x1)  =  p_1_in_a
if_p_1_in_1_a2(x1, x2)  =  if_p_1_in_1_a1(x2)
if_p_1_in_2_a3(x1, x2, x3)  =  if_p_1_in_2_a1(x3)
p_1_out_g1(x1)  =  p_1_out_g
p_1_out_a1(x1)  =  p_1_out_a
if_p_1_in_3_a2(x1, x2)  =  if_p_1_in_3_a1(x2)
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_3_A2(x1, x2)  =  IF_P_1_IN_3_A1(x2)
IF_P_1_IN_1_A2(x1, x2)  =  IF_P_1_IN_1_A1(x2)
IF_P_1_IN_1_G2(x1, x2)  =  IF_P_1_IN_1_G1(x2)
IF_P_1_IN_3_G2(x1, x2)  =  IF_P_1_IN_3_G1(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 1 SCC with 11 less nodes.

↳ PROLOG
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ PiDP
              ↳ UsableRulesProof
  ↳ PrologToPiTRSProof

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

P_1_IN_A1(g_11(X)) -> P_1_IN_A1(X)

The TRS R consists of the following rules:

p_1_in_g1(X) -> if_p_1_in_1_g2(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_g2(X, q_1_out_a1(f_11(Y))) -> if_p_1_in_2_g3(X, Y, p_1_in_g1(Y))
p_1_in_g1(g_11(X)) -> if_p_1_in_3_g2(X, p_1_in_a1(X))
p_1_in_a1(X) -> if_p_1_in_1_a2(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_2_a3(X, Y, p_1_out_g1(Y)) -> p_1_out_a1(X)
p_1_in_a1(g_11(X)) -> if_p_1_in_3_a2(X, p_1_in_a1(X))
if_p_1_in_3_a2(X, p_1_out_a1(X)) -> p_1_out_a1(g_11(X))
if_p_1_in_3_g2(X, p_1_out_a1(X)) -> p_1_out_g1(g_11(X))
if_p_1_in_2_g3(X, Y, p_1_out_g1(Y)) -> p_1_out_g1(X)

The argument filtering Pi contains the following mapping:
p_1_in_g1(x1)  =  p_1_in_g1(x1)
f_11(x1)  =  f_11(x1)
g_11(x1)  =  g_1
if_p_1_in_1_g2(x1, x2)  =  if_p_1_in_1_g1(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_g3(x1, x2, x3)  =  if_p_1_in_2_g1(x3)
if_p_1_in_3_g2(x1, x2)  =  if_p_1_in_3_g1(x2)
p_1_in_a1(x1)  =  p_1_in_a
if_p_1_in_1_a2(x1, x2)  =  if_p_1_in_1_a1(x2)
if_p_1_in_2_a3(x1, x2, x3)  =  if_p_1_in_2_a1(x3)
p_1_out_g1(x1)  =  p_1_out_g
p_1_out_a1(x1)  =  p_1_out_a
if_p_1_in_3_a2(x1, x2)  =  if_p_1_in_3_a1(x2)
P_1_IN_A1(x1)  =  P_1_IN_A

We have to consider all (P,R,Pi)-chains
For (infinitary) constructor rewriting 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_1_IN_A1(g_11(X)) -> P_1_IN_A1(X)

R is empty.
The argument filtering Pi contains the following mapping:
g_11(x1)  =  g_1
P_1_IN_A1(x1)  =  P_1_IN_A

We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem into ordinary QDP problem by application of Pi.

↳ PROLOG
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ PiDP
              ↳ UsableRulesProof
                ↳ PiDP
                  ↳ PiDPToQDPProof
                    ↳ QDP
  ↳ PrologToPiTRSProof

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

P_1_IN_A -> P_1_IN_A

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {P_1_IN_A}.
With regard to the inferred argument filtering the predicates were used in the following modes:
p1: (b) (f)
Transforming PROLOG into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

p_1_in_g1(X) -> if_p_1_in_1_g2(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_g2(X, q_1_out_a1(f_11(Y))) -> if_p_1_in_2_g3(X, Y, p_1_in_g1(Y))
p_1_in_g1(g_11(X)) -> if_p_1_in_3_g2(X, p_1_in_a1(X))
p_1_in_a1(X) -> if_p_1_in_1_a2(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_2_a3(X, Y, p_1_out_g1(Y)) -> p_1_out_a1(X)
p_1_in_a1(g_11(X)) -> if_p_1_in_3_a2(X, p_1_in_a1(X))
if_p_1_in_3_a2(X, p_1_out_a1(X)) -> p_1_out_a1(g_11(X))
if_p_1_in_3_g2(X, p_1_out_a1(X)) -> p_1_out_g1(g_11(X))
if_p_1_in_2_g3(X, Y, p_1_out_g1(Y)) -> p_1_out_g1(X)

The argument filtering Pi contains the following mapping:
p_1_in_g1(x1)  =  p_1_in_g1(x1)
f_11(x1)  =  f_11(x1)
g_11(x1)  =  g_1
if_p_1_in_1_g2(x1, x2)  =  if_p_1_in_1_g2(x1, 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_g3(x1, x2, x3)  =  if_p_1_in_2_g2(x1, x3)
if_p_1_in_3_g2(x1, x2)  =  if_p_1_in_3_g1(x2)
p_1_in_a1(x1)  =  p_1_in_a
if_p_1_in_1_a2(x1, x2)  =  if_p_1_in_1_a1(x2)
if_p_1_in_2_a3(x1, x2, x3)  =  if_p_1_in_2_a1(x3)
p_1_out_g1(x1)  =  p_1_out_g1(x1)
p_1_out_a1(x1)  =  p_1_out_a
if_p_1_in_3_a2(x1, x2)  =  if_p_1_in_3_a1(x2)

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:

p_1_in_g1(X) -> if_p_1_in_1_g2(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_g2(X, q_1_out_a1(f_11(Y))) -> if_p_1_in_2_g3(X, Y, p_1_in_g1(Y))
p_1_in_g1(g_11(X)) -> if_p_1_in_3_g2(X, p_1_in_a1(X))
p_1_in_a1(X) -> if_p_1_in_1_a2(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_2_a3(X, Y, p_1_out_g1(Y)) -> p_1_out_a1(X)
p_1_in_a1(g_11(X)) -> if_p_1_in_3_a2(X, p_1_in_a1(X))
if_p_1_in_3_a2(X, p_1_out_a1(X)) -> p_1_out_a1(g_11(X))
if_p_1_in_3_g2(X, p_1_out_a1(X)) -> p_1_out_g1(g_11(X))
if_p_1_in_2_g3(X, Y, p_1_out_g1(Y)) -> p_1_out_g1(X)

The argument filtering Pi contains the following mapping:
p_1_in_g1(x1)  =  p_1_in_g1(x1)
f_11(x1)  =  f_11(x1)
g_11(x1)  =  g_1
if_p_1_in_1_g2(x1, x2)  =  if_p_1_in_1_g2(x1, 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_g3(x1, x2, x3)  =  if_p_1_in_2_g2(x1, x3)
if_p_1_in_3_g2(x1, x2)  =  if_p_1_in_3_g1(x2)
p_1_in_a1(x1)  =  p_1_in_a
if_p_1_in_1_a2(x1, x2)  =  if_p_1_in_1_a1(x2)
if_p_1_in_2_a3(x1, x2, x3)  =  if_p_1_in_2_a1(x3)
p_1_out_g1(x1)  =  p_1_out_g1(x1)
p_1_out_a1(x1)  =  p_1_out_a
if_p_1_in_3_a2(x1, x2)  =  if_p_1_in_3_a1(x2)


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

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)
P_1_IN_G1(g_11(X)) -> IF_P_1_IN_3_G2(X, p_1_in_a1(X))
P_1_IN_G1(g_11(X)) -> P_1_IN_A1(X)
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_A1(g_11(X)) -> IF_P_1_IN_3_A2(X, p_1_in_a1(X))
P_1_IN_A1(g_11(X)) -> P_1_IN_A1(X)

The TRS R consists of the following rules:

p_1_in_g1(X) -> if_p_1_in_1_g2(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_g2(X, q_1_out_a1(f_11(Y))) -> if_p_1_in_2_g3(X, Y, p_1_in_g1(Y))
p_1_in_g1(g_11(X)) -> if_p_1_in_3_g2(X, p_1_in_a1(X))
p_1_in_a1(X) -> if_p_1_in_1_a2(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_2_a3(X, Y, p_1_out_g1(Y)) -> p_1_out_a1(X)
p_1_in_a1(g_11(X)) -> if_p_1_in_3_a2(X, p_1_in_a1(X))
if_p_1_in_3_a2(X, p_1_out_a1(X)) -> p_1_out_a1(g_11(X))
if_p_1_in_3_g2(X, p_1_out_a1(X)) -> p_1_out_g1(g_11(X))
if_p_1_in_2_g3(X, Y, p_1_out_g1(Y)) -> p_1_out_g1(X)

The argument filtering Pi contains the following mapping:
p_1_in_g1(x1)  =  p_1_in_g1(x1)
f_11(x1)  =  f_11(x1)
g_11(x1)  =  g_1
if_p_1_in_1_g2(x1, x2)  =  if_p_1_in_1_g2(x1, 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_g3(x1, x2, x3)  =  if_p_1_in_2_g2(x1, x3)
if_p_1_in_3_g2(x1, x2)  =  if_p_1_in_3_g1(x2)
p_1_in_a1(x1)  =  p_1_in_a
if_p_1_in_1_a2(x1, x2)  =  if_p_1_in_1_a1(x2)
if_p_1_in_2_a3(x1, x2, x3)  =  if_p_1_in_2_a1(x3)
p_1_out_g1(x1)  =  p_1_out_g1(x1)
p_1_out_a1(x1)  =  p_1_out_a
if_p_1_in_3_a2(x1, x2)  =  if_p_1_in_3_a1(x2)
IF_P_1_IN_2_G3(x1, x2, x3)  =  IF_P_1_IN_2_G2(x1, 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_3_A2(x1, x2)  =  IF_P_1_IN_3_A1(x2)
IF_P_1_IN_1_A2(x1, x2)  =  IF_P_1_IN_1_A1(x2)
IF_P_1_IN_1_G2(x1, x2)  =  IF_P_1_IN_1_G2(x1, x2)
IF_P_1_IN_3_G2(x1, x2)  =  IF_P_1_IN_3_G1(x2)
P_1_IN_G1(x1)  =  P_1_IN_G1(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:

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)
P_1_IN_G1(g_11(X)) -> IF_P_1_IN_3_G2(X, p_1_in_a1(X))
P_1_IN_G1(g_11(X)) -> P_1_IN_A1(X)
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_A1(g_11(X)) -> IF_P_1_IN_3_A2(X, p_1_in_a1(X))
P_1_IN_A1(g_11(X)) -> P_1_IN_A1(X)

The TRS R consists of the following rules:

p_1_in_g1(X) -> if_p_1_in_1_g2(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_g2(X, q_1_out_a1(f_11(Y))) -> if_p_1_in_2_g3(X, Y, p_1_in_g1(Y))
p_1_in_g1(g_11(X)) -> if_p_1_in_3_g2(X, p_1_in_a1(X))
p_1_in_a1(X) -> if_p_1_in_1_a2(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_2_a3(X, Y, p_1_out_g1(Y)) -> p_1_out_a1(X)
p_1_in_a1(g_11(X)) -> if_p_1_in_3_a2(X, p_1_in_a1(X))
if_p_1_in_3_a2(X, p_1_out_a1(X)) -> p_1_out_a1(g_11(X))
if_p_1_in_3_g2(X, p_1_out_a1(X)) -> p_1_out_g1(g_11(X))
if_p_1_in_2_g3(X, Y, p_1_out_g1(Y)) -> p_1_out_g1(X)

The argument filtering Pi contains the following mapping:
p_1_in_g1(x1)  =  p_1_in_g1(x1)
f_11(x1)  =  f_11(x1)
g_11(x1)  =  g_1
if_p_1_in_1_g2(x1, x2)  =  if_p_1_in_1_g2(x1, 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_g3(x1, x2, x3)  =  if_p_1_in_2_g2(x1, x3)
if_p_1_in_3_g2(x1, x2)  =  if_p_1_in_3_g1(x2)
p_1_in_a1(x1)  =  p_1_in_a
if_p_1_in_1_a2(x1, x2)  =  if_p_1_in_1_a1(x2)
if_p_1_in_2_a3(x1, x2, x3)  =  if_p_1_in_2_a1(x3)
p_1_out_g1(x1)  =  p_1_out_g1(x1)
p_1_out_a1(x1)  =  p_1_out_a
if_p_1_in_3_a2(x1, x2)  =  if_p_1_in_3_a1(x2)
IF_P_1_IN_2_G3(x1, x2, x3)  =  IF_P_1_IN_2_G2(x1, 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_3_A2(x1, x2)  =  IF_P_1_IN_3_A1(x2)
IF_P_1_IN_1_A2(x1, x2)  =  IF_P_1_IN_1_A1(x2)
IF_P_1_IN_1_G2(x1, x2)  =  IF_P_1_IN_1_G2(x1, x2)
IF_P_1_IN_3_G2(x1, x2)  =  IF_P_1_IN_3_G1(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 1 SCC with 11 less nodes.

↳ PROLOG
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ PiDP
              ↳ UsableRulesProof

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

P_1_IN_A1(g_11(X)) -> P_1_IN_A1(X)

The TRS R consists of the following rules:

p_1_in_g1(X) -> if_p_1_in_1_g2(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_g2(X, q_1_out_a1(f_11(Y))) -> if_p_1_in_2_g3(X, Y, p_1_in_g1(Y))
p_1_in_g1(g_11(X)) -> if_p_1_in_3_g2(X, p_1_in_a1(X))
p_1_in_a1(X) -> if_p_1_in_1_a2(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_2_a3(X, Y, p_1_out_g1(Y)) -> p_1_out_a1(X)
p_1_in_a1(g_11(X)) -> if_p_1_in_3_a2(X, p_1_in_a1(X))
if_p_1_in_3_a2(X, p_1_out_a1(X)) -> p_1_out_a1(g_11(X))
if_p_1_in_3_g2(X, p_1_out_a1(X)) -> p_1_out_g1(g_11(X))
if_p_1_in_2_g3(X, Y, p_1_out_g1(Y)) -> p_1_out_g1(X)

The argument filtering Pi contains the following mapping:
p_1_in_g1(x1)  =  p_1_in_g1(x1)
f_11(x1)  =  f_11(x1)
g_11(x1)  =  g_1
if_p_1_in_1_g2(x1, x2)  =  if_p_1_in_1_g2(x1, 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_g3(x1, x2, x3)  =  if_p_1_in_2_g2(x1, x3)
if_p_1_in_3_g2(x1, x2)  =  if_p_1_in_3_g1(x2)
p_1_in_a1(x1)  =  p_1_in_a
if_p_1_in_1_a2(x1, x2)  =  if_p_1_in_1_a1(x2)
if_p_1_in_2_a3(x1, x2, x3)  =  if_p_1_in_2_a1(x3)
p_1_out_g1(x1)  =  p_1_out_g1(x1)
p_1_out_a1(x1)  =  p_1_out_a
if_p_1_in_3_a2(x1, x2)  =  if_p_1_in_3_a1(x2)
P_1_IN_A1(x1)  =  P_1_IN_A

We have to consider all (P,R,Pi)-chains
For (infinitary) constructor rewriting 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_1_IN_A1(g_11(X)) -> P_1_IN_A1(X)

R is empty.
The argument filtering Pi contains the following mapping:
g_11(x1)  =  g_1
P_1_IN_A1(x1)  =  P_1_IN_A

We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem into ordinary QDP problem by application of Pi.

↳ PROLOG
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ PiDP
              ↳ UsableRulesProof
                ↳ PiDP
                  ↳ PiDPToQDPProof
                    ↳ QDP

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

P_1_IN_A -> P_1_IN_A

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {P_1_IN_A}.