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



PROLOG
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof

p1(X) :- l1(X), q1(X).
q1(.2(A, {}0)).
r1(10).
l1({}0).
l1(.2(H, T)) :- r1(H), l1(T).


With regard to the inferred argument filtering the predicates were used in the following modes:
p1: (f)
l1: (f)
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, l_1_in_a1(X))
l_1_in_a1([]_0) -> l_1_out_a1([]_0)
l_1_in_a1(._22(H, T)) -> if_l_1_in_1_a3(H, T, r_1_in_a1(H))
r_1_in_a1(1_0) -> r_1_out_a1(1_0)
if_l_1_in_1_a3(H, T, r_1_out_a1(H)) -> if_l_1_in_2_a3(H, T, l_1_in_a1(T))
if_l_1_in_2_a3(H, T, l_1_out_a1(T)) -> l_1_out_a1(._22(H, T))
if_p_1_in_1_a2(X, l_1_out_a1(X)) -> if_p_1_in_2_a2(X, q_1_in_g1(X))
q_1_in_g1(._22(A, []_0)) -> q_1_out_g1(._22(A, []_0))
if_p_1_in_2_a2(X, q_1_out_g1(X)) -> p_1_out_a1(X)

The argument filtering Pi contains the following mapping:
p_1_in_a1(x1)  =  p_1_in_a
._22(x1, x2)  =  ._22(x1, x2)
[]_0  =  []_0
1_0  =  1_0
if_p_1_in_1_a2(x1, x2)  =  if_p_1_in_1_a1(x2)
l_1_in_a1(x1)  =  l_1_in_a
l_1_out_a1(x1)  =  l_1_out_a1(x1)
if_l_1_in_1_a3(x1, x2, x3)  =  if_l_1_in_1_a1(x3)
r_1_in_a1(x1)  =  r_1_in_a
r_1_out_a1(x1)  =  r_1_out_a1(x1)
if_l_1_in_2_a3(x1, x2, x3)  =  if_l_1_in_2_a2(x1, x3)
if_p_1_in_2_a2(x1, x2)  =  if_p_1_in_2_a2(x1, x2)
q_1_in_g1(x1)  =  q_1_in_g1(x1)
q_1_out_g1(x1)  =  q_1_out_g
p_1_out_a1(x1)  =  p_1_out_a1(x1)

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_a1(X) -> if_p_1_in_1_a2(X, l_1_in_a1(X))
l_1_in_a1([]_0) -> l_1_out_a1([]_0)
l_1_in_a1(._22(H, T)) -> if_l_1_in_1_a3(H, T, r_1_in_a1(H))
r_1_in_a1(1_0) -> r_1_out_a1(1_0)
if_l_1_in_1_a3(H, T, r_1_out_a1(H)) -> if_l_1_in_2_a3(H, T, l_1_in_a1(T))
if_l_1_in_2_a3(H, T, l_1_out_a1(T)) -> l_1_out_a1(._22(H, T))
if_p_1_in_1_a2(X, l_1_out_a1(X)) -> if_p_1_in_2_a2(X, q_1_in_g1(X))
q_1_in_g1(._22(A, []_0)) -> q_1_out_g1(._22(A, []_0))
if_p_1_in_2_a2(X, q_1_out_g1(X)) -> p_1_out_a1(X)

The argument filtering Pi contains the following mapping:
p_1_in_a1(x1)  =  p_1_in_a
._22(x1, x2)  =  ._22(x1, x2)
[]_0  =  []_0
1_0  =  1_0
if_p_1_in_1_a2(x1, x2)  =  if_p_1_in_1_a1(x2)
l_1_in_a1(x1)  =  l_1_in_a
l_1_out_a1(x1)  =  l_1_out_a1(x1)
if_l_1_in_1_a3(x1, x2, x3)  =  if_l_1_in_1_a1(x3)
r_1_in_a1(x1)  =  r_1_in_a
r_1_out_a1(x1)  =  r_1_out_a1(x1)
if_l_1_in_2_a3(x1, x2, x3)  =  if_l_1_in_2_a2(x1, x3)
if_p_1_in_2_a2(x1, x2)  =  if_p_1_in_2_a2(x1, x2)
q_1_in_g1(x1)  =  q_1_in_g1(x1)
q_1_out_g1(x1)  =  q_1_out_g
p_1_out_a1(x1)  =  p_1_out_a1(x1)


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

P_1_IN_A1(X) -> IF_P_1_IN_1_A2(X, l_1_in_a1(X))
P_1_IN_A1(X) -> L_1_IN_A1(X)
L_1_IN_A1(._22(H, T)) -> IF_L_1_IN_1_A3(H, T, r_1_in_a1(H))
L_1_IN_A1(._22(H, T)) -> R_1_IN_A1(H)
IF_L_1_IN_1_A3(H, T, r_1_out_a1(H)) -> IF_L_1_IN_2_A3(H, T, l_1_in_a1(T))
IF_L_1_IN_1_A3(H, T, r_1_out_a1(H)) -> L_1_IN_A1(T)
IF_P_1_IN_1_A2(X, l_1_out_a1(X)) -> IF_P_1_IN_2_A2(X, q_1_in_g1(X))
IF_P_1_IN_1_A2(X, l_1_out_a1(X)) -> Q_1_IN_G1(X)

The TRS R consists of the following rules:

p_1_in_a1(X) -> if_p_1_in_1_a2(X, l_1_in_a1(X))
l_1_in_a1([]_0) -> l_1_out_a1([]_0)
l_1_in_a1(._22(H, T)) -> if_l_1_in_1_a3(H, T, r_1_in_a1(H))
r_1_in_a1(1_0) -> r_1_out_a1(1_0)
if_l_1_in_1_a3(H, T, r_1_out_a1(H)) -> if_l_1_in_2_a3(H, T, l_1_in_a1(T))
if_l_1_in_2_a3(H, T, l_1_out_a1(T)) -> l_1_out_a1(._22(H, T))
if_p_1_in_1_a2(X, l_1_out_a1(X)) -> if_p_1_in_2_a2(X, q_1_in_g1(X))
q_1_in_g1(._22(A, []_0)) -> q_1_out_g1(._22(A, []_0))
if_p_1_in_2_a2(X, q_1_out_g1(X)) -> p_1_out_a1(X)

The argument filtering Pi contains the following mapping:
p_1_in_a1(x1)  =  p_1_in_a
._22(x1, x2)  =  ._22(x1, x2)
[]_0  =  []_0
1_0  =  1_0
if_p_1_in_1_a2(x1, x2)  =  if_p_1_in_1_a1(x2)
l_1_in_a1(x1)  =  l_1_in_a
l_1_out_a1(x1)  =  l_1_out_a1(x1)
if_l_1_in_1_a3(x1, x2, x3)  =  if_l_1_in_1_a1(x3)
r_1_in_a1(x1)  =  r_1_in_a
r_1_out_a1(x1)  =  r_1_out_a1(x1)
if_l_1_in_2_a3(x1, x2, x3)  =  if_l_1_in_2_a2(x1, x3)
if_p_1_in_2_a2(x1, x2)  =  if_p_1_in_2_a2(x1, x2)
q_1_in_g1(x1)  =  q_1_in_g1(x1)
q_1_out_g1(x1)  =  q_1_out_g
p_1_out_a1(x1)  =  p_1_out_a1(x1)
L_1_IN_A1(x1)  =  L_1_IN_A
Q_1_IN_G1(x1)  =  Q_1_IN_G1(x1)
P_1_IN_A1(x1)  =  P_1_IN_A
IF_P_1_IN_2_A2(x1, x2)  =  IF_P_1_IN_2_A2(x1, x2)
IF_L_1_IN_2_A3(x1, x2, x3)  =  IF_L_1_IN_2_A2(x1, x3)
IF_P_1_IN_1_A2(x1, x2)  =  IF_P_1_IN_1_A1(x2)
IF_L_1_IN_1_A3(x1, x2, x3)  =  IF_L_1_IN_1_A1(x3)
R_1_IN_A1(x1)  =  R_1_IN_A

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_A1(X) -> IF_P_1_IN_1_A2(X, l_1_in_a1(X))
P_1_IN_A1(X) -> L_1_IN_A1(X)
L_1_IN_A1(._22(H, T)) -> IF_L_1_IN_1_A3(H, T, r_1_in_a1(H))
L_1_IN_A1(._22(H, T)) -> R_1_IN_A1(H)
IF_L_1_IN_1_A3(H, T, r_1_out_a1(H)) -> IF_L_1_IN_2_A3(H, T, l_1_in_a1(T))
IF_L_1_IN_1_A3(H, T, r_1_out_a1(H)) -> L_1_IN_A1(T)
IF_P_1_IN_1_A2(X, l_1_out_a1(X)) -> IF_P_1_IN_2_A2(X, q_1_in_g1(X))
IF_P_1_IN_1_A2(X, l_1_out_a1(X)) -> Q_1_IN_G1(X)

The TRS R consists of the following rules:

p_1_in_a1(X) -> if_p_1_in_1_a2(X, l_1_in_a1(X))
l_1_in_a1([]_0) -> l_1_out_a1([]_0)
l_1_in_a1(._22(H, T)) -> if_l_1_in_1_a3(H, T, r_1_in_a1(H))
r_1_in_a1(1_0) -> r_1_out_a1(1_0)
if_l_1_in_1_a3(H, T, r_1_out_a1(H)) -> if_l_1_in_2_a3(H, T, l_1_in_a1(T))
if_l_1_in_2_a3(H, T, l_1_out_a1(T)) -> l_1_out_a1(._22(H, T))
if_p_1_in_1_a2(X, l_1_out_a1(X)) -> if_p_1_in_2_a2(X, q_1_in_g1(X))
q_1_in_g1(._22(A, []_0)) -> q_1_out_g1(._22(A, []_0))
if_p_1_in_2_a2(X, q_1_out_g1(X)) -> p_1_out_a1(X)

The argument filtering Pi contains the following mapping:
p_1_in_a1(x1)  =  p_1_in_a
._22(x1, x2)  =  ._22(x1, x2)
[]_0  =  []_0
1_0  =  1_0
if_p_1_in_1_a2(x1, x2)  =  if_p_1_in_1_a1(x2)
l_1_in_a1(x1)  =  l_1_in_a
l_1_out_a1(x1)  =  l_1_out_a1(x1)
if_l_1_in_1_a3(x1, x2, x3)  =  if_l_1_in_1_a1(x3)
r_1_in_a1(x1)  =  r_1_in_a
r_1_out_a1(x1)  =  r_1_out_a1(x1)
if_l_1_in_2_a3(x1, x2, x3)  =  if_l_1_in_2_a2(x1, x3)
if_p_1_in_2_a2(x1, x2)  =  if_p_1_in_2_a2(x1, x2)
q_1_in_g1(x1)  =  q_1_in_g1(x1)
q_1_out_g1(x1)  =  q_1_out_g
p_1_out_a1(x1)  =  p_1_out_a1(x1)
L_1_IN_A1(x1)  =  L_1_IN_A
Q_1_IN_G1(x1)  =  Q_1_IN_G1(x1)
P_1_IN_A1(x1)  =  P_1_IN_A
IF_P_1_IN_2_A2(x1, x2)  =  IF_P_1_IN_2_A2(x1, x2)
IF_L_1_IN_2_A3(x1, x2, x3)  =  IF_L_1_IN_2_A2(x1, x3)
IF_P_1_IN_1_A2(x1, x2)  =  IF_P_1_IN_1_A1(x2)
IF_L_1_IN_1_A3(x1, x2, x3)  =  IF_L_1_IN_1_A1(x3)
R_1_IN_A1(x1)  =  R_1_IN_A

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph contains 1 SCC with 6 less nodes.

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

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

L_1_IN_A1(._22(H, T)) -> IF_L_1_IN_1_A3(H, T, r_1_in_a1(H))
IF_L_1_IN_1_A3(H, T, r_1_out_a1(H)) -> L_1_IN_A1(T)

The TRS R consists of the following rules:

p_1_in_a1(X) -> if_p_1_in_1_a2(X, l_1_in_a1(X))
l_1_in_a1([]_0) -> l_1_out_a1([]_0)
l_1_in_a1(._22(H, T)) -> if_l_1_in_1_a3(H, T, r_1_in_a1(H))
r_1_in_a1(1_0) -> r_1_out_a1(1_0)
if_l_1_in_1_a3(H, T, r_1_out_a1(H)) -> if_l_1_in_2_a3(H, T, l_1_in_a1(T))
if_l_1_in_2_a3(H, T, l_1_out_a1(T)) -> l_1_out_a1(._22(H, T))
if_p_1_in_1_a2(X, l_1_out_a1(X)) -> if_p_1_in_2_a2(X, q_1_in_g1(X))
q_1_in_g1(._22(A, []_0)) -> q_1_out_g1(._22(A, []_0))
if_p_1_in_2_a2(X, q_1_out_g1(X)) -> p_1_out_a1(X)

The argument filtering Pi contains the following mapping:
p_1_in_a1(x1)  =  p_1_in_a
._22(x1, x2)  =  ._22(x1, x2)
[]_0  =  []_0
1_0  =  1_0
if_p_1_in_1_a2(x1, x2)  =  if_p_1_in_1_a1(x2)
l_1_in_a1(x1)  =  l_1_in_a
l_1_out_a1(x1)  =  l_1_out_a1(x1)
if_l_1_in_1_a3(x1, x2, x3)  =  if_l_1_in_1_a1(x3)
r_1_in_a1(x1)  =  r_1_in_a
r_1_out_a1(x1)  =  r_1_out_a1(x1)
if_l_1_in_2_a3(x1, x2, x3)  =  if_l_1_in_2_a2(x1, x3)
if_p_1_in_2_a2(x1, x2)  =  if_p_1_in_2_a2(x1, x2)
q_1_in_g1(x1)  =  q_1_in_g1(x1)
q_1_out_g1(x1)  =  q_1_out_g
p_1_out_a1(x1)  =  p_1_out_a1(x1)
L_1_IN_A1(x1)  =  L_1_IN_A
IF_L_1_IN_1_A3(x1, x2, x3)  =  IF_L_1_IN_1_A1(x3)

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:

L_1_IN_A1(._22(H, T)) -> IF_L_1_IN_1_A3(H, T, r_1_in_a1(H))
IF_L_1_IN_1_A3(H, T, r_1_out_a1(H)) -> L_1_IN_A1(T)

The TRS R consists of the following rules:

r_1_in_a1(1_0) -> r_1_out_a1(1_0)

The argument filtering Pi contains the following mapping:
._22(x1, x2)  =  ._22(x1, x2)
1_0  =  1_0
r_1_in_a1(x1)  =  r_1_in_a
r_1_out_a1(x1)  =  r_1_out_a1(x1)
L_1_IN_A1(x1)  =  L_1_IN_A
IF_L_1_IN_1_A3(x1, x2, x3)  =  IF_L_1_IN_1_A1(x3)

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:

L_1_IN_A -> IF_L_1_IN_1_A1(r_1_in_a)
IF_L_1_IN_1_A1(r_1_out_a1(H)) -> L_1_IN_A

The TRS R consists of the following rules:

r_1_in_a -> r_1_out_a1(1_0)

The set Q consists of the following terms:

r_1_in_a

We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {IF_L_1_IN_1_A1, L_1_IN_A}.
With regard to the inferred argument filtering the predicates were used in the following modes:
p1: (f)
l1: (f)
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, l_1_in_a1(X))
l_1_in_a1([]_0) -> l_1_out_a1([]_0)
l_1_in_a1(._22(H, T)) -> if_l_1_in_1_a3(H, T, r_1_in_a1(H))
r_1_in_a1(1_0) -> r_1_out_a1(1_0)
if_l_1_in_1_a3(H, T, r_1_out_a1(H)) -> if_l_1_in_2_a3(H, T, l_1_in_a1(T))
if_l_1_in_2_a3(H, T, l_1_out_a1(T)) -> l_1_out_a1(._22(H, T))
if_p_1_in_1_a2(X, l_1_out_a1(X)) -> if_p_1_in_2_a2(X, q_1_in_g1(X))
q_1_in_g1(._22(A, []_0)) -> q_1_out_g1(._22(A, []_0))
if_p_1_in_2_a2(X, q_1_out_g1(X)) -> p_1_out_a1(X)

The argument filtering Pi contains the following mapping:
p_1_in_a1(x1)  =  p_1_in_a
._22(x1, x2)  =  ._22(x1, x2)
[]_0  =  []_0
1_0  =  1_0
if_p_1_in_1_a2(x1, x2)  =  if_p_1_in_1_a1(x2)
l_1_in_a1(x1)  =  l_1_in_a
l_1_out_a1(x1)  =  l_1_out_a1(x1)
if_l_1_in_1_a3(x1, x2, x3)  =  if_l_1_in_1_a1(x3)
r_1_in_a1(x1)  =  r_1_in_a
r_1_out_a1(x1)  =  r_1_out_a1(x1)
if_l_1_in_2_a3(x1, x2, x3)  =  if_l_1_in_2_a2(x1, x3)
if_p_1_in_2_a2(x1, x2)  =  if_p_1_in_2_a2(x1, x2)
q_1_in_g1(x1)  =  q_1_in_g1(x1)
q_1_out_g1(x1)  =  q_1_out_g1(x1)
p_1_out_a1(x1)  =  p_1_out_a1(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:

p_1_in_a1(X) -> if_p_1_in_1_a2(X, l_1_in_a1(X))
l_1_in_a1([]_0) -> l_1_out_a1([]_0)
l_1_in_a1(._22(H, T)) -> if_l_1_in_1_a3(H, T, r_1_in_a1(H))
r_1_in_a1(1_0) -> r_1_out_a1(1_0)
if_l_1_in_1_a3(H, T, r_1_out_a1(H)) -> if_l_1_in_2_a3(H, T, l_1_in_a1(T))
if_l_1_in_2_a3(H, T, l_1_out_a1(T)) -> l_1_out_a1(._22(H, T))
if_p_1_in_1_a2(X, l_1_out_a1(X)) -> if_p_1_in_2_a2(X, q_1_in_g1(X))
q_1_in_g1(._22(A, []_0)) -> q_1_out_g1(._22(A, []_0))
if_p_1_in_2_a2(X, q_1_out_g1(X)) -> p_1_out_a1(X)

The argument filtering Pi contains the following mapping:
p_1_in_a1(x1)  =  p_1_in_a
._22(x1, x2)  =  ._22(x1, x2)
[]_0  =  []_0
1_0  =  1_0
if_p_1_in_1_a2(x1, x2)  =  if_p_1_in_1_a1(x2)
l_1_in_a1(x1)  =  l_1_in_a
l_1_out_a1(x1)  =  l_1_out_a1(x1)
if_l_1_in_1_a3(x1, x2, x3)  =  if_l_1_in_1_a1(x3)
r_1_in_a1(x1)  =  r_1_in_a
r_1_out_a1(x1)  =  r_1_out_a1(x1)
if_l_1_in_2_a3(x1, x2, x3)  =  if_l_1_in_2_a2(x1, x3)
if_p_1_in_2_a2(x1, x2)  =  if_p_1_in_2_a2(x1, x2)
q_1_in_g1(x1)  =  q_1_in_g1(x1)
q_1_out_g1(x1)  =  q_1_out_g1(x1)
p_1_out_a1(x1)  =  p_1_out_a1(x1)


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

P_1_IN_A1(X) -> IF_P_1_IN_1_A2(X, l_1_in_a1(X))
P_1_IN_A1(X) -> L_1_IN_A1(X)
L_1_IN_A1(._22(H, T)) -> IF_L_1_IN_1_A3(H, T, r_1_in_a1(H))
L_1_IN_A1(._22(H, T)) -> R_1_IN_A1(H)
IF_L_1_IN_1_A3(H, T, r_1_out_a1(H)) -> IF_L_1_IN_2_A3(H, T, l_1_in_a1(T))
IF_L_1_IN_1_A3(H, T, r_1_out_a1(H)) -> L_1_IN_A1(T)
IF_P_1_IN_1_A2(X, l_1_out_a1(X)) -> IF_P_1_IN_2_A2(X, q_1_in_g1(X))
IF_P_1_IN_1_A2(X, l_1_out_a1(X)) -> Q_1_IN_G1(X)

The TRS R consists of the following rules:

p_1_in_a1(X) -> if_p_1_in_1_a2(X, l_1_in_a1(X))
l_1_in_a1([]_0) -> l_1_out_a1([]_0)
l_1_in_a1(._22(H, T)) -> if_l_1_in_1_a3(H, T, r_1_in_a1(H))
r_1_in_a1(1_0) -> r_1_out_a1(1_0)
if_l_1_in_1_a3(H, T, r_1_out_a1(H)) -> if_l_1_in_2_a3(H, T, l_1_in_a1(T))
if_l_1_in_2_a3(H, T, l_1_out_a1(T)) -> l_1_out_a1(._22(H, T))
if_p_1_in_1_a2(X, l_1_out_a1(X)) -> if_p_1_in_2_a2(X, q_1_in_g1(X))
q_1_in_g1(._22(A, []_0)) -> q_1_out_g1(._22(A, []_0))
if_p_1_in_2_a2(X, q_1_out_g1(X)) -> p_1_out_a1(X)

The argument filtering Pi contains the following mapping:
p_1_in_a1(x1)  =  p_1_in_a
._22(x1, x2)  =  ._22(x1, x2)
[]_0  =  []_0
1_0  =  1_0
if_p_1_in_1_a2(x1, x2)  =  if_p_1_in_1_a1(x2)
l_1_in_a1(x1)  =  l_1_in_a
l_1_out_a1(x1)  =  l_1_out_a1(x1)
if_l_1_in_1_a3(x1, x2, x3)  =  if_l_1_in_1_a1(x3)
r_1_in_a1(x1)  =  r_1_in_a
r_1_out_a1(x1)  =  r_1_out_a1(x1)
if_l_1_in_2_a3(x1, x2, x3)  =  if_l_1_in_2_a2(x1, x3)
if_p_1_in_2_a2(x1, x2)  =  if_p_1_in_2_a2(x1, x2)
q_1_in_g1(x1)  =  q_1_in_g1(x1)
q_1_out_g1(x1)  =  q_1_out_g1(x1)
p_1_out_a1(x1)  =  p_1_out_a1(x1)
L_1_IN_A1(x1)  =  L_1_IN_A
Q_1_IN_G1(x1)  =  Q_1_IN_G1(x1)
P_1_IN_A1(x1)  =  P_1_IN_A
IF_P_1_IN_2_A2(x1, x2)  =  IF_P_1_IN_2_A2(x1, x2)
IF_L_1_IN_2_A3(x1, x2, x3)  =  IF_L_1_IN_2_A2(x1, x3)
IF_P_1_IN_1_A2(x1, x2)  =  IF_P_1_IN_1_A1(x2)
IF_L_1_IN_1_A3(x1, x2, x3)  =  IF_L_1_IN_1_A1(x3)
R_1_IN_A1(x1)  =  R_1_IN_A

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_A1(X) -> IF_P_1_IN_1_A2(X, l_1_in_a1(X))
P_1_IN_A1(X) -> L_1_IN_A1(X)
L_1_IN_A1(._22(H, T)) -> IF_L_1_IN_1_A3(H, T, r_1_in_a1(H))
L_1_IN_A1(._22(H, T)) -> R_1_IN_A1(H)
IF_L_1_IN_1_A3(H, T, r_1_out_a1(H)) -> IF_L_1_IN_2_A3(H, T, l_1_in_a1(T))
IF_L_1_IN_1_A3(H, T, r_1_out_a1(H)) -> L_1_IN_A1(T)
IF_P_1_IN_1_A2(X, l_1_out_a1(X)) -> IF_P_1_IN_2_A2(X, q_1_in_g1(X))
IF_P_1_IN_1_A2(X, l_1_out_a1(X)) -> Q_1_IN_G1(X)

The TRS R consists of the following rules:

p_1_in_a1(X) -> if_p_1_in_1_a2(X, l_1_in_a1(X))
l_1_in_a1([]_0) -> l_1_out_a1([]_0)
l_1_in_a1(._22(H, T)) -> if_l_1_in_1_a3(H, T, r_1_in_a1(H))
r_1_in_a1(1_0) -> r_1_out_a1(1_0)
if_l_1_in_1_a3(H, T, r_1_out_a1(H)) -> if_l_1_in_2_a3(H, T, l_1_in_a1(T))
if_l_1_in_2_a3(H, T, l_1_out_a1(T)) -> l_1_out_a1(._22(H, T))
if_p_1_in_1_a2(X, l_1_out_a1(X)) -> if_p_1_in_2_a2(X, q_1_in_g1(X))
q_1_in_g1(._22(A, []_0)) -> q_1_out_g1(._22(A, []_0))
if_p_1_in_2_a2(X, q_1_out_g1(X)) -> p_1_out_a1(X)

The argument filtering Pi contains the following mapping:
p_1_in_a1(x1)  =  p_1_in_a
._22(x1, x2)  =  ._22(x1, x2)
[]_0  =  []_0
1_0  =  1_0
if_p_1_in_1_a2(x1, x2)  =  if_p_1_in_1_a1(x2)
l_1_in_a1(x1)  =  l_1_in_a
l_1_out_a1(x1)  =  l_1_out_a1(x1)
if_l_1_in_1_a3(x1, x2, x3)  =  if_l_1_in_1_a1(x3)
r_1_in_a1(x1)  =  r_1_in_a
r_1_out_a1(x1)  =  r_1_out_a1(x1)
if_l_1_in_2_a3(x1, x2, x3)  =  if_l_1_in_2_a2(x1, x3)
if_p_1_in_2_a2(x1, x2)  =  if_p_1_in_2_a2(x1, x2)
q_1_in_g1(x1)  =  q_1_in_g1(x1)
q_1_out_g1(x1)  =  q_1_out_g1(x1)
p_1_out_a1(x1)  =  p_1_out_a1(x1)
L_1_IN_A1(x1)  =  L_1_IN_A
Q_1_IN_G1(x1)  =  Q_1_IN_G1(x1)
P_1_IN_A1(x1)  =  P_1_IN_A
IF_P_1_IN_2_A2(x1, x2)  =  IF_P_1_IN_2_A2(x1, x2)
IF_L_1_IN_2_A3(x1, x2, x3)  =  IF_L_1_IN_2_A2(x1, x3)
IF_P_1_IN_1_A2(x1, x2)  =  IF_P_1_IN_1_A1(x2)
IF_L_1_IN_1_A3(x1, x2, x3)  =  IF_L_1_IN_1_A1(x3)
R_1_IN_A1(x1)  =  R_1_IN_A

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph contains 1 SCC with 6 less nodes.

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

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

L_1_IN_A1(._22(H, T)) -> IF_L_1_IN_1_A3(H, T, r_1_in_a1(H))
IF_L_1_IN_1_A3(H, T, r_1_out_a1(H)) -> L_1_IN_A1(T)

The TRS R consists of the following rules:

p_1_in_a1(X) -> if_p_1_in_1_a2(X, l_1_in_a1(X))
l_1_in_a1([]_0) -> l_1_out_a1([]_0)
l_1_in_a1(._22(H, T)) -> if_l_1_in_1_a3(H, T, r_1_in_a1(H))
r_1_in_a1(1_0) -> r_1_out_a1(1_0)
if_l_1_in_1_a3(H, T, r_1_out_a1(H)) -> if_l_1_in_2_a3(H, T, l_1_in_a1(T))
if_l_1_in_2_a3(H, T, l_1_out_a1(T)) -> l_1_out_a1(._22(H, T))
if_p_1_in_1_a2(X, l_1_out_a1(X)) -> if_p_1_in_2_a2(X, q_1_in_g1(X))
q_1_in_g1(._22(A, []_0)) -> q_1_out_g1(._22(A, []_0))
if_p_1_in_2_a2(X, q_1_out_g1(X)) -> p_1_out_a1(X)

The argument filtering Pi contains the following mapping:
p_1_in_a1(x1)  =  p_1_in_a
._22(x1, x2)  =  ._22(x1, x2)
[]_0  =  []_0
1_0  =  1_0
if_p_1_in_1_a2(x1, x2)  =  if_p_1_in_1_a1(x2)
l_1_in_a1(x1)  =  l_1_in_a
l_1_out_a1(x1)  =  l_1_out_a1(x1)
if_l_1_in_1_a3(x1, x2, x3)  =  if_l_1_in_1_a1(x3)
r_1_in_a1(x1)  =  r_1_in_a
r_1_out_a1(x1)  =  r_1_out_a1(x1)
if_l_1_in_2_a3(x1, x2, x3)  =  if_l_1_in_2_a2(x1, x3)
if_p_1_in_2_a2(x1, x2)  =  if_p_1_in_2_a2(x1, x2)
q_1_in_g1(x1)  =  q_1_in_g1(x1)
q_1_out_g1(x1)  =  q_1_out_g1(x1)
p_1_out_a1(x1)  =  p_1_out_a1(x1)
L_1_IN_A1(x1)  =  L_1_IN_A
IF_L_1_IN_1_A3(x1, x2, x3)  =  IF_L_1_IN_1_A1(x3)

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:

L_1_IN_A1(._22(H, T)) -> IF_L_1_IN_1_A3(H, T, r_1_in_a1(H))
IF_L_1_IN_1_A3(H, T, r_1_out_a1(H)) -> L_1_IN_A1(T)

The TRS R consists of the following rules:

r_1_in_a1(1_0) -> r_1_out_a1(1_0)

The argument filtering Pi contains the following mapping:
._22(x1, x2)  =  ._22(x1, x2)
1_0  =  1_0
r_1_in_a1(x1)  =  r_1_in_a
r_1_out_a1(x1)  =  r_1_out_a1(x1)
L_1_IN_A1(x1)  =  L_1_IN_A
IF_L_1_IN_1_A3(x1, x2, x3)  =  IF_L_1_IN_1_A1(x3)

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:

L_1_IN_A -> IF_L_1_IN_1_A1(r_1_in_a)
IF_L_1_IN_1_A1(r_1_out_a1(H)) -> L_1_IN_A

The TRS R consists of the following rules:

r_1_in_a -> r_1_out_a1(1_0)

The set Q consists of the following terms:

r_1_in_a

We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {IF_L_1_IN_1_A1, L_1_IN_A}.