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



↳ PROLOG
  ↳ PrologToPiTRSProof

p1(.2(X, {}0)).
p1(.2(s12 (X), .2(Y, Xs))) :- p1(.2(X, .2(Y, Xs))), mult3(X, Y, Z), p1(.2(Z, Xs)).
p1(.2(00, Xs)) :- p1(Xs).
sum3(X, 00, X).
sum3(X, s1(Y), s1(Z)) :- sum3(X, Y, Z).
mult3(underscore, 00, 00).
mult3(X, s1(Y), Z) :- mult3(X, Y, W), sum3(W, X, Z).


With regard to the inferred argument filtering the predicates were used in the following modes:
p1: (b)
mult3: (b,b,f)
sum3: (b,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(._22(X, []_0)) -> p_1_out_g1(._22(X, []_0))
p_1_in_g1(._22(s_11(s_11(X)), ._22(Y, Xs))) -> if_p_1_in_1_g4(X, Y, Xs, p_1_in_g1(._22(X, ._22(Y, Xs))))
p_1_in_g1(._22(0_0, Xs)) -> if_p_1_in_4_g2(Xs, p_1_in_g1(Xs))
if_p_1_in_4_g2(Xs, p_1_out_g1(Xs)) -> p_1_out_g1(._22(0_0, Xs))
if_p_1_in_1_g4(X, Y, Xs, p_1_out_g1(._22(X, ._22(Y, Xs)))) -> if_p_1_in_2_g4(X, Y, Xs, mult_3_in_gga3(X, Y, Z))
mult_3_in_gga3(underscore, 0_0, 0_0) -> mult_3_out_gga3(underscore, 0_0, 0_0)
mult_3_in_gga3(X, s_11(Y), Z) -> if_mult_3_in_1_gga4(X, Y, Z, mult_3_in_gga3(X, Y, W))
if_mult_3_in_1_gga4(X, Y, Z, mult_3_out_gga3(X, Y, W)) -> if_mult_3_in_2_gga5(X, Y, Z, W, sum_3_in_gga3(W, X, Z))
sum_3_in_gga3(X, 0_0, X) -> sum_3_out_gga3(X, 0_0, X)
sum_3_in_gga3(X, s_11(Y), s_11(Z)) -> if_sum_3_in_1_gga4(X, Y, Z, sum_3_in_gga3(X, Y, Z))
if_sum_3_in_1_gga4(X, Y, Z, sum_3_out_gga3(X, Y, Z)) -> sum_3_out_gga3(X, s_11(Y), s_11(Z))
if_mult_3_in_2_gga5(X, Y, Z, W, sum_3_out_gga3(W, X, Z)) -> mult_3_out_gga3(X, s_11(Y), Z)
if_p_1_in_2_g4(X, Y, Xs, mult_3_out_gga3(X, Y, Z)) -> if_p_1_in_3_g5(X, Y, Xs, Z, p_1_in_g1(._22(Z, Xs)))
if_p_1_in_3_g5(X, Y, Xs, Z, p_1_out_g1(._22(Z, Xs))) -> p_1_out_g1(._22(s_11(s_11(X)), ._22(Y, Xs)))

The argument filtering Pi contains the following mapping:
p_1_in_g1(x1)  =  p_1_in_g1(x1)
._22(x1, x2)  =  ._22(x1, x2)
[]_0  =  []_0
s_11(x1)  =  s_11(x1)
0_0  =  0_0
p_1_out_g1(x1)  =  p_1_out_g
if_p_1_in_1_g4(x1, x2, x3, x4)  =  if_p_1_in_1_g4(x1, x2, x3, x4)
if_p_1_in_4_g2(x1, x2)  =  if_p_1_in_4_g1(x2)
if_p_1_in_2_g4(x1, x2, x3, x4)  =  if_p_1_in_2_g2(x3, x4)
mult_3_in_gga3(x1, x2, x3)  =  mult_3_in_gga2(x1, x2)
mult_3_out_gga3(x1, x2, x3)  =  mult_3_out_gga1(x3)
if_mult_3_in_1_gga4(x1, x2, x3, x4)  =  if_mult_3_in_1_gga2(x1, x4)
if_mult_3_in_2_gga5(x1, x2, x3, x4, x5)  =  if_mult_3_in_2_gga1(x5)
sum_3_in_gga3(x1, x2, x3)  =  sum_3_in_gga2(x1, x2)
sum_3_out_gga3(x1, x2, x3)  =  sum_3_out_gga1(x3)
if_sum_3_in_1_gga4(x1, x2, x3, x4)  =  if_sum_3_in_1_gga1(x4)
if_p_1_in_3_g5(x1, x2, x3, x4, x5)  =  if_p_1_in_3_g1(x5)

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_g1(._22(X, []_0)) -> p_1_out_g1(._22(X, []_0))
p_1_in_g1(._22(s_11(s_11(X)), ._22(Y, Xs))) -> if_p_1_in_1_g4(X, Y, Xs, p_1_in_g1(._22(X, ._22(Y, Xs))))
p_1_in_g1(._22(0_0, Xs)) -> if_p_1_in_4_g2(Xs, p_1_in_g1(Xs))
if_p_1_in_4_g2(Xs, p_1_out_g1(Xs)) -> p_1_out_g1(._22(0_0, Xs))
if_p_1_in_1_g4(X, Y, Xs, p_1_out_g1(._22(X, ._22(Y, Xs)))) -> if_p_1_in_2_g4(X, Y, Xs, mult_3_in_gga3(X, Y, Z))
mult_3_in_gga3(underscore, 0_0, 0_0) -> mult_3_out_gga3(underscore, 0_0, 0_0)
mult_3_in_gga3(X, s_11(Y), Z) -> if_mult_3_in_1_gga4(X, Y, Z, mult_3_in_gga3(X, Y, W))
if_mult_3_in_1_gga4(X, Y, Z, mult_3_out_gga3(X, Y, W)) -> if_mult_3_in_2_gga5(X, Y, Z, W, sum_3_in_gga3(W, X, Z))
sum_3_in_gga3(X, 0_0, X) -> sum_3_out_gga3(X, 0_0, X)
sum_3_in_gga3(X, s_11(Y), s_11(Z)) -> if_sum_3_in_1_gga4(X, Y, Z, sum_3_in_gga3(X, Y, Z))
if_sum_3_in_1_gga4(X, Y, Z, sum_3_out_gga3(X, Y, Z)) -> sum_3_out_gga3(X, s_11(Y), s_11(Z))
if_mult_3_in_2_gga5(X, Y, Z, W, sum_3_out_gga3(W, X, Z)) -> mult_3_out_gga3(X, s_11(Y), Z)
if_p_1_in_2_g4(X, Y, Xs, mult_3_out_gga3(X, Y, Z)) -> if_p_1_in_3_g5(X, Y, Xs, Z, p_1_in_g1(._22(Z, Xs)))
if_p_1_in_3_g5(X, Y, Xs, Z, p_1_out_g1(._22(Z, Xs))) -> p_1_out_g1(._22(s_11(s_11(X)), ._22(Y, Xs)))

The argument filtering Pi contains the following mapping:
p_1_in_g1(x1)  =  p_1_in_g1(x1)
._22(x1, x2)  =  ._22(x1, x2)
[]_0  =  []_0
s_11(x1)  =  s_11(x1)
0_0  =  0_0
p_1_out_g1(x1)  =  p_1_out_g
if_p_1_in_1_g4(x1, x2, x3, x4)  =  if_p_1_in_1_g4(x1, x2, x3, x4)
if_p_1_in_4_g2(x1, x2)  =  if_p_1_in_4_g1(x2)
if_p_1_in_2_g4(x1, x2, x3, x4)  =  if_p_1_in_2_g2(x3, x4)
mult_3_in_gga3(x1, x2, x3)  =  mult_3_in_gga2(x1, x2)
mult_3_out_gga3(x1, x2, x3)  =  mult_3_out_gga1(x3)
if_mult_3_in_1_gga4(x1, x2, x3, x4)  =  if_mult_3_in_1_gga2(x1, x4)
if_mult_3_in_2_gga5(x1, x2, x3, x4, x5)  =  if_mult_3_in_2_gga1(x5)
sum_3_in_gga3(x1, x2, x3)  =  sum_3_in_gga2(x1, x2)
sum_3_out_gga3(x1, x2, x3)  =  sum_3_out_gga1(x3)
if_sum_3_in_1_gga4(x1, x2, x3, x4)  =  if_sum_3_in_1_gga1(x4)
if_p_1_in_3_g5(x1, x2, x3, x4, x5)  =  if_p_1_in_3_g1(x5)


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

P_1_IN_G1(._22(s_11(s_11(X)), ._22(Y, Xs))) -> IF_P_1_IN_1_G4(X, Y, Xs, p_1_in_g1(._22(X, ._22(Y, Xs))))
P_1_IN_G1(._22(s_11(s_11(X)), ._22(Y, Xs))) -> P_1_IN_G1(._22(X, ._22(Y, Xs)))
P_1_IN_G1(._22(0_0, Xs)) -> IF_P_1_IN_4_G2(Xs, p_1_in_g1(Xs))
P_1_IN_G1(._22(0_0, Xs)) -> P_1_IN_G1(Xs)
IF_P_1_IN_1_G4(X, Y, Xs, p_1_out_g1(._22(X, ._22(Y, Xs)))) -> IF_P_1_IN_2_G4(X, Y, Xs, mult_3_in_gga3(X, Y, Z))
IF_P_1_IN_1_G4(X, Y, Xs, p_1_out_g1(._22(X, ._22(Y, Xs)))) -> MULT_3_IN_GGA3(X, Y, Z)
MULT_3_IN_GGA3(X, s_11(Y), Z) -> IF_MULT_3_IN_1_GGA4(X, Y, Z, mult_3_in_gga3(X, Y, W))
MULT_3_IN_GGA3(X, s_11(Y), Z) -> MULT_3_IN_GGA3(X, Y, W)
IF_MULT_3_IN_1_GGA4(X, Y, Z, mult_3_out_gga3(X, Y, W)) -> IF_MULT_3_IN_2_GGA5(X, Y, Z, W, sum_3_in_gga3(W, X, Z))
IF_MULT_3_IN_1_GGA4(X, Y, Z, mult_3_out_gga3(X, Y, W)) -> SUM_3_IN_GGA3(W, X, Z)
SUM_3_IN_GGA3(X, s_11(Y), s_11(Z)) -> IF_SUM_3_IN_1_GGA4(X, Y, Z, sum_3_in_gga3(X, Y, Z))
SUM_3_IN_GGA3(X, s_11(Y), s_11(Z)) -> SUM_3_IN_GGA3(X, Y, Z)
IF_P_1_IN_2_G4(X, Y, Xs, mult_3_out_gga3(X, Y, Z)) -> IF_P_1_IN_3_G5(X, Y, Xs, Z, p_1_in_g1(._22(Z, Xs)))
IF_P_1_IN_2_G4(X, Y, Xs, mult_3_out_gga3(X, Y, Z)) -> P_1_IN_G1(._22(Z, Xs))

The TRS R consists of the following rules:

p_1_in_g1(._22(X, []_0)) -> p_1_out_g1(._22(X, []_0))
p_1_in_g1(._22(s_11(s_11(X)), ._22(Y, Xs))) -> if_p_1_in_1_g4(X, Y, Xs, p_1_in_g1(._22(X, ._22(Y, Xs))))
p_1_in_g1(._22(0_0, Xs)) -> if_p_1_in_4_g2(Xs, p_1_in_g1(Xs))
if_p_1_in_4_g2(Xs, p_1_out_g1(Xs)) -> p_1_out_g1(._22(0_0, Xs))
if_p_1_in_1_g4(X, Y, Xs, p_1_out_g1(._22(X, ._22(Y, Xs)))) -> if_p_1_in_2_g4(X, Y, Xs, mult_3_in_gga3(X, Y, Z))
mult_3_in_gga3(underscore, 0_0, 0_0) -> mult_3_out_gga3(underscore, 0_0, 0_0)
mult_3_in_gga3(X, s_11(Y), Z) -> if_mult_3_in_1_gga4(X, Y, Z, mult_3_in_gga3(X, Y, W))
if_mult_3_in_1_gga4(X, Y, Z, mult_3_out_gga3(X, Y, W)) -> if_mult_3_in_2_gga5(X, Y, Z, W, sum_3_in_gga3(W, X, Z))
sum_3_in_gga3(X, 0_0, X) -> sum_3_out_gga3(X, 0_0, X)
sum_3_in_gga3(X, s_11(Y), s_11(Z)) -> if_sum_3_in_1_gga4(X, Y, Z, sum_3_in_gga3(X, Y, Z))
if_sum_3_in_1_gga4(X, Y, Z, sum_3_out_gga3(X, Y, Z)) -> sum_3_out_gga3(X, s_11(Y), s_11(Z))
if_mult_3_in_2_gga5(X, Y, Z, W, sum_3_out_gga3(W, X, Z)) -> mult_3_out_gga3(X, s_11(Y), Z)
if_p_1_in_2_g4(X, Y, Xs, mult_3_out_gga3(X, Y, Z)) -> if_p_1_in_3_g5(X, Y, Xs, Z, p_1_in_g1(._22(Z, Xs)))
if_p_1_in_3_g5(X, Y, Xs, Z, p_1_out_g1(._22(Z, Xs))) -> p_1_out_g1(._22(s_11(s_11(X)), ._22(Y, Xs)))

The argument filtering Pi contains the following mapping:
p_1_in_g1(x1)  =  p_1_in_g1(x1)
._22(x1, x2)  =  ._22(x1, x2)
[]_0  =  []_0
s_11(x1)  =  s_11(x1)
0_0  =  0_0
p_1_out_g1(x1)  =  p_1_out_g
if_p_1_in_1_g4(x1, x2, x3, x4)  =  if_p_1_in_1_g4(x1, x2, x3, x4)
if_p_1_in_4_g2(x1, x2)  =  if_p_1_in_4_g1(x2)
if_p_1_in_2_g4(x1, x2, x3, x4)  =  if_p_1_in_2_g2(x3, x4)
mult_3_in_gga3(x1, x2, x3)  =  mult_3_in_gga2(x1, x2)
mult_3_out_gga3(x1, x2, x3)  =  mult_3_out_gga1(x3)
if_mult_3_in_1_gga4(x1, x2, x3, x4)  =  if_mult_3_in_1_gga2(x1, x4)
if_mult_3_in_2_gga5(x1, x2, x3, x4, x5)  =  if_mult_3_in_2_gga1(x5)
sum_3_in_gga3(x1, x2, x3)  =  sum_3_in_gga2(x1, x2)
sum_3_out_gga3(x1, x2, x3)  =  sum_3_out_gga1(x3)
if_sum_3_in_1_gga4(x1, x2, x3, x4)  =  if_sum_3_in_1_gga1(x4)
if_p_1_in_3_g5(x1, x2, x3, x4, x5)  =  if_p_1_in_3_g1(x5)
IF_SUM_3_IN_1_GGA4(x1, x2, x3, x4)  =  IF_SUM_3_IN_1_GGA1(x4)
IF_MULT_3_IN_2_GGA5(x1, x2, x3, x4, x5)  =  IF_MULT_3_IN_2_GGA1(x5)
MULT_3_IN_GGA3(x1, x2, x3)  =  MULT_3_IN_GGA2(x1, x2)
IF_P_1_IN_2_G4(x1, x2, x3, x4)  =  IF_P_1_IN_2_G2(x3, x4)
SUM_3_IN_GGA3(x1, x2, x3)  =  SUM_3_IN_GGA2(x1, x2)
IF_MULT_3_IN_1_GGA4(x1, x2, x3, x4)  =  IF_MULT_3_IN_1_GGA2(x1, x4)
IF_P_1_IN_4_G2(x1, x2)  =  IF_P_1_IN_4_G1(x2)
IF_P_1_IN_3_G5(x1, x2, x3, x4, x5)  =  IF_P_1_IN_3_G1(x5)
IF_P_1_IN_1_G4(x1, x2, x3, x4)  =  IF_P_1_IN_1_G4(x1, x2, x3, x4)
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_G1(._22(s_11(s_11(X)), ._22(Y, Xs))) -> IF_P_1_IN_1_G4(X, Y, Xs, p_1_in_g1(._22(X, ._22(Y, Xs))))
P_1_IN_G1(._22(s_11(s_11(X)), ._22(Y, Xs))) -> P_1_IN_G1(._22(X, ._22(Y, Xs)))
P_1_IN_G1(._22(0_0, Xs)) -> IF_P_1_IN_4_G2(Xs, p_1_in_g1(Xs))
P_1_IN_G1(._22(0_0, Xs)) -> P_1_IN_G1(Xs)
IF_P_1_IN_1_G4(X, Y, Xs, p_1_out_g1(._22(X, ._22(Y, Xs)))) -> IF_P_1_IN_2_G4(X, Y, Xs, mult_3_in_gga3(X, Y, Z))
IF_P_1_IN_1_G4(X, Y, Xs, p_1_out_g1(._22(X, ._22(Y, Xs)))) -> MULT_3_IN_GGA3(X, Y, Z)
MULT_3_IN_GGA3(X, s_11(Y), Z) -> IF_MULT_3_IN_1_GGA4(X, Y, Z, mult_3_in_gga3(X, Y, W))
MULT_3_IN_GGA3(X, s_11(Y), Z) -> MULT_3_IN_GGA3(X, Y, W)
IF_MULT_3_IN_1_GGA4(X, Y, Z, mult_3_out_gga3(X, Y, W)) -> IF_MULT_3_IN_2_GGA5(X, Y, Z, W, sum_3_in_gga3(W, X, Z))
IF_MULT_3_IN_1_GGA4(X, Y, Z, mult_3_out_gga3(X, Y, W)) -> SUM_3_IN_GGA3(W, X, Z)
SUM_3_IN_GGA3(X, s_11(Y), s_11(Z)) -> IF_SUM_3_IN_1_GGA4(X, Y, Z, sum_3_in_gga3(X, Y, Z))
SUM_3_IN_GGA3(X, s_11(Y), s_11(Z)) -> SUM_3_IN_GGA3(X, Y, Z)
IF_P_1_IN_2_G4(X, Y, Xs, mult_3_out_gga3(X, Y, Z)) -> IF_P_1_IN_3_G5(X, Y, Xs, Z, p_1_in_g1(._22(Z, Xs)))
IF_P_1_IN_2_G4(X, Y, Xs, mult_3_out_gga3(X, Y, Z)) -> P_1_IN_G1(._22(Z, Xs))

The TRS R consists of the following rules:

p_1_in_g1(._22(X, []_0)) -> p_1_out_g1(._22(X, []_0))
p_1_in_g1(._22(s_11(s_11(X)), ._22(Y, Xs))) -> if_p_1_in_1_g4(X, Y, Xs, p_1_in_g1(._22(X, ._22(Y, Xs))))
p_1_in_g1(._22(0_0, Xs)) -> if_p_1_in_4_g2(Xs, p_1_in_g1(Xs))
if_p_1_in_4_g2(Xs, p_1_out_g1(Xs)) -> p_1_out_g1(._22(0_0, Xs))
if_p_1_in_1_g4(X, Y, Xs, p_1_out_g1(._22(X, ._22(Y, Xs)))) -> if_p_1_in_2_g4(X, Y, Xs, mult_3_in_gga3(X, Y, Z))
mult_3_in_gga3(underscore, 0_0, 0_0) -> mult_3_out_gga3(underscore, 0_0, 0_0)
mult_3_in_gga3(X, s_11(Y), Z) -> if_mult_3_in_1_gga4(X, Y, Z, mult_3_in_gga3(X, Y, W))
if_mult_3_in_1_gga4(X, Y, Z, mult_3_out_gga3(X, Y, W)) -> if_mult_3_in_2_gga5(X, Y, Z, W, sum_3_in_gga3(W, X, Z))
sum_3_in_gga3(X, 0_0, X) -> sum_3_out_gga3(X, 0_0, X)
sum_3_in_gga3(X, s_11(Y), s_11(Z)) -> if_sum_3_in_1_gga4(X, Y, Z, sum_3_in_gga3(X, Y, Z))
if_sum_3_in_1_gga4(X, Y, Z, sum_3_out_gga3(X, Y, Z)) -> sum_3_out_gga3(X, s_11(Y), s_11(Z))
if_mult_3_in_2_gga5(X, Y, Z, W, sum_3_out_gga3(W, X, Z)) -> mult_3_out_gga3(X, s_11(Y), Z)
if_p_1_in_2_g4(X, Y, Xs, mult_3_out_gga3(X, Y, Z)) -> if_p_1_in_3_g5(X, Y, Xs, Z, p_1_in_g1(._22(Z, Xs)))
if_p_1_in_3_g5(X, Y, Xs, Z, p_1_out_g1(._22(Z, Xs))) -> p_1_out_g1(._22(s_11(s_11(X)), ._22(Y, Xs)))

The argument filtering Pi contains the following mapping:
p_1_in_g1(x1)  =  p_1_in_g1(x1)
._22(x1, x2)  =  ._22(x1, x2)
[]_0  =  []_0
s_11(x1)  =  s_11(x1)
0_0  =  0_0
p_1_out_g1(x1)  =  p_1_out_g
if_p_1_in_1_g4(x1, x2, x3, x4)  =  if_p_1_in_1_g4(x1, x2, x3, x4)
if_p_1_in_4_g2(x1, x2)  =  if_p_1_in_4_g1(x2)
if_p_1_in_2_g4(x1, x2, x3, x4)  =  if_p_1_in_2_g2(x3, x4)
mult_3_in_gga3(x1, x2, x3)  =  mult_3_in_gga2(x1, x2)
mult_3_out_gga3(x1, x2, x3)  =  mult_3_out_gga1(x3)
if_mult_3_in_1_gga4(x1, x2, x3, x4)  =  if_mult_3_in_1_gga2(x1, x4)
if_mult_3_in_2_gga5(x1, x2, x3, x4, x5)  =  if_mult_3_in_2_gga1(x5)
sum_3_in_gga3(x1, x2, x3)  =  sum_3_in_gga2(x1, x2)
sum_3_out_gga3(x1, x2, x3)  =  sum_3_out_gga1(x3)
if_sum_3_in_1_gga4(x1, x2, x3, x4)  =  if_sum_3_in_1_gga1(x4)
if_p_1_in_3_g5(x1, x2, x3, x4, x5)  =  if_p_1_in_3_g1(x5)
IF_SUM_3_IN_1_GGA4(x1, x2, x3, x4)  =  IF_SUM_3_IN_1_GGA1(x4)
IF_MULT_3_IN_2_GGA5(x1, x2, x3, x4, x5)  =  IF_MULT_3_IN_2_GGA1(x5)
MULT_3_IN_GGA3(x1, x2, x3)  =  MULT_3_IN_GGA2(x1, x2)
IF_P_1_IN_2_G4(x1, x2, x3, x4)  =  IF_P_1_IN_2_G2(x3, x4)
SUM_3_IN_GGA3(x1, x2, x3)  =  SUM_3_IN_GGA2(x1, x2)
IF_MULT_3_IN_1_GGA4(x1, x2, x3, x4)  =  IF_MULT_3_IN_1_GGA2(x1, x4)
IF_P_1_IN_4_G2(x1, x2)  =  IF_P_1_IN_4_G1(x2)
IF_P_1_IN_3_G5(x1, x2, x3, x4, x5)  =  IF_P_1_IN_3_G1(x5)
IF_P_1_IN_1_G4(x1, x2, x3, x4)  =  IF_P_1_IN_1_G4(x1, x2, x3, x4)
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 3 SCCs with 7 less nodes.

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

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

SUM_3_IN_GGA3(X, s_11(Y), s_11(Z)) -> SUM_3_IN_GGA3(X, Y, Z)

The TRS R consists of the following rules:

p_1_in_g1(._22(X, []_0)) -> p_1_out_g1(._22(X, []_0))
p_1_in_g1(._22(s_11(s_11(X)), ._22(Y, Xs))) -> if_p_1_in_1_g4(X, Y, Xs, p_1_in_g1(._22(X, ._22(Y, Xs))))
p_1_in_g1(._22(0_0, Xs)) -> if_p_1_in_4_g2(Xs, p_1_in_g1(Xs))
if_p_1_in_4_g2(Xs, p_1_out_g1(Xs)) -> p_1_out_g1(._22(0_0, Xs))
if_p_1_in_1_g4(X, Y, Xs, p_1_out_g1(._22(X, ._22(Y, Xs)))) -> if_p_1_in_2_g4(X, Y, Xs, mult_3_in_gga3(X, Y, Z))
mult_3_in_gga3(underscore, 0_0, 0_0) -> mult_3_out_gga3(underscore, 0_0, 0_0)
mult_3_in_gga3(X, s_11(Y), Z) -> if_mult_3_in_1_gga4(X, Y, Z, mult_3_in_gga3(X, Y, W))
if_mult_3_in_1_gga4(X, Y, Z, mult_3_out_gga3(X, Y, W)) -> if_mult_3_in_2_gga5(X, Y, Z, W, sum_3_in_gga3(W, X, Z))
sum_3_in_gga3(X, 0_0, X) -> sum_3_out_gga3(X, 0_0, X)
sum_3_in_gga3(X, s_11(Y), s_11(Z)) -> if_sum_3_in_1_gga4(X, Y, Z, sum_3_in_gga3(X, Y, Z))
if_sum_3_in_1_gga4(X, Y, Z, sum_3_out_gga3(X, Y, Z)) -> sum_3_out_gga3(X, s_11(Y), s_11(Z))
if_mult_3_in_2_gga5(X, Y, Z, W, sum_3_out_gga3(W, X, Z)) -> mult_3_out_gga3(X, s_11(Y), Z)
if_p_1_in_2_g4(X, Y, Xs, mult_3_out_gga3(X, Y, Z)) -> if_p_1_in_3_g5(X, Y, Xs, Z, p_1_in_g1(._22(Z, Xs)))
if_p_1_in_3_g5(X, Y, Xs, Z, p_1_out_g1(._22(Z, Xs))) -> p_1_out_g1(._22(s_11(s_11(X)), ._22(Y, Xs)))

The argument filtering Pi contains the following mapping:
p_1_in_g1(x1)  =  p_1_in_g1(x1)
._22(x1, x2)  =  ._22(x1, x2)
[]_0  =  []_0
s_11(x1)  =  s_11(x1)
0_0  =  0_0
p_1_out_g1(x1)  =  p_1_out_g
if_p_1_in_1_g4(x1, x2, x3, x4)  =  if_p_1_in_1_g4(x1, x2, x3, x4)
if_p_1_in_4_g2(x1, x2)  =  if_p_1_in_4_g1(x2)
if_p_1_in_2_g4(x1, x2, x3, x4)  =  if_p_1_in_2_g2(x3, x4)
mult_3_in_gga3(x1, x2, x3)  =  mult_3_in_gga2(x1, x2)
mult_3_out_gga3(x1, x2, x3)  =  mult_3_out_gga1(x3)
if_mult_3_in_1_gga4(x1, x2, x3, x4)  =  if_mult_3_in_1_gga2(x1, x4)
if_mult_3_in_2_gga5(x1, x2, x3, x4, x5)  =  if_mult_3_in_2_gga1(x5)
sum_3_in_gga3(x1, x2, x3)  =  sum_3_in_gga2(x1, x2)
sum_3_out_gga3(x1, x2, x3)  =  sum_3_out_gga1(x3)
if_sum_3_in_1_gga4(x1, x2, x3, x4)  =  if_sum_3_in_1_gga1(x4)
if_p_1_in_3_g5(x1, x2, x3, x4, x5)  =  if_p_1_in_3_g1(x5)
SUM_3_IN_GGA3(x1, x2, x3)  =  SUM_3_IN_GGA2(x1, x2)

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
            ↳ AND
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
              ↳ PiDP
              ↳ PiDP

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

SUM_3_IN_GGA3(X, s_11(Y), s_11(Z)) -> SUM_3_IN_GGA3(X, Y, Z)

R is empty.
The argument filtering Pi contains the following mapping:
s_11(x1)  =  s_11(x1)
SUM_3_IN_GGA3(x1, x2, x3)  =  SUM_3_IN_GGA2(x1, x2)

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
            ↳ AND
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ QDPSizeChangeProof
              ↳ PiDP
              ↳ PiDP

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

SUM_3_IN_GGA2(X, s_11(Y)) -> SUM_3_IN_GGA2(X, Y)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {SUM_3_IN_GGA2}.
By using the subterm criterion together with the size-change analysis we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs: 



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

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

MULT_3_IN_GGA3(X, s_11(Y), Z) -> MULT_3_IN_GGA3(X, Y, W)

The TRS R consists of the following rules:

p_1_in_g1(._22(X, []_0)) -> p_1_out_g1(._22(X, []_0))
p_1_in_g1(._22(s_11(s_11(X)), ._22(Y, Xs))) -> if_p_1_in_1_g4(X, Y, Xs, p_1_in_g1(._22(X, ._22(Y, Xs))))
p_1_in_g1(._22(0_0, Xs)) -> if_p_1_in_4_g2(Xs, p_1_in_g1(Xs))
if_p_1_in_4_g2(Xs, p_1_out_g1(Xs)) -> p_1_out_g1(._22(0_0, Xs))
if_p_1_in_1_g4(X, Y, Xs, p_1_out_g1(._22(X, ._22(Y, Xs)))) -> if_p_1_in_2_g4(X, Y, Xs, mult_3_in_gga3(X, Y, Z))
mult_3_in_gga3(underscore, 0_0, 0_0) -> mult_3_out_gga3(underscore, 0_0, 0_0)
mult_3_in_gga3(X, s_11(Y), Z) -> if_mult_3_in_1_gga4(X, Y, Z, mult_3_in_gga3(X, Y, W))
if_mult_3_in_1_gga4(X, Y, Z, mult_3_out_gga3(X, Y, W)) -> if_mult_3_in_2_gga5(X, Y, Z, W, sum_3_in_gga3(W, X, Z))
sum_3_in_gga3(X, 0_0, X) -> sum_3_out_gga3(X, 0_0, X)
sum_3_in_gga3(X, s_11(Y), s_11(Z)) -> if_sum_3_in_1_gga4(X, Y, Z, sum_3_in_gga3(X, Y, Z))
if_sum_3_in_1_gga4(X, Y, Z, sum_3_out_gga3(X, Y, Z)) -> sum_3_out_gga3(X, s_11(Y), s_11(Z))
if_mult_3_in_2_gga5(X, Y, Z, W, sum_3_out_gga3(W, X, Z)) -> mult_3_out_gga3(X, s_11(Y), Z)
if_p_1_in_2_g4(X, Y, Xs, mult_3_out_gga3(X, Y, Z)) -> if_p_1_in_3_g5(X, Y, Xs, Z, p_1_in_g1(._22(Z, Xs)))
if_p_1_in_3_g5(X, Y, Xs, Z, p_1_out_g1(._22(Z, Xs))) -> p_1_out_g1(._22(s_11(s_11(X)), ._22(Y, Xs)))

The argument filtering Pi contains the following mapping:
p_1_in_g1(x1)  =  p_1_in_g1(x1)
._22(x1, x2)  =  ._22(x1, x2)
[]_0  =  []_0
s_11(x1)  =  s_11(x1)
0_0  =  0_0
p_1_out_g1(x1)  =  p_1_out_g
if_p_1_in_1_g4(x1, x2, x3, x4)  =  if_p_1_in_1_g4(x1, x2, x3, x4)
if_p_1_in_4_g2(x1, x2)  =  if_p_1_in_4_g1(x2)
if_p_1_in_2_g4(x1, x2, x3, x4)  =  if_p_1_in_2_g2(x3, x4)
mult_3_in_gga3(x1, x2, x3)  =  mult_3_in_gga2(x1, x2)
mult_3_out_gga3(x1, x2, x3)  =  mult_3_out_gga1(x3)
if_mult_3_in_1_gga4(x1, x2, x3, x4)  =  if_mult_3_in_1_gga2(x1, x4)
if_mult_3_in_2_gga5(x1, x2, x3, x4, x5)  =  if_mult_3_in_2_gga1(x5)
sum_3_in_gga3(x1, x2, x3)  =  sum_3_in_gga2(x1, x2)
sum_3_out_gga3(x1, x2, x3)  =  sum_3_out_gga1(x3)
if_sum_3_in_1_gga4(x1, x2, x3, x4)  =  if_sum_3_in_1_gga1(x4)
if_p_1_in_3_g5(x1, x2, x3, x4, x5)  =  if_p_1_in_3_g1(x5)
MULT_3_IN_GGA3(x1, x2, x3)  =  MULT_3_IN_GGA2(x1, x2)

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
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
              ↳ PiDP

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

MULT_3_IN_GGA3(X, s_11(Y), Z) -> MULT_3_IN_GGA3(X, Y, W)

R is empty.
The argument filtering Pi contains the following mapping:
s_11(x1)  =  s_11(x1)
MULT_3_IN_GGA3(x1, x2, x3)  =  MULT_3_IN_GGA2(x1, x2)

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
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ QDPSizeChangeProof
              ↳ PiDP

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

MULT_3_IN_GGA2(X, s_11(Y)) -> MULT_3_IN_GGA2(X, Y)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {MULT_3_IN_GGA2}.
By using the subterm criterion together with the size-change analysis we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs: 



↳ PROLOG
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
              ↳ PiDP
                ↳ PiDPToQDPProof

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

P_1_IN_G1(._22(s_11(s_11(X)), ._22(Y, Xs))) -> IF_P_1_IN_1_G4(X, Y, Xs, p_1_in_g1(._22(X, ._22(Y, Xs))))
IF_P_1_IN_2_G4(X, Y, Xs, mult_3_out_gga3(X, Y, Z)) -> P_1_IN_G1(._22(Z, Xs))
P_1_IN_G1(._22(0_0, Xs)) -> P_1_IN_G1(Xs)
P_1_IN_G1(._22(s_11(s_11(X)), ._22(Y, Xs))) -> P_1_IN_G1(._22(X, ._22(Y, Xs)))
IF_P_1_IN_1_G4(X, Y, Xs, p_1_out_g1(._22(X, ._22(Y, Xs)))) -> IF_P_1_IN_2_G4(X, Y, Xs, mult_3_in_gga3(X, Y, Z))

The TRS R consists of the following rules:

p_1_in_g1(._22(X, []_0)) -> p_1_out_g1(._22(X, []_0))
p_1_in_g1(._22(s_11(s_11(X)), ._22(Y, Xs))) -> if_p_1_in_1_g4(X, Y, Xs, p_1_in_g1(._22(X, ._22(Y, Xs))))
p_1_in_g1(._22(0_0, Xs)) -> if_p_1_in_4_g2(Xs, p_1_in_g1(Xs))
if_p_1_in_4_g2(Xs, p_1_out_g1(Xs)) -> p_1_out_g1(._22(0_0, Xs))
if_p_1_in_1_g4(X, Y, Xs, p_1_out_g1(._22(X, ._22(Y, Xs)))) -> if_p_1_in_2_g4(X, Y, Xs, mult_3_in_gga3(X, Y, Z))
mult_3_in_gga3(underscore, 0_0, 0_0) -> mult_3_out_gga3(underscore, 0_0, 0_0)
mult_3_in_gga3(X, s_11(Y), Z) -> if_mult_3_in_1_gga4(X, Y, Z, mult_3_in_gga3(X, Y, W))
if_mult_3_in_1_gga4(X, Y, Z, mult_3_out_gga3(X, Y, W)) -> if_mult_3_in_2_gga5(X, Y, Z, W, sum_3_in_gga3(W, X, Z))
sum_3_in_gga3(X, 0_0, X) -> sum_3_out_gga3(X, 0_0, X)
sum_3_in_gga3(X, s_11(Y), s_11(Z)) -> if_sum_3_in_1_gga4(X, Y, Z, sum_3_in_gga3(X, Y, Z))
if_sum_3_in_1_gga4(X, Y, Z, sum_3_out_gga3(X, Y, Z)) -> sum_3_out_gga3(X, s_11(Y), s_11(Z))
if_mult_3_in_2_gga5(X, Y, Z, W, sum_3_out_gga3(W, X, Z)) -> mult_3_out_gga3(X, s_11(Y), Z)
if_p_1_in_2_g4(X, Y, Xs, mult_3_out_gga3(X, Y, Z)) -> if_p_1_in_3_g5(X, Y, Xs, Z, p_1_in_g1(._22(Z, Xs)))
if_p_1_in_3_g5(X, Y, Xs, Z, p_1_out_g1(._22(Z, Xs))) -> p_1_out_g1(._22(s_11(s_11(X)), ._22(Y, Xs)))

The argument filtering Pi contains the following mapping:
p_1_in_g1(x1)  =  p_1_in_g1(x1)
._22(x1, x2)  =  ._22(x1, x2)
[]_0  =  []_0
s_11(x1)  =  s_11(x1)
0_0  =  0_0
p_1_out_g1(x1)  =  p_1_out_g
if_p_1_in_1_g4(x1, x2, x3, x4)  =  if_p_1_in_1_g4(x1, x2, x3, x4)
if_p_1_in_4_g2(x1, x2)  =  if_p_1_in_4_g1(x2)
if_p_1_in_2_g4(x1, x2, x3, x4)  =  if_p_1_in_2_g2(x3, x4)
mult_3_in_gga3(x1, x2, x3)  =  mult_3_in_gga2(x1, x2)
mult_3_out_gga3(x1, x2, x3)  =  mult_3_out_gga1(x3)
if_mult_3_in_1_gga4(x1, x2, x3, x4)  =  if_mult_3_in_1_gga2(x1, x4)
if_mult_3_in_2_gga5(x1, x2, x3, x4, x5)  =  if_mult_3_in_2_gga1(x5)
sum_3_in_gga3(x1, x2, x3)  =  sum_3_in_gga2(x1, x2)
sum_3_out_gga3(x1, x2, x3)  =  sum_3_out_gga1(x3)
if_sum_3_in_1_gga4(x1, x2, x3, x4)  =  if_sum_3_in_1_gga1(x4)
if_p_1_in_3_g5(x1, x2, x3, x4, x5)  =  if_p_1_in_3_g1(x5)
IF_P_1_IN_2_G4(x1, x2, x3, x4)  =  IF_P_1_IN_2_G2(x3, x4)
IF_P_1_IN_1_G4(x1, x2, x3, x4)  =  IF_P_1_IN_1_G4(x1, x2, x3, x4)
P_1_IN_G1(x1)  =  P_1_IN_G1(x1)

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
            ↳ AND
              ↳ PiDP
              ↳ PiDP
              ↳ PiDP
                ↳ PiDPToQDPProof
                  ↳ QDP
                    ↳ QDPPoloProof

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

P_1_IN_G1(._22(s_11(s_11(X)), ._22(Y, Xs))) -> IF_P_1_IN_1_G4(X, Y, Xs, p_1_in_g1(._22(X, ._22(Y, Xs))))
IF_P_1_IN_2_G2(Xs, mult_3_out_gga1(Z)) -> P_1_IN_G1(._22(Z, Xs))
P_1_IN_G1(._22(0_0, Xs)) -> P_1_IN_G1(Xs)
P_1_IN_G1(._22(s_11(s_11(X)), ._22(Y, Xs))) -> P_1_IN_G1(._22(X, ._22(Y, Xs)))
IF_P_1_IN_1_G4(X, Y, Xs, p_1_out_g) -> IF_P_1_IN_2_G2(Xs, mult_3_in_gga2(X, Y))

The TRS R consists of the following rules:

p_1_in_g1(._22(X, []_0)) -> p_1_out_g
p_1_in_g1(._22(s_11(s_11(X)), ._22(Y, Xs))) -> if_p_1_in_1_g4(X, Y, Xs, p_1_in_g1(._22(X, ._22(Y, Xs))))
p_1_in_g1(._22(0_0, Xs)) -> if_p_1_in_4_g1(p_1_in_g1(Xs))
if_p_1_in_4_g1(p_1_out_g) -> p_1_out_g
if_p_1_in_1_g4(X, Y, Xs, p_1_out_g) -> if_p_1_in_2_g2(Xs, mult_3_in_gga2(X, Y))
mult_3_in_gga2(underscore, 0_0) -> mult_3_out_gga1(0_0)
mult_3_in_gga2(X, s_11(Y)) -> if_mult_3_in_1_gga2(X, mult_3_in_gga2(X, Y))
if_mult_3_in_1_gga2(X, mult_3_out_gga1(W)) -> if_mult_3_in_2_gga1(sum_3_in_gga2(W, X))
sum_3_in_gga2(X, 0_0) -> sum_3_out_gga1(X)
sum_3_in_gga2(X, s_11(Y)) -> if_sum_3_in_1_gga1(sum_3_in_gga2(X, Y))
if_sum_3_in_1_gga1(sum_3_out_gga1(Z)) -> sum_3_out_gga1(s_11(Z))
if_mult_3_in_2_gga1(sum_3_out_gga1(Z)) -> mult_3_out_gga1(Z)
if_p_1_in_2_g2(Xs, mult_3_out_gga1(Z)) -> if_p_1_in_3_g1(p_1_in_g1(._22(Z, Xs)))
if_p_1_in_3_g1(p_1_out_g) -> p_1_out_g

The set Q consists of the following terms:

p_1_in_g1(x0)
if_p_1_in_4_g1(x0)
if_p_1_in_1_g4(x0, x1, x2, x3)
mult_3_in_gga2(x0, x1)
if_mult_3_in_1_gga2(x0, x1)
sum_3_in_gga2(x0, x1)
if_sum_3_in_1_gga1(x0)
if_mult_3_in_2_gga1(x0)
if_p_1_in_2_g2(x0, x1)
if_p_1_in_3_g1(x0)

We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {IF_P_1_IN_1_G4, P_1_IN_G1, IF_P_1_IN_2_G2}.
By using a polynomial ordering, the following set of Dependency Pairs of this DP problem can be strictly oriented.

P_1_IN_G1(._22(s_11(s_11(X)), ._22(Y, Xs))) -> IF_P_1_IN_1_G4(X, Y, Xs, p_1_in_g1(._22(X, ._22(Y, Xs))))
P_1_IN_G1(._22(0_0, Xs)) -> P_1_IN_G1(Xs)
The remaining Dependency Pairs were at least non-strictly be oriented.

IF_P_1_IN_2_G2(Xs, mult_3_out_gga1(Z)) -> P_1_IN_G1(._22(Z, Xs))
P_1_IN_G1(._22(s_11(s_11(X)), ._22(Y, Xs))) -> P_1_IN_G1(._22(X, ._22(Y, Xs)))
IF_P_1_IN_1_G4(X, Y, Xs, p_1_out_g) -> IF_P_1_IN_2_G2(Xs, mult_3_in_gga2(X, Y))
With the implicit AFS there is no usable rule.

Used ordering: POLO with Polynomial interpretation:


POL(if_mult_3_in_2_gga1(x1)) = 0   
POL(0_0) = 0   
POL(IF_P_1_IN_1_G4(x1, x2, x3, x4)) = 1 + x3   
POL(if_p_1_in_2_g2(x1, x2)) = 0   
POL(sum_3_out_gga1(x1)) = 0   
POL(mult_3_in_gga2(x1, x2)) = 0   
POL(if_p_1_in_3_g1(x1)) = 0   
POL([]_0) = 0   
POL(p_1_in_g1(x1)) = 0   
POL(P_1_IN_G1(x1)) = x1   
POL(p_1_out_g) = 0   
POL(if_mult_3_in_1_gga2(x1, x2)) = 0   
POL(._22(x1, x2)) = 1 + x2   
POL(if_p_1_in_4_g1(x1)) = 0   
POL(if_sum_3_in_1_gga1(x1)) = 0   
POL(IF_P_1_IN_2_G2(x1, x2)) = 1 + x1   
POL(sum_3_in_gga2(x1, x2)) = 0   
POL(s_11(x1)) = 0   
POL(if_p_1_in_1_g4(x1, x2, x3, x4)) = 0   
POL(mult_3_out_gga1(x1)) = 0   



↳ PROLOG
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
              ↳ PiDP
                ↳ PiDPToQDPProof
                  ↳ QDP
                    ↳ QDPPoloProof
                      ↳ QDP
                        ↳ DependencyGraphProof

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

IF_P_1_IN_2_G2(Xs, mult_3_out_gga1(Z)) -> P_1_IN_G1(._22(Z, Xs))
P_1_IN_G1(._22(s_11(s_11(X)), ._22(Y, Xs))) -> P_1_IN_G1(._22(X, ._22(Y, Xs)))
IF_P_1_IN_1_G4(X, Y, Xs, p_1_out_g) -> IF_P_1_IN_2_G2(Xs, mult_3_in_gga2(X, Y))

The TRS R consists of the following rules:

p_1_in_g1(._22(X, []_0)) -> p_1_out_g
p_1_in_g1(._22(s_11(s_11(X)), ._22(Y, Xs))) -> if_p_1_in_1_g4(X, Y, Xs, p_1_in_g1(._22(X, ._22(Y, Xs))))
p_1_in_g1(._22(0_0, Xs)) -> if_p_1_in_4_g1(p_1_in_g1(Xs))
if_p_1_in_4_g1(p_1_out_g) -> p_1_out_g
if_p_1_in_1_g4(X, Y, Xs, p_1_out_g) -> if_p_1_in_2_g2(Xs, mult_3_in_gga2(X, Y))
mult_3_in_gga2(underscore, 0_0) -> mult_3_out_gga1(0_0)
mult_3_in_gga2(X, s_11(Y)) -> if_mult_3_in_1_gga2(X, mult_3_in_gga2(X, Y))
if_mult_3_in_1_gga2(X, mult_3_out_gga1(W)) -> if_mult_3_in_2_gga1(sum_3_in_gga2(W, X))
sum_3_in_gga2(X, 0_0) -> sum_3_out_gga1(X)
sum_3_in_gga2(X, s_11(Y)) -> if_sum_3_in_1_gga1(sum_3_in_gga2(X, Y))
if_sum_3_in_1_gga1(sum_3_out_gga1(Z)) -> sum_3_out_gga1(s_11(Z))
if_mult_3_in_2_gga1(sum_3_out_gga1(Z)) -> mult_3_out_gga1(Z)
if_p_1_in_2_g2(Xs, mult_3_out_gga1(Z)) -> if_p_1_in_3_g1(p_1_in_g1(._22(Z, Xs)))
if_p_1_in_3_g1(p_1_out_g) -> p_1_out_g

The set Q consists of the following terms:

p_1_in_g1(x0)
if_p_1_in_4_g1(x0)
if_p_1_in_1_g4(x0, x1, x2, x3)
mult_3_in_gga2(x0, x1)
if_mult_3_in_1_gga2(x0, x1)
sum_3_in_gga2(x0, x1)
if_sum_3_in_1_gga1(x0)
if_mult_3_in_2_gga1(x0)
if_p_1_in_2_g2(x0, x1)
if_p_1_in_3_g1(x0)

We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {P_1_IN_G1, IF_P_1_IN_2_G2, IF_P_1_IN_1_G4}.
The approximation of the Dependency Graph contains 1 SCC with 2 less nodes.

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

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

P_1_IN_G1(._22(s_11(s_11(X)), ._22(Y, Xs))) -> P_1_IN_G1(._22(X, ._22(Y, Xs)))

The TRS R consists of the following rules:

p_1_in_g1(._22(X, []_0)) -> p_1_out_g
p_1_in_g1(._22(s_11(s_11(X)), ._22(Y, Xs))) -> if_p_1_in_1_g4(X, Y, Xs, p_1_in_g1(._22(X, ._22(Y, Xs))))
p_1_in_g1(._22(0_0, Xs)) -> if_p_1_in_4_g1(p_1_in_g1(Xs))
if_p_1_in_4_g1(p_1_out_g) -> p_1_out_g
if_p_1_in_1_g4(X, Y, Xs, p_1_out_g) -> if_p_1_in_2_g2(Xs, mult_3_in_gga2(X, Y))
mult_3_in_gga2(underscore, 0_0) -> mult_3_out_gga1(0_0)
mult_3_in_gga2(X, s_11(Y)) -> if_mult_3_in_1_gga2(X, mult_3_in_gga2(X, Y))
if_mult_3_in_1_gga2(X, mult_3_out_gga1(W)) -> if_mult_3_in_2_gga1(sum_3_in_gga2(W, X))
sum_3_in_gga2(X, 0_0) -> sum_3_out_gga1(X)
sum_3_in_gga2(X, s_11(Y)) -> if_sum_3_in_1_gga1(sum_3_in_gga2(X, Y))
if_sum_3_in_1_gga1(sum_3_out_gga1(Z)) -> sum_3_out_gga1(s_11(Z))
if_mult_3_in_2_gga1(sum_3_out_gga1(Z)) -> mult_3_out_gga1(Z)
if_p_1_in_2_g2(Xs, mult_3_out_gga1(Z)) -> if_p_1_in_3_g1(p_1_in_g1(._22(Z, Xs)))
if_p_1_in_3_g1(p_1_out_g) -> p_1_out_g

The set Q consists of the following terms:

p_1_in_g1(x0)
if_p_1_in_4_g1(x0)
if_p_1_in_1_g4(x0, x1, x2, x3)
mult_3_in_gga2(x0, x1)
if_mult_3_in_1_gga2(x0, x1)
sum_3_in_gga2(x0, x1)
if_sum_3_in_1_gga1(x0)
if_mult_3_in_2_gga1(x0)
if_p_1_in_2_g2(x0, x1)
if_p_1_in_3_g1(x0)

We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {P_1_IN_G1}.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, we can delete all non-usable rules from R.

↳ PROLOG
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
              ↳ PiDP
                ↳ PiDPToQDPProof
                  ↳ QDP
                    ↳ QDPPoloProof
                      ↳ QDP
                        ↳ DependencyGraphProof
                          ↳ QDP
                            ↳ UsableRulesProof
                              ↳ QDP
                                ↳ QDPSizeChangeProof

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

P_1_IN_G1(._22(s_11(s_11(X)), ._22(Y, Xs))) -> P_1_IN_G1(._22(X, ._22(Y, Xs)))

R is empty.
The set Q consists of the following terms:

p_1_in_g1(x0)
if_p_1_in_4_g1(x0)
if_p_1_in_1_g4(x0, x1, x2, x3)
mult_3_in_gga2(x0, x1)
if_mult_3_in_1_gga2(x0, x1)
sum_3_in_gga2(x0, x1)
if_sum_3_in_1_gga1(x0)
if_mult_3_in_2_gga1(x0)
if_p_1_in_2_g2(x0, x1)
if_p_1_in_3_g1(x0)

We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {P_1_IN_G1}.
We used the following order and afs together with the size-change analysis to show that there are no infinite chains for this DP problem.

Order:Homeomorphic Embedding Order

AFS:
s_11(x1)  =  s_11(x1)
._22(x1, x2)  =  x1

From the DPs we obtained the following set of size-change graphs:

We oriented the following set of usable rules. none