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



↳ PROLOG
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof

rem3(X, Y, R) :- notZero1(Y), sub3(X, Y, Z), rem3(Z, Y, R).
rem3(X, Y, X) :- notZero1(Y), geq2(X, Y).
sub3(s1(X), s1(Y), Z) :- sub3(X, Y, Z).
sub3(X, 00, X).
notZero1(s1(X)).
geq2(s1(X), s1(Y)) :- geq2(X, Y).
geq2(X, 00).


With regard to the inferred argument filtering the predicates were used in the following modes:
rem3: (b,b,f)
sub3: (b,b,f)
geq2: (b,b)
Transforming PROLOG into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:


rem_3_in_gga3(X, Y, R) -> if_rem_3_in_1_gga4(X, Y, R, notZero_1_in_g1(Y))
notZero_1_in_g1(s_11(X)) -> notZero_1_out_g1(s_11(X))
if_rem_3_in_1_gga4(X, Y, R, notZero_1_out_g1(Y)) -> if_rem_3_in_2_gga4(X, Y, R, sub_3_in_gga3(X, Y, Z))
sub_3_in_gga3(s_11(X), s_11(Y), Z) -> if_sub_3_in_1_gga4(X, Y, Z, sub_3_in_gga3(X, Y, Z))
sub_3_in_gga3(X, 0_0, X) -> sub_3_out_gga3(X, 0_0, X)
if_sub_3_in_1_gga4(X, Y, Z, sub_3_out_gga3(X, Y, Z)) -> sub_3_out_gga3(s_11(X), s_11(Y), Z)
if_rem_3_in_2_gga4(X, Y, R, sub_3_out_gga3(X, Y, Z)) -> if_rem_3_in_3_gga5(X, Y, R, Z, rem_3_in_gga3(Z, Y, R))
rem_3_in_gga3(X, Y, X) -> if_rem_3_in_4_gga3(X, Y, notZero_1_in_g1(Y))
if_rem_3_in_4_gga3(X, Y, notZero_1_out_g1(Y)) -> if_rem_3_in_5_gga3(X, Y, geq_2_in_gg2(X, Y))
geq_2_in_gg2(s_11(X), s_11(Y)) -> if_geq_2_in_1_gg3(X, Y, geq_2_in_gg2(X, Y))
geq_2_in_gg2(X, 0_0) -> geq_2_out_gg2(X, 0_0)
if_geq_2_in_1_gg3(X, Y, geq_2_out_gg2(X, Y)) -> geq_2_out_gg2(s_11(X), s_11(Y))
if_rem_3_in_5_gga3(X, Y, geq_2_out_gg2(X, Y)) -> rem_3_out_gga3(X, Y, X)
if_rem_3_in_3_gga5(X, Y, R, Z, rem_3_out_gga3(Z, Y, R)) -> rem_3_out_gga3(X, Y, R)

The argument filtering Pi contains the following mapping:
rem_3_in_gga3(x1, x2, x3)  =  rem_3_in_gga2(x1, x2)
s_11(x1)  =  s_11(x1)
0_0  =  0_0
if_rem_3_in_1_gga4(x1, x2, x3, x4)  =  if_rem_3_in_1_gga3(x1, x2, x4)
notZero_1_in_g1(x1)  =  notZero_1_in_g1(x1)
notZero_1_out_g1(x1)  =  notZero_1_out_g
if_rem_3_in_2_gga4(x1, x2, x3, x4)  =  if_rem_3_in_2_gga2(x2, x4)
sub_3_in_gga3(x1, x2, x3)  =  sub_3_in_gga2(x1, x2)
if_sub_3_in_1_gga4(x1, x2, x3, x4)  =  if_sub_3_in_1_gga1(x4)
sub_3_out_gga3(x1, x2, x3)  =  sub_3_out_gga1(x3)
if_rem_3_in_3_gga5(x1, x2, x3, x4, x5)  =  if_rem_3_in_3_gga1(x5)
if_rem_3_in_4_gga3(x1, x2, x3)  =  if_rem_3_in_4_gga3(x1, x2, x3)
if_rem_3_in_5_gga3(x1, x2, x3)  =  if_rem_3_in_5_gga2(x1, x3)
geq_2_in_gg2(x1, x2)  =  geq_2_in_gg2(x1, x2)
if_geq_2_in_1_gg3(x1, x2, x3)  =  if_geq_2_in_1_gg1(x3)
geq_2_out_gg2(x1, x2)  =  geq_2_out_gg
rem_3_out_gga3(x1, x2, x3)  =  rem_3_out_gga1(x3)

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:

rem_3_in_gga3(X, Y, R) -> if_rem_3_in_1_gga4(X, Y, R, notZero_1_in_g1(Y))
notZero_1_in_g1(s_11(X)) -> notZero_1_out_g1(s_11(X))
if_rem_3_in_1_gga4(X, Y, R, notZero_1_out_g1(Y)) -> if_rem_3_in_2_gga4(X, Y, R, sub_3_in_gga3(X, Y, Z))
sub_3_in_gga3(s_11(X), s_11(Y), Z) -> if_sub_3_in_1_gga4(X, Y, Z, sub_3_in_gga3(X, Y, Z))
sub_3_in_gga3(X, 0_0, X) -> sub_3_out_gga3(X, 0_0, X)
if_sub_3_in_1_gga4(X, Y, Z, sub_3_out_gga3(X, Y, Z)) -> sub_3_out_gga3(s_11(X), s_11(Y), Z)
if_rem_3_in_2_gga4(X, Y, R, sub_3_out_gga3(X, Y, Z)) -> if_rem_3_in_3_gga5(X, Y, R, Z, rem_3_in_gga3(Z, Y, R))
rem_3_in_gga3(X, Y, X) -> if_rem_3_in_4_gga3(X, Y, notZero_1_in_g1(Y))
if_rem_3_in_4_gga3(X, Y, notZero_1_out_g1(Y)) -> if_rem_3_in_5_gga3(X, Y, geq_2_in_gg2(X, Y))
geq_2_in_gg2(s_11(X), s_11(Y)) -> if_geq_2_in_1_gg3(X, Y, geq_2_in_gg2(X, Y))
geq_2_in_gg2(X, 0_0) -> geq_2_out_gg2(X, 0_0)
if_geq_2_in_1_gg3(X, Y, geq_2_out_gg2(X, Y)) -> geq_2_out_gg2(s_11(X), s_11(Y))
if_rem_3_in_5_gga3(X, Y, geq_2_out_gg2(X, Y)) -> rem_3_out_gga3(X, Y, X)
if_rem_3_in_3_gga5(X, Y, R, Z, rem_3_out_gga3(Z, Y, R)) -> rem_3_out_gga3(X, Y, R)

The argument filtering Pi contains the following mapping:
rem_3_in_gga3(x1, x2, x3)  =  rem_3_in_gga2(x1, x2)
s_11(x1)  =  s_11(x1)
0_0  =  0_0
if_rem_3_in_1_gga4(x1, x2, x3, x4)  =  if_rem_3_in_1_gga3(x1, x2, x4)
notZero_1_in_g1(x1)  =  notZero_1_in_g1(x1)
notZero_1_out_g1(x1)  =  notZero_1_out_g
if_rem_3_in_2_gga4(x1, x2, x3, x4)  =  if_rem_3_in_2_gga2(x2, x4)
sub_3_in_gga3(x1, x2, x3)  =  sub_3_in_gga2(x1, x2)
if_sub_3_in_1_gga4(x1, x2, x3, x4)  =  if_sub_3_in_1_gga1(x4)
sub_3_out_gga3(x1, x2, x3)  =  sub_3_out_gga1(x3)
if_rem_3_in_3_gga5(x1, x2, x3, x4, x5)  =  if_rem_3_in_3_gga1(x5)
if_rem_3_in_4_gga3(x1, x2, x3)  =  if_rem_3_in_4_gga3(x1, x2, x3)
if_rem_3_in_5_gga3(x1, x2, x3)  =  if_rem_3_in_5_gga2(x1, x3)
geq_2_in_gg2(x1, x2)  =  geq_2_in_gg2(x1, x2)
if_geq_2_in_1_gg3(x1, x2, x3)  =  if_geq_2_in_1_gg1(x3)
geq_2_out_gg2(x1, x2)  =  geq_2_out_gg
rem_3_out_gga3(x1, x2, x3)  =  rem_3_out_gga1(x3)


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

REM_3_IN_GGA3(X, Y, R) -> IF_REM_3_IN_1_GGA4(X, Y, R, notZero_1_in_g1(Y))
REM_3_IN_GGA3(X, Y, R) -> NOTZERO_1_IN_G1(Y)
IF_REM_3_IN_1_GGA4(X, Y, R, notZero_1_out_g1(Y)) -> IF_REM_3_IN_2_GGA4(X, Y, R, sub_3_in_gga3(X, Y, Z))
IF_REM_3_IN_1_GGA4(X, Y, R, notZero_1_out_g1(Y)) -> SUB_3_IN_GGA3(X, Y, Z)
SUB_3_IN_GGA3(s_11(X), s_11(Y), Z) -> IF_SUB_3_IN_1_GGA4(X, Y, Z, sub_3_in_gga3(X, Y, Z))
SUB_3_IN_GGA3(s_11(X), s_11(Y), Z) -> SUB_3_IN_GGA3(X, Y, Z)
IF_REM_3_IN_2_GGA4(X, Y, R, sub_3_out_gga3(X, Y, Z)) -> IF_REM_3_IN_3_GGA5(X, Y, R, Z, rem_3_in_gga3(Z, Y, R))
IF_REM_3_IN_2_GGA4(X, Y, R, sub_3_out_gga3(X, Y, Z)) -> REM_3_IN_GGA3(Z, Y, R)
REM_3_IN_GGA3(X, Y, X) -> IF_REM_3_IN_4_GGA3(X, Y, notZero_1_in_g1(Y))
REM_3_IN_GGA3(X, Y, X) -> NOTZERO_1_IN_G1(Y)
IF_REM_3_IN_4_GGA3(X, Y, notZero_1_out_g1(Y)) -> IF_REM_3_IN_5_GGA3(X, Y, geq_2_in_gg2(X, Y))
IF_REM_3_IN_4_GGA3(X, Y, notZero_1_out_g1(Y)) -> GEQ_2_IN_GG2(X, Y)
GEQ_2_IN_GG2(s_11(X), s_11(Y)) -> IF_GEQ_2_IN_1_GG3(X, Y, geq_2_in_gg2(X, Y))
GEQ_2_IN_GG2(s_11(X), s_11(Y)) -> GEQ_2_IN_GG2(X, Y)

The TRS R consists of the following rules:

rem_3_in_gga3(X, Y, R) -> if_rem_3_in_1_gga4(X, Y, R, notZero_1_in_g1(Y))
notZero_1_in_g1(s_11(X)) -> notZero_1_out_g1(s_11(X))
if_rem_3_in_1_gga4(X, Y, R, notZero_1_out_g1(Y)) -> if_rem_3_in_2_gga4(X, Y, R, sub_3_in_gga3(X, Y, Z))
sub_3_in_gga3(s_11(X), s_11(Y), Z) -> if_sub_3_in_1_gga4(X, Y, Z, sub_3_in_gga3(X, Y, Z))
sub_3_in_gga3(X, 0_0, X) -> sub_3_out_gga3(X, 0_0, X)
if_sub_3_in_1_gga4(X, Y, Z, sub_3_out_gga3(X, Y, Z)) -> sub_3_out_gga3(s_11(X), s_11(Y), Z)
if_rem_3_in_2_gga4(X, Y, R, sub_3_out_gga3(X, Y, Z)) -> if_rem_3_in_3_gga5(X, Y, R, Z, rem_3_in_gga3(Z, Y, R))
rem_3_in_gga3(X, Y, X) -> if_rem_3_in_4_gga3(X, Y, notZero_1_in_g1(Y))
if_rem_3_in_4_gga3(X, Y, notZero_1_out_g1(Y)) -> if_rem_3_in_5_gga3(X, Y, geq_2_in_gg2(X, Y))
geq_2_in_gg2(s_11(X), s_11(Y)) -> if_geq_2_in_1_gg3(X, Y, geq_2_in_gg2(X, Y))
geq_2_in_gg2(X, 0_0) -> geq_2_out_gg2(X, 0_0)
if_geq_2_in_1_gg3(X, Y, geq_2_out_gg2(X, Y)) -> geq_2_out_gg2(s_11(X), s_11(Y))
if_rem_3_in_5_gga3(X, Y, geq_2_out_gg2(X, Y)) -> rem_3_out_gga3(X, Y, X)
if_rem_3_in_3_gga5(X, Y, R, Z, rem_3_out_gga3(Z, Y, R)) -> rem_3_out_gga3(X, Y, R)

The argument filtering Pi contains the following mapping:
rem_3_in_gga3(x1, x2, x3)  =  rem_3_in_gga2(x1, x2)
s_11(x1)  =  s_11(x1)
0_0  =  0_0
if_rem_3_in_1_gga4(x1, x2, x3, x4)  =  if_rem_3_in_1_gga3(x1, x2, x4)
notZero_1_in_g1(x1)  =  notZero_1_in_g1(x1)
notZero_1_out_g1(x1)  =  notZero_1_out_g
if_rem_3_in_2_gga4(x1, x2, x3, x4)  =  if_rem_3_in_2_gga2(x2, x4)
sub_3_in_gga3(x1, x2, x3)  =  sub_3_in_gga2(x1, x2)
if_sub_3_in_1_gga4(x1, x2, x3, x4)  =  if_sub_3_in_1_gga1(x4)
sub_3_out_gga3(x1, x2, x3)  =  sub_3_out_gga1(x3)
if_rem_3_in_3_gga5(x1, x2, x3, x4, x5)  =  if_rem_3_in_3_gga1(x5)
if_rem_3_in_4_gga3(x1, x2, x3)  =  if_rem_3_in_4_gga3(x1, x2, x3)
if_rem_3_in_5_gga3(x1, x2, x3)  =  if_rem_3_in_5_gga2(x1, x3)
geq_2_in_gg2(x1, x2)  =  geq_2_in_gg2(x1, x2)
if_geq_2_in_1_gg3(x1, x2, x3)  =  if_geq_2_in_1_gg1(x3)
geq_2_out_gg2(x1, x2)  =  geq_2_out_gg
rem_3_out_gga3(x1, x2, x3)  =  rem_3_out_gga1(x3)
IF_REM_3_IN_5_GGA3(x1, x2, x3)  =  IF_REM_3_IN_5_GGA2(x1, x3)
REM_3_IN_GGA3(x1, x2, x3)  =  REM_3_IN_GGA2(x1, x2)
GEQ_2_IN_GG2(x1, x2)  =  GEQ_2_IN_GG2(x1, x2)
IF_SUB_3_IN_1_GGA4(x1, x2, x3, x4)  =  IF_SUB_3_IN_1_GGA1(x4)
IF_REM_3_IN_2_GGA4(x1, x2, x3, x4)  =  IF_REM_3_IN_2_GGA2(x2, x4)
NOTZERO_1_IN_G1(x1)  =  NOTZERO_1_IN_G1(x1)
IF_REM_3_IN_4_GGA3(x1, x2, x3)  =  IF_REM_3_IN_4_GGA3(x1, x2, x3)
IF_REM_3_IN_3_GGA5(x1, x2, x3, x4, x5)  =  IF_REM_3_IN_3_GGA1(x5)
IF_REM_3_IN_1_GGA4(x1, x2, x3, x4)  =  IF_REM_3_IN_1_GGA3(x1, x2, x4)
IF_GEQ_2_IN_1_GG3(x1, x2, x3)  =  IF_GEQ_2_IN_1_GG1(x3)
SUB_3_IN_GGA3(x1, x2, x3)  =  SUB_3_IN_GGA2(x1, x2)

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:

REM_3_IN_GGA3(X, Y, R) -> IF_REM_3_IN_1_GGA4(X, Y, R, notZero_1_in_g1(Y))
REM_3_IN_GGA3(X, Y, R) -> NOTZERO_1_IN_G1(Y)
IF_REM_3_IN_1_GGA4(X, Y, R, notZero_1_out_g1(Y)) -> IF_REM_3_IN_2_GGA4(X, Y, R, sub_3_in_gga3(X, Y, Z))
IF_REM_3_IN_1_GGA4(X, Y, R, notZero_1_out_g1(Y)) -> SUB_3_IN_GGA3(X, Y, Z)
SUB_3_IN_GGA3(s_11(X), s_11(Y), Z) -> IF_SUB_3_IN_1_GGA4(X, Y, Z, sub_3_in_gga3(X, Y, Z))
SUB_3_IN_GGA3(s_11(X), s_11(Y), Z) -> SUB_3_IN_GGA3(X, Y, Z)
IF_REM_3_IN_2_GGA4(X, Y, R, sub_3_out_gga3(X, Y, Z)) -> IF_REM_3_IN_3_GGA5(X, Y, R, Z, rem_3_in_gga3(Z, Y, R))
IF_REM_3_IN_2_GGA4(X, Y, R, sub_3_out_gga3(X, Y, Z)) -> REM_3_IN_GGA3(Z, Y, R)
REM_3_IN_GGA3(X, Y, X) -> IF_REM_3_IN_4_GGA3(X, Y, notZero_1_in_g1(Y))
REM_3_IN_GGA3(X, Y, X) -> NOTZERO_1_IN_G1(Y)
IF_REM_3_IN_4_GGA3(X, Y, notZero_1_out_g1(Y)) -> IF_REM_3_IN_5_GGA3(X, Y, geq_2_in_gg2(X, Y))
IF_REM_3_IN_4_GGA3(X, Y, notZero_1_out_g1(Y)) -> GEQ_2_IN_GG2(X, Y)
GEQ_2_IN_GG2(s_11(X), s_11(Y)) -> IF_GEQ_2_IN_1_GG3(X, Y, geq_2_in_gg2(X, Y))
GEQ_2_IN_GG2(s_11(X), s_11(Y)) -> GEQ_2_IN_GG2(X, Y)

The TRS R consists of the following rules:

rem_3_in_gga3(X, Y, R) -> if_rem_3_in_1_gga4(X, Y, R, notZero_1_in_g1(Y))
notZero_1_in_g1(s_11(X)) -> notZero_1_out_g1(s_11(X))
if_rem_3_in_1_gga4(X, Y, R, notZero_1_out_g1(Y)) -> if_rem_3_in_2_gga4(X, Y, R, sub_3_in_gga3(X, Y, Z))
sub_3_in_gga3(s_11(X), s_11(Y), Z) -> if_sub_3_in_1_gga4(X, Y, Z, sub_3_in_gga3(X, Y, Z))
sub_3_in_gga3(X, 0_0, X) -> sub_3_out_gga3(X, 0_0, X)
if_sub_3_in_1_gga4(X, Y, Z, sub_3_out_gga3(X, Y, Z)) -> sub_3_out_gga3(s_11(X), s_11(Y), Z)
if_rem_3_in_2_gga4(X, Y, R, sub_3_out_gga3(X, Y, Z)) -> if_rem_3_in_3_gga5(X, Y, R, Z, rem_3_in_gga3(Z, Y, R))
rem_3_in_gga3(X, Y, X) -> if_rem_3_in_4_gga3(X, Y, notZero_1_in_g1(Y))
if_rem_3_in_4_gga3(X, Y, notZero_1_out_g1(Y)) -> if_rem_3_in_5_gga3(X, Y, geq_2_in_gg2(X, Y))
geq_2_in_gg2(s_11(X), s_11(Y)) -> if_geq_2_in_1_gg3(X, Y, geq_2_in_gg2(X, Y))
geq_2_in_gg2(X, 0_0) -> geq_2_out_gg2(X, 0_0)
if_geq_2_in_1_gg3(X, Y, geq_2_out_gg2(X, Y)) -> geq_2_out_gg2(s_11(X), s_11(Y))
if_rem_3_in_5_gga3(X, Y, geq_2_out_gg2(X, Y)) -> rem_3_out_gga3(X, Y, X)
if_rem_3_in_3_gga5(X, Y, R, Z, rem_3_out_gga3(Z, Y, R)) -> rem_3_out_gga3(X, Y, R)

The argument filtering Pi contains the following mapping:
rem_3_in_gga3(x1, x2, x3)  =  rem_3_in_gga2(x1, x2)
s_11(x1)  =  s_11(x1)
0_0  =  0_0
if_rem_3_in_1_gga4(x1, x2, x3, x4)  =  if_rem_3_in_1_gga3(x1, x2, x4)
notZero_1_in_g1(x1)  =  notZero_1_in_g1(x1)
notZero_1_out_g1(x1)  =  notZero_1_out_g
if_rem_3_in_2_gga4(x1, x2, x3, x4)  =  if_rem_3_in_2_gga2(x2, x4)
sub_3_in_gga3(x1, x2, x3)  =  sub_3_in_gga2(x1, x2)
if_sub_3_in_1_gga4(x1, x2, x3, x4)  =  if_sub_3_in_1_gga1(x4)
sub_3_out_gga3(x1, x2, x3)  =  sub_3_out_gga1(x3)
if_rem_3_in_3_gga5(x1, x2, x3, x4, x5)  =  if_rem_3_in_3_gga1(x5)
if_rem_3_in_4_gga3(x1, x2, x3)  =  if_rem_3_in_4_gga3(x1, x2, x3)
if_rem_3_in_5_gga3(x1, x2, x3)  =  if_rem_3_in_5_gga2(x1, x3)
geq_2_in_gg2(x1, x2)  =  geq_2_in_gg2(x1, x2)
if_geq_2_in_1_gg3(x1, x2, x3)  =  if_geq_2_in_1_gg1(x3)
geq_2_out_gg2(x1, x2)  =  geq_2_out_gg
rem_3_out_gga3(x1, x2, x3)  =  rem_3_out_gga1(x3)
IF_REM_3_IN_5_GGA3(x1, x2, x3)  =  IF_REM_3_IN_5_GGA2(x1, x3)
REM_3_IN_GGA3(x1, x2, x3)  =  REM_3_IN_GGA2(x1, x2)
GEQ_2_IN_GG2(x1, x2)  =  GEQ_2_IN_GG2(x1, x2)
IF_SUB_3_IN_1_GGA4(x1, x2, x3, x4)  =  IF_SUB_3_IN_1_GGA1(x4)
IF_REM_3_IN_2_GGA4(x1, x2, x3, x4)  =  IF_REM_3_IN_2_GGA2(x2, x4)
NOTZERO_1_IN_G1(x1)  =  NOTZERO_1_IN_G1(x1)
IF_REM_3_IN_4_GGA3(x1, x2, x3)  =  IF_REM_3_IN_4_GGA3(x1, x2, x3)
IF_REM_3_IN_3_GGA5(x1, x2, x3, x4, x5)  =  IF_REM_3_IN_3_GGA1(x5)
IF_REM_3_IN_1_GGA4(x1, x2, x3, x4)  =  IF_REM_3_IN_1_GGA3(x1, x2, x4)
IF_GEQ_2_IN_1_GG3(x1, x2, x3)  =  IF_GEQ_2_IN_1_GG1(x3)
SUB_3_IN_GGA3(x1, x2, x3)  =  SUB_3_IN_GGA2(x1, x2)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph contains 3 SCCs with 9 less nodes.

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

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

GEQ_2_IN_GG2(s_11(X), s_11(Y)) -> GEQ_2_IN_GG2(X, Y)

The TRS R consists of the following rules:

rem_3_in_gga3(X, Y, R) -> if_rem_3_in_1_gga4(X, Y, R, notZero_1_in_g1(Y))
notZero_1_in_g1(s_11(X)) -> notZero_1_out_g1(s_11(X))
if_rem_3_in_1_gga4(X, Y, R, notZero_1_out_g1(Y)) -> if_rem_3_in_2_gga4(X, Y, R, sub_3_in_gga3(X, Y, Z))
sub_3_in_gga3(s_11(X), s_11(Y), Z) -> if_sub_3_in_1_gga4(X, Y, Z, sub_3_in_gga3(X, Y, Z))
sub_3_in_gga3(X, 0_0, X) -> sub_3_out_gga3(X, 0_0, X)
if_sub_3_in_1_gga4(X, Y, Z, sub_3_out_gga3(X, Y, Z)) -> sub_3_out_gga3(s_11(X), s_11(Y), Z)
if_rem_3_in_2_gga4(X, Y, R, sub_3_out_gga3(X, Y, Z)) -> if_rem_3_in_3_gga5(X, Y, R, Z, rem_3_in_gga3(Z, Y, R))
rem_3_in_gga3(X, Y, X) -> if_rem_3_in_4_gga3(X, Y, notZero_1_in_g1(Y))
if_rem_3_in_4_gga3(X, Y, notZero_1_out_g1(Y)) -> if_rem_3_in_5_gga3(X, Y, geq_2_in_gg2(X, Y))
geq_2_in_gg2(s_11(X), s_11(Y)) -> if_geq_2_in_1_gg3(X, Y, geq_2_in_gg2(X, Y))
geq_2_in_gg2(X, 0_0) -> geq_2_out_gg2(X, 0_0)
if_geq_2_in_1_gg3(X, Y, geq_2_out_gg2(X, Y)) -> geq_2_out_gg2(s_11(X), s_11(Y))
if_rem_3_in_5_gga3(X, Y, geq_2_out_gg2(X, Y)) -> rem_3_out_gga3(X, Y, X)
if_rem_3_in_3_gga5(X, Y, R, Z, rem_3_out_gga3(Z, Y, R)) -> rem_3_out_gga3(X, Y, R)

The argument filtering Pi contains the following mapping:
rem_3_in_gga3(x1, x2, x3)  =  rem_3_in_gga2(x1, x2)
s_11(x1)  =  s_11(x1)
0_0  =  0_0
if_rem_3_in_1_gga4(x1, x2, x3, x4)  =  if_rem_3_in_1_gga3(x1, x2, x4)
notZero_1_in_g1(x1)  =  notZero_1_in_g1(x1)
notZero_1_out_g1(x1)  =  notZero_1_out_g
if_rem_3_in_2_gga4(x1, x2, x3, x4)  =  if_rem_3_in_2_gga2(x2, x4)
sub_3_in_gga3(x1, x2, x3)  =  sub_3_in_gga2(x1, x2)
if_sub_3_in_1_gga4(x1, x2, x3, x4)  =  if_sub_3_in_1_gga1(x4)
sub_3_out_gga3(x1, x2, x3)  =  sub_3_out_gga1(x3)
if_rem_3_in_3_gga5(x1, x2, x3, x4, x5)  =  if_rem_3_in_3_gga1(x5)
if_rem_3_in_4_gga3(x1, x2, x3)  =  if_rem_3_in_4_gga3(x1, x2, x3)
if_rem_3_in_5_gga3(x1, x2, x3)  =  if_rem_3_in_5_gga2(x1, x3)
geq_2_in_gg2(x1, x2)  =  geq_2_in_gg2(x1, x2)
if_geq_2_in_1_gg3(x1, x2, x3)  =  if_geq_2_in_1_gg1(x3)
geq_2_out_gg2(x1, x2)  =  geq_2_out_gg
rem_3_out_gga3(x1, x2, x3)  =  rem_3_out_gga1(x3)
GEQ_2_IN_GG2(x1, x2)  =  GEQ_2_IN_GG2(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
  ↳ PrologToPiTRSProof

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

GEQ_2_IN_GG2(s_11(X), s_11(Y)) -> GEQ_2_IN_GG2(X, Y)

R is empty.
Pi is empty.
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
  ↳ PrologToPiTRSProof

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

GEQ_2_IN_GG2(s_11(X), s_11(Y)) -> GEQ_2_IN_GG2(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 {GEQ_2_IN_GG2}.
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
  ↳ PrologToPiTRSProof

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

SUB_3_IN_GGA3(s_11(X), s_11(Y), Z) -> SUB_3_IN_GGA3(X, Y, Z)

The TRS R consists of the following rules:

rem_3_in_gga3(X, Y, R) -> if_rem_3_in_1_gga4(X, Y, R, notZero_1_in_g1(Y))
notZero_1_in_g1(s_11(X)) -> notZero_1_out_g1(s_11(X))
if_rem_3_in_1_gga4(X, Y, R, notZero_1_out_g1(Y)) -> if_rem_3_in_2_gga4(X, Y, R, sub_3_in_gga3(X, Y, Z))
sub_3_in_gga3(s_11(X), s_11(Y), Z) -> if_sub_3_in_1_gga4(X, Y, Z, sub_3_in_gga3(X, Y, Z))
sub_3_in_gga3(X, 0_0, X) -> sub_3_out_gga3(X, 0_0, X)
if_sub_3_in_1_gga4(X, Y, Z, sub_3_out_gga3(X, Y, Z)) -> sub_3_out_gga3(s_11(X), s_11(Y), Z)
if_rem_3_in_2_gga4(X, Y, R, sub_3_out_gga3(X, Y, Z)) -> if_rem_3_in_3_gga5(X, Y, R, Z, rem_3_in_gga3(Z, Y, R))
rem_3_in_gga3(X, Y, X) -> if_rem_3_in_4_gga3(X, Y, notZero_1_in_g1(Y))
if_rem_3_in_4_gga3(X, Y, notZero_1_out_g1(Y)) -> if_rem_3_in_5_gga3(X, Y, geq_2_in_gg2(X, Y))
geq_2_in_gg2(s_11(X), s_11(Y)) -> if_geq_2_in_1_gg3(X, Y, geq_2_in_gg2(X, Y))
geq_2_in_gg2(X, 0_0) -> geq_2_out_gg2(X, 0_0)
if_geq_2_in_1_gg3(X, Y, geq_2_out_gg2(X, Y)) -> geq_2_out_gg2(s_11(X), s_11(Y))
if_rem_3_in_5_gga3(X, Y, geq_2_out_gg2(X, Y)) -> rem_3_out_gga3(X, Y, X)
if_rem_3_in_3_gga5(X, Y, R, Z, rem_3_out_gga3(Z, Y, R)) -> rem_3_out_gga3(X, Y, R)

The argument filtering Pi contains the following mapping:
rem_3_in_gga3(x1, x2, x3)  =  rem_3_in_gga2(x1, x2)
s_11(x1)  =  s_11(x1)
0_0  =  0_0
if_rem_3_in_1_gga4(x1, x2, x3, x4)  =  if_rem_3_in_1_gga3(x1, x2, x4)
notZero_1_in_g1(x1)  =  notZero_1_in_g1(x1)
notZero_1_out_g1(x1)  =  notZero_1_out_g
if_rem_3_in_2_gga4(x1, x2, x3, x4)  =  if_rem_3_in_2_gga2(x2, x4)
sub_3_in_gga3(x1, x2, x3)  =  sub_3_in_gga2(x1, x2)
if_sub_3_in_1_gga4(x1, x2, x3, x4)  =  if_sub_3_in_1_gga1(x4)
sub_3_out_gga3(x1, x2, x3)  =  sub_3_out_gga1(x3)
if_rem_3_in_3_gga5(x1, x2, x3, x4, x5)  =  if_rem_3_in_3_gga1(x5)
if_rem_3_in_4_gga3(x1, x2, x3)  =  if_rem_3_in_4_gga3(x1, x2, x3)
if_rem_3_in_5_gga3(x1, x2, x3)  =  if_rem_3_in_5_gga2(x1, x3)
geq_2_in_gg2(x1, x2)  =  geq_2_in_gg2(x1, x2)
if_geq_2_in_1_gg3(x1, x2, x3)  =  if_geq_2_in_1_gg1(x3)
geq_2_out_gg2(x1, x2)  =  geq_2_out_gg
rem_3_out_gga3(x1, x2, x3)  =  rem_3_out_gga1(x3)
SUB_3_IN_GGA3(x1, x2, x3)  =  SUB_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
  ↳ PrologToPiTRSProof

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

SUB_3_IN_GGA3(s_11(X), s_11(Y), Z) -> SUB_3_IN_GGA3(X, Y, Z)

R is empty.
The argument filtering Pi contains the following mapping:
s_11(x1)  =  s_11(x1)
SUB_3_IN_GGA3(x1, x2, x3)  =  SUB_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
  ↳ PrologToPiTRSProof

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

SUB_3_IN_GGA2(s_11(X), s_11(Y)) -> SUB_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 {SUB_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
                ↳ UsableRulesProof
  ↳ PrologToPiTRSProof

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

REM_3_IN_GGA3(X, Y, R) -> IF_REM_3_IN_1_GGA4(X, Y, R, notZero_1_in_g1(Y))
IF_REM_3_IN_2_GGA4(X, Y, R, sub_3_out_gga3(X, Y, Z)) -> REM_3_IN_GGA3(Z, Y, R)
IF_REM_3_IN_1_GGA4(X, Y, R, notZero_1_out_g1(Y)) -> IF_REM_3_IN_2_GGA4(X, Y, R, sub_3_in_gga3(X, Y, Z))

The TRS R consists of the following rules:

rem_3_in_gga3(X, Y, R) -> if_rem_3_in_1_gga4(X, Y, R, notZero_1_in_g1(Y))
notZero_1_in_g1(s_11(X)) -> notZero_1_out_g1(s_11(X))
if_rem_3_in_1_gga4(X, Y, R, notZero_1_out_g1(Y)) -> if_rem_3_in_2_gga4(X, Y, R, sub_3_in_gga3(X, Y, Z))
sub_3_in_gga3(s_11(X), s_11(Y), Z) -> if_sub_3_in_1_gga4(X, Y, Z, sub_3_in_gga3(X, Y, Z))
sub_3_in_gga3(X, 0_0, X) -> sub_3_out_gga3(X, 0_0, X)
if_sub_3_in_1_gga4(X, Y, Z, sub_3_out_gga3(X, Y, Z)) -> sub_3_out_gga3(s_11(X), s_11(Y), Z)
if_rem_3_in_2_gga4(X, Y, R, sub_3_out_gga3(X, Y, Z)) -> if_rem_3_in_3_gga5(X, Y, R, Z, rem_3_in_gga3(Z, Y, R))
rem_3_in_gga3(X, Y, X) -> if_rem_3_in_4_gga3(X, Y, notZero_1_in_g1(Y))
if_rem_3_in_4_gga3(X, Y, notZero_1_out_g1(Y)) -> if_rem_3_in_5_gga3(X, Y, geq_2_in_gg2(X, Y))
geq_2_in_gg2(s_11(X), s_11(Y)) -> if_geq_2_in_1_gg3(X, Y, geq_2_in_gg2(X, Y))
geq_2_in_gg2(X, 0_0) -> geq_2_out_gg2(X, 0_0)
if_geq_2_in_1_gg3(X, Y, geq_2_out_gg2(X, Y)) -> geq_2_out_gg2(s_11(X), s_11(Y))
if_rem_3_in_5_gga3(X, Y, geq_2_out_gg2(X, Y)) -> rem_3_out_gga3(X, Y, X)
if_rem_3_in_3_gga5(X, Y, R, Z, rem_3_out_gga3(Z, Y, R)) -> rem_3_out_gga3(X, Y, R)

The argument filtering Pi contains the following mapping:
rem_3_in_gga3(x1, x2, x3)  =  rem_3_in_gga2(x1, x2)
s_11(x1)  =  s_11(x1)
0_0  =  0_0
if_rem_3_in_1_gga4(x1, x2, x3, x4)  =  if_rem_3_in_1_gga3(x1, x2, x4)
notZero_1_in_g1(x1)  =  notZero_1_in_g1(x1)
notZero_1_out_g1(x1)  =  notZero_1_out_g
if_rem_3_in_2_gga4(x1, x2, x3, x4)  =  if_rem_3_in_2_gga2(x2, x4)
sub_3_in_gga3(x1, x2, x3)  =  sub_3_in_gga2(x1, x2)
if_sub_3_in_1_gga4(x1, x2, x3, x4)  =  if_sub_3_in_1_gga1(x4)
sub_3_out_gga3(x1, x2, x3)  =  sub_3_out_gga1(x3)
if_rem_3_in_3_gga5(x1, x2, x3, x4, x5)  =  if_rem_3_in_3_gga1(x5)
if_rem_3_in_4_gga3(x1, x2, x3)  =  if_rem_3_in_4_gga3(x1, x2, x3)
if_rem_3_in_5_gga3(x1, x2, x3)  =  if_rem_3_in_5_gga2(x1, x3)
geq_2_in_gg2(x1, x2)  =  geq_2_in_gg2(x1, x2)
if_geq_2_in_1_gg3(x1, x2, x3)  =  if_geq_2_in_1_gg1(x3)
geq_2_out_gg2(x1, x2)  =  geq_2_out_gg
rem_3_out_gga3(x1, x2, x3)  =  rem_3_out_gga1(x3)
REM_3_IN_GGA3(x1, x2, x3)  =  REM_3_IN_GGA2(x1, x2)
IF_REM_3_IN_2_GGA4(x1, x2, x3, x4)  =  IF_REM_3_IN_2_GGA2(x2, x4)
IF_REM_3_IN_1_GGA4(x1, x2, x3, x4)  =  IF_REM_3_IN_1_GGA3(x1, x2, x4)

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
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
  ↳ PrologToPiTRSProof

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

REM_3_IN_GGA3(X, Y, R) -> IF_REM_3_IN_1_GGA4(X, Y, R, notZero_1_in_g1(Y))
IF_REM_3_IN_2_GGA4(X, Y, R, sub_3_out_gga3(X, Y, Z)) -> REM_3_IN_GGA3(Z, Y, R)
IF_REM_3_IN_1_GGA4(X, Y, R, notZero_1_out_g1(Y)) -> IF_REM_3_IN_2_GGA4(X, Y, R, sub_3_in_gga3(X, Y, Z))

The TRS R consists of the following rules:

notZero_1_in_g1(s_11(X)) -> notZero_1_out_g1(s_11(X))
sub_3_in_gga3(s_11(X), s_11(Y), Z) -> if_sub_3_in_1_gga4(X, Y, Z, sub_3_in_gga3(X, Y, Z))
sub_3_in_gga3(X, 0_0, X) -> sub_3_out_gga3(X, 0_0, X)
if_sub_3_in_1_gga4(X, Y, Z, sub_3_out_gga3(X, Y, Z)) -> sub_3_out_gga3(s_11(X), s_11(Y), Z)

The argument filtering Pi contains the following mapping:
s_11(x1)  =  s_11(x1)
0_0  =  0_0
notZero_1_in_g1(x1)  =  notZero_1_in_g1(x1)
notZero_1_out_g1(x1)  =  notZero_1_out_g
sub_3_in_gga3(x1, x2, x3)  =  sub_3_in_gga2(x1, x2)
if_sub_3_in_1_gga4(x1, x2, x3, x4)  =  if_sub_3_in_1_gga1(x4)
sub_3_out_gga3(x1, x2, x3)  =  sub_3_out_gga1(x3)
REM_3_IN_GGA3(x1, x2, x3)  =  REM_3_IN_GGA2(x1, x2)
IF_REM_3_IN_2_GGA4(x1, x2, x3, x4)  =  IF_REM_3_IN_2_GGA2(x2, x4)
IF_REM_3_IN_1_GGA4(x1, x2, x3, x4)  =  IF_REM_3_IN_1_GGA3(x1, x2, x4)

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
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
  ↳ PrologToPiTRSProof

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

REM_3_IN_GGA2(X, Y) -> IF_REM_3_IN_1_GGA3(X, Y, notZero_1_in_g1(Y))
IF_REM_3_IN_2_GGA2(Y, sub_3_out_gga1(Z)) -> REM_3_IN_GGA2(Z, Y)
IF_REM_3_IN_1_GGA3(X, Y, notZero_1_out_g) -> IF_REM_3_IN_2_GGA2(Y, sub_3_in_gga2(X, Y))

The TRS R consists of the following rules:

notZero_1_in_g1(s_11(X)) -> notZero_1_out_g
sub_3_in_gga2(s_11(X), s_11(Y)) -> if_sub_3_in_1_gga1(sub_3_in_gga2(X, Y))
sub_3_in_gga2(X, 0_0) -> sub_3_out_gga1(X)
if_sub_3_in_1_gga1(sub_3_out_gga1(Z)) -> sub_3_out_gga1(Z)

The set Q consists of the following terms:

notZero_1_in_g1(x0)
sub_3_in_gga2(x0, x1)
if_sub_3_in_1_gga1(x0)

We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {IF_REM_3_IN_1_GGA3, REM_3_IN_GGA2, IF_REM_3_IN_2_GGA2}.
With regard to the inferred argument filtering the predicates were used in the following modes:
rem3: (b,b,f)
sub3: (b,b,f)
geq2: (b,b)
Transforming PROLOG into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

rem_3_in_gga3(X, Y, R) -> if_rem_3_in_1_gga4(X, Y, R, notZero_1_in_g1(Y))
notZero_1_in_g1(s_11(X)) -> notZero_1_out_g1(s_11(X))
if_rem_3_in_1_gga4(X, Y, R, notZero_1_out_g1(Y)) -> if_rem_3_in_2_gga4(X, Y, R, sub_3_in_gga3(X, Y, Z))
sub_3_in_gga3(s_11(X), s_11(Y), Z) -> if_sub_3_in_1_gga4(X, Y, Z, sub_3_in_gga3(X, Y, Z))
sub_3_in_gga3(X, 0_0, X) -> sub_3_out_gga3(X, 0_0, X)
if_sub_3_in_1_gga4(X, Y, Z, sub_3_out_gga3(X, Y, Z)) -> sub_3_out_gga3(s_11(X), s_11(Y), Z)
if_rem_3_in_2_gga4(X, Y, R, sub_3_out_gga3(X, Y, Z)) -> if_rem_3_in_3_gga5(X, Y, R, Z, rem_3_in_gga3(Z, Y, R))
rem_3_in_gga3(X, Y, X) -> if_rem_3_in_4_gga3(X, Y, notZero_1_in_g1(Y))
if_rem_3_in_4_gga3(X, Y, notZero_1_out_g1(Y)) -> if_rem_3_in_5_gga3(X, Y, geq_2_in_gg2(X, Y))
geq_2_in_gg2(s_11(X), s_11(Y)) -> if_geq_2_in_1_gg3(X, Y, geq_2_in_gg2(X, Y))
geq_2_in_gg2(X, 0_0) -> geq_2_out_gg2(X, 0_0)
if_geq_2_in_1_gg3(X, Y, geq_2_out_gg2(X, Y)) -> geq_2_out_gg2(s_11(X), s_11(Y))
if_rem_3_in_5_gga3(X, Y, geq_2_out_gg2(X, Y)) -> rem_3_out_gga3(X, Y, X)
if_rem_3_in_3_gga5(X, Y, R, Z, rem_3_out_gga3(Z, Y, R)) -> rem_3_out_gga3(X, Y, R)

The argument filtering Pi contains the following mapping:
rem_3_in_gga3(x1, x2, x3)  =  rem_3_in_gga2(x1, x2)
s_11(x1)  =  s_11(x1)
0_0  =  0_0
if_rem_3_in_1_gga4(x1, x2, x3, x4)  =  if_rem_3_in_1_gga3(x1, x2, x4)
notZero_1_in_g1(x1)  =  notZero_1_in_g1(x1)
notZero_1_out_g1(x1)  =  notZero_1_out_g1(x1)
if_rem_3_in_2_gga4(x1, x2, x3, x4)  =  if_rem_3_in_2_gga3(x1, x2, x4)
sub_3_in_gga3(x1, x2, x3)  =  sub_3_in_gga2(x1, x2)
if_sub_3_in_1_gga4(x1, x2, x3, x4)  =  if_sub_3_in_1_gga3(x1, x2, x4)
sub_3_out_gga3(x1, x2, x3)  =  sub_3_out_gga3(x1, x2, x3)
if_rem_3_in_3_gga5(x1, x2, x3, x4, x5)  =  if_rem_3_in_3_gga3(x1, x2, x5)
if_rem_3_in_4_gga3(x1, x2, x3)  =  if_rem_3_in_4_gga3(x1, x2, x3)
if_rem_3_in_5_gga3(x1, x2, x3)  =  if_rem_3_in_5_gga3(x1, x2, x3)
geq_2_in_gg2(x1, x2)  =  geq_2_in_gg2(x1, x2)
if_geq_2_in_1_gg3(x1, x2, x3)  =  if_geq_2_in_1_gg3(x1, x2, x3)
geq_2_out_gg2(x1, x2)  =  geq_2_out_gg2(x1, x2)
rem_3_out_gga3(x1, x2, x3)  =  rem_3_out_gga3(x1, x2, x3)

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:

rem_3_in_gga3(X, Y, R) -> if_rem_3_in_1_gga4(X, Y, R, notZero_1_in_g1(Y))
notZero_1_in_g1(s_11(X)) -> notZero_1_out_g1(s_11(X))
if_rem_3_in_1_gga4(X, Y, R, notZero_1_out_g1(Y)) -> if_rem_3_in_2_gga4(X, Y, R, sub_3_in_gga3(X, Y, Z))
sub_3_in_gga3(s_11(X), s_11(Y), Z) -> if_sub_3_in_1_gga4(X, Y, Z, sub_3_in_gga3(X, Y, Z))
sub_3_in_gga3(X, 0_0, X) -> sub_3_out_gga3(X, 0_0, X)
if_sub_3_in_1_gga4(X, Y, Z, sub_3_out_gga3(X, Y, Z)) -> sub_3_out_gga3(s_11(X), s_11(Y), Z)
if_rem_3_in_2_gga4(X, Y, R, sub_3_out_gga3(X, Y, Z)) -> if_rem_3_in_3_gga5(X, Y, R, Z, rem_3_in_gga3(Z, Y, R))
rem_3_in_gga3(X, Y, X) -> if_rem_3_in_4_gga3(X, Y, notZero_1_in_g1(Y))
if_rem_3_in_4_gga3(X, Y, notZero_1_out_g1(Y)) -> if_rem_3_in_5_gga3(X, Y, geq_2_in_gg2(X, Y))
geq_2_in_gg2(s_11(X), s_11(Y)) -> if_geq_2_in_1_gg3(X, Y, geq_2_in_gg2(X, Y))
geq_2_in_gg2(X, 0_0) -> geq_2_out_gg2(X, 0_0)
if_geq_2_in_1_gg3(X, Y, geq_2_out_gg2(X, Y)) -> geq_2_out_gg2(s_11(X), s_11(Y))
if_rem_3_in_5_gga3(X, Y, geq_2_out_gg2(X, Y)) -> rem_3_out_gga3(X, Y, X)
if_rem_3_in_3_gga5(X, Y, R, Z, rem_3_out_gga3(Z, Y, R)) -> rem_3_out_gga3(X, Y, R)

The argument filtering Pi contains the following mapping:
rem_3_in_gga3(x1, x2, x3)  =  rem_3_in_gga2(x1, x2)
s_11(x1)  =  s_11(x1)
0_0  =  0_0
if_rem_3_in_1_gga4(x1, x2, x3, x4)  =  if_rem_3_in_1_gga3(x1, x2, x4)
notZero_1_in_g1(x1)  =  notZero_1_in_g1(x1)
notZero_1_out_g1(x1)  =  notZero_1_out_g1(x1)
if_rem_3_in_2_gga4(x1, x2, x3, x4)  =  if_rem_3_in_2_gga3(x1, x2, x4)
sub_3_in_gga3(x1, x2, x3)  =  sub_3_in_gga2(x1, x2)
if_sub_3_in_1_gga4(x1, x2, x3, x4)  =  if_sub_3_in_1_gga3(x1, x2, x4)
sub_3_out_gga3(x1, x2, x3)  =  sub_3_out_gga3(x1, x2, x3)
if_rem_3_in_3_gga5(x1, x2, x3, x4, x5)  =  if_rem_3_in_3_gga3(x1, x2, x5)
if_rem_3_in_4_gga3(x1, x2, x3)  =  if_rem_3_in_4_gga3(x1, x2, x3)
if_rem_3_in_5_gga3(x1, x2, x3)  =  if_rem_3_in_5_gga3(x1, x2, x3)
geq_2_in_gg2(x1, x2)  =  geq_2_in_gg2(x1, x2)
if_geq_2_in_1_gg3(x1, x2, x3)  =  if_geq_2_in_1_gg3(x1, x2, x3)
geq_2_out_gg2(x1, x2)  =  geq_2_out_gg2(x1, x2)
rem_3_out_gga3(x1, x2, x3)  =  rem_3_out_gga3(x1, x2, x3)


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

REM_3_IN_GGA3(X, Y, R) -> IF_REM_3_IN_1_GGA4(X, Y, R, notZero_1_in_g1(Y))
REM_3_IN_GGA3(X, Y, R) -> NOTZERO_1_IN_G1(Y)
IF_REM_3_IN_1_GGA4(X, Y, R, notZero_1_out_g1(Y)) -> IF_REM_3_IN_2_GGA4(X, Y, R, sub_3_in_gga3(X, Y, Z))
IF_REM_3_IN_1_GGA4(X, Y, R, notZero_1_out_g1(Y)) -> SUB_3_IN_GGA3(X, Y, Z)
SUB_3_IN_GGA3(s_11(X), s_11(Y), Z) -> IF_SUB_3_IN_1_GGA4(X, Y, Z, sub_3_in_gga3(X, Y, Z))
SUB_3_IN_GGA3(s_11(X), s_11(Y), Z) -> SUB_3_IN_GGA3(X, Y, Z)
IF_REM_3_IN_2_GGA4(X, Y, R, sub_3_out_gga3(X, Y, Z)) -> IF_REM_3_IN_3_GGA5(X, Y, R, Z, rem_3_in_gga3(Z, Y, R))
IF_REM_3_IN_2_GGA4(X, Y, R, sub_3_out_gga3(X, Y, Z)) -> REM_3_IN_GGA3(Z, Y, R)
REM_3_IN_GGA3(X, Y, X) -> IF_REM_3_IN_4_GGA3(X, Y, notZero_1_in_g1(Y))
REM_3_IN_GGA3(X, Y, X) -> NOTZERO_1_IN_G1(Y)
IF_REM_3_IN_4_GGA3(X, Y, notZero_1_out_g1(Y)) -> IF_REM_3_IN_5_GGA3(X, Y, geq_2_in_gg2(X, Y))
IF_REM_3_IN_4_GGA3(X, Y, notZero_1_out_g1(Y)) -> GEQ_2_IN_GG2(X, Y)
GEQ_2_IN_GG2(s_11(X), s_11(Y)) -> IF_GEQ_2_IN_1_GG3(X, Y, geq_2_in_gg2(X, Y))
GEQ_2_IN_GG2(s_11(X), s_11(Y)) -> GEQ_2_IN_GG2(X, Y)

The TRS R consists of the following rules:

rem_3_in_gga3(X, Y, R) -> if_rem_3_in_1_gga4(X, Y, R, notZero_1_in_g1(Y))
notZero_1_in_g1(s_11(X)) -> notZero_1_out_g1(s_11(X))
if_rem_3_in_1_gga4(X, Y, R, notZero_1_out_g1(Y)) -> if_rem_3_in_2_gga4(X, Y, R, sub_3_in_gga3(X, Y, Z))
sub_3_in_gga3(s_11(X), s_11(Y), Z) -> if_sub_3_in_1_gga4(X, Y, Z, sub_3_in_gga3(X, Y, Z))
sub_3_in_gga3(X, 0_0, X) -> sub_3_out_gga3(X, 0_0, X)
if_sub_3_in_1_gga4(X, Y, Z, sub_3_out_gga3(X, Y, Z)) -> sub_3_out_gga3(s_11(X), s_11(Y), Z)
if_rem_3_in_2_gga4(X, Y, R, sub_3_out_gga3(X, Y, Z)) -> if_rem_3_in_3_gga5(X, Y, R, Z, rem_3_in_gga3(Z, Y, R))
rem_3_in_gga3(X, Y, X) -> if_rem_3_in_4_gga3(X, Y, notZero_1_in_g1(Y))
if_rem_3_in_4_gga3(X, Y, notZero_1_out_g1(Y)) -> if_rem_3_in_5_gga3(X, Y, geq_2_in_gg2(X, Y))
geq_2_in_gg2(s_11(X), s_11(Y)) -> if_geq_2_in_1_gg3(X, Y, geq_2_in_gg2(X, Y))
geq_2_in_gg2(X, 0_0) -> geq_2_out_gg2(X, 0_0)
if_geq_2_in_1_gg3(X, Y, geq_2_out_gg2(X, Y)) -> geq_2_out_gg2(s_11(X), s_11(Y))
if_rem_3_in_5_gga3(X, Y, geq_2_out_gg2(X, Y)) -> rem_3_out_gga3(X, Y, X)
if_rem_3_in_3_gga5(X, Y, R, Z, rem_3_out_gga3(Z, Y, R)) -> rem_3_out_gga3(X, Y, R)

The argument filtering Pi contains the following mapping:
rem_3_in_gga3(x1, x2, x3)  =  rem_3_in_gga2(x1, x2)
s_11(x1)  =  s_11(x1)
0_0  =  0_0
if_rem_3_in_1_gga4(x1, x2, x3, x4)  =  if_rem_3_in_1_gga3(x1, x2, x4)
notZero_1_in_g1(x1)  =  notZero_1_in_g1(x1)
notZero_1_out_g1(x1)  =  notZero_1_out_g1(x1)
if_rem_3_in_2_gga4(x1, x2, x3, x4)  =  if_rem_3_in_2_gga3(x1, x2, x4)
sub_3_in_gga3(x1, x2, x3)  =  sub_3_in_gga2(x1, x2)
if_sub_3_in_1_gga4(x1, x2, x3, x4)  =  if_sub_3_in_1_gga3(x1, x2, x4)
sub_3_out_gga3(x1, x2, x3)  =  sub_3_out_gga3(x1, x2, x3)
if_rem_3_in_3_gga5(x1, x2, x3, x4, x5)  =  if_rem_3_in_3_gga3(x1, x2, x5)
if_rem_3_in_4_gga3(x1, x2, x3)  =  if_rem_3_in_4_gga3(x1, x2, x3)
if_rem_3_in_5_gga3(x1, x2, x3)  =  if_rem_3_in_5_gga3(x1, x2, x3)
geq_2_in_gg2(x1, x2)  =  geq_2_in_gg2(x1, x2)
if_geq_2_in_1_gg3(x1, x2, x3)  =  if_geq_2_in_1_gg3(x1, x2, x3)
geq_2_out_gg2(x1, x2)  =  geq_2_out_gg2(x1, x2)
rem_3_out_gga3(x1, x2, x3)  =  rem_3_out_gga3(x1, x2, x3)
IF_REM_3_IN_5_GGA3(x1, x2, x3)  =  IF_REM_3_IN_5_GGA3(x1, x2, x3)
REM_3_IN_GGA3(x1, x2, x3)  =  REM_3_IN_GGA2(x1, x2)
GEQ_2_IN_GG2(x1, x2)  =  GEQ_2_IN_GG2(x1, x2)
IF_SUB_3_IN_1_GGA4(x1, x2, x3, x4)  =  IF_SUB_3_IN_1_GGA3(x1, x2, x4)
IF_REM_3_IN_2_GGA4(x1, x2, x3, x4)  =  IF_REM_3_IN_2_GGA3(x1, x2, x4)
NOTZERO_1_IN_G1(x1)  =  NOTZERO_1_IN_G1(x1)
IF_REM_3_IN_4_GGA3(x1, x2, x3)  =  IF_REM_3_IN_4_GGA3(x1, x2, x3)
IF_REM_3_IN_3_GGA5(x1, x2, x3, x4, x5)  =  IF_REM_3_IN_3_GGA3(x1, x2, x5)
IF_REM_3_IN_1_GGA4(x1, x2, x3, x4)  =  IF_REM_3_IN_1_GGA3(x1, x2, x4)
IF_GEQ_2_IN_1_GG3(x1, x2, x3)  =  IF_GEQ_2_IN_1_GG3(x1, x2, x3)
SUB_3_IN_GGA3(x1, x2, x3)  =  SUB_3_IN_GGA2(x1, x2)

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:

REM_3_IN_GGA3(X, Y, R) -> IF_REM_3_IN_1_GGA4(X, Y, R, notZero_1_in_g1(Y))
REM_3_IN_GGA3(X, Y, R) -> NOTZERO_1_IN_G1(Y)
IF_REM_3_IN_1_GGA4(X, Y, R, notZero_1_out_g1(Y)) -> IF_REM_3_IN_2_GGA4(X, Y, R, sub_3_in_gga3(X, Y, Z))
IF_REM_3_IN_1_GGA4(X, Y, R, notZero_1_out_g1(Y)) -> SUB_3_IN_GGA3(X, Y, Z)
SUB_3_IN_GGA3(s_11(X), s_11(Y), Z) -> IF_SUB_3_IN_1_GGA4(X, Y, Z, sub_3_in_gga3(X, Y, Z))
SUB_3_IN_GGA3(s_11(X), s_11(Y), Z) -> SUB_3_IN_GGA3(X, Y, Z)
IF_REM_3_IN_2_GGA4(X, Y, R, sub_3_out_gga3(X, Y, Z)) -> IF_REM_3_IN_3_GGA5(X, Y, R, Z, rem_3_in_gga3(Z, Y, R))
IF_REM_3_IN_2_GGA4(X, Y, R, sub_3_out_gga3(X, Y, Z)) -> REM_3_IN_GGA3(Z, Y, R)
REM_3_IN_GGA3(X, Y, X) -> IF_REM_3_IN_4_GGA3(X, Y, notZero_1_in_g1(Y))
REM_3_IN_GGA3(X, Y, X) -> NOTZERO_1_IN_G1(Y)
IF_REM_3_IN_4_GGA3(X, Y, notZero_1_out_g1(Y)) -> IF_REM_3_IN_5_GGA3(X, Y, geq_2_in_gg2(X, Y))
IF_REM_3_IN_4_GGA3(X, Y, notZero_1_out_g1(Y)) -> GEQ_2_IN_GG2(X, Y)
GEQ_2_IN_GG2(s_11(X), s_11(Y)) -> IF_GEQ_2_IN_1_GG3(X, Y, geq_2_in_gg2(X, Y))
GEQ_2_IN_GG2(s_11(X), s_11(Y)) -> GEQ_2_IN_GG2(X, Y)

The TRS R consists of the following rules:

rem_3_in_gga3(X, Y, R) -> if_rem_3_in_1_gga4(X, Y, R, notZero_1_in_g1(Y))
notZero_1_in_g1(s_11(X)) -> notZero_1_out_g1(s_11(X))
if_rem_3_in_1_gga4(X, Y, R, notZero_1_out_g1(Y)) -> if_rem_3_in_2_gga4(X, Y, R, sub_3_in_gga3(X, Y, Z))
sub_3_in_gga3(s_11(X), s_11(Y), Z) -> if_sub_3_in_1_gga4(X, Y, Z, sub_3_in_gga3(X, Y, Z))
sub_3_in_gga3(X, 0_0, X) -> sub_3_out_gga3(X, 0_0, X)
if_sub_3_in_1_gga4(X, Y, Z, sub_3_out_gga3(X, Y, Z)) -> sub_3_out_gga3(s_11(X), s_11(Y), Z)
if_rem_3_in_2_gga4(X, Y, R, sub_3_out_gga3(X, Y, Z)) -> if_rem_3_in_3_gga5(X, Y, R, Z, rem_3_in_gga3(Z, Y, R))
rem_3_in_gga3(X, Y, X) -> if_rem_3_in_4_gga3(X, Y, notZero_1_in_g1(Y))
if_rem_3_in_4_gga3(X, Y, notZero_1_out_g1(Y)) -> if_rem_3_in_5_gga3(X, Y, geq_2_in_gg2(X, Y))
geq_2_in_gg2(s_11(X), s_11(Y)) -> if_geq_2_in_1_gg3(X, Y, geq_2_in_gg2(X, Y))
geq_2_in_gg2(X, 0_0) -> geq_2_out_gg2(X, 0_0)
if_geq_2_in_1_gg3(X, Y, geq_2_out_gg2(X, Y)) -> geq_2_out_gg2(s_11(X), s_11(Y))
if_rem_3_in_5_gga3(X, Y, geq_2_out_gg2(X, Y)) -> rem_3_out_gga3(X, Y, X)
if_rem_3_in_3_gga5(X, Y, R, Z, rem_3_out_gga3(Z, Y, R)) -> rem_3_out_gga3(X, Y, R)

The argument filtering Pi contains the following mapping:
rem_3_in_gga3(x1, x2, x3)  =  rem_3_in_gga2(x1, x2)
s_11(x1)  =  s_11(x1)
0_0  =  0_0
if_rem_3_in_1_gga4(x1, x2, x3, x4)  =  if_rem_3_in_1_gga3(x1, x2, x4)
notZero_1_in_g1(x1)  =  notZero_1_in_g1(x1)
notZero_1_out_g1(x1)  =  notZero_1_out_g1(x1)
if_rem_3_in_2_gga4(x1, x2, x3, x4)  =  if_rem_3_in_2_gga3(x1, x2, x4)
sub_3_in_gga3(x1, x2, x3)  =  sub_3_in_gga2(x1, x2)
if_sub_3_in_1_gga4(x1, x2, x3, x4)  =  if_sub_3_in_1_gga3(x1, x2, x4)
sub_3_out_gga3(x1, x2, x3)  =  sub_3_out_gga3(x1, x2, x3)
if_rem_3_in_3_gga5(x1, x2, x3, x4, x5)  =  if_rem_3_in_3_gga3(x1, x2, x5)
if_rem_3_in_4_gga3(x1, x2, x3)  =  if_rem_3_in_4_gga3(x1, x2, x3)
if_rem_3_in_5_gga3(x1, x2, x3)  =  if_rem_3_in_5_gga3(x1, x2, x3)
geq_2_in_gg2(x1, x2)  =  geq_2_in_gg2(x1, x2)
if_geq_2_in_1_gg3(x1, x2, x3)  =  if_geq_2_in_1_gg3(x1, x2, x3)
geq_2_out_gg2(x1, x2)  =  geq_2_out_gg2(x1, x2)
rem_3_out_gga3(x1, x2, x3)  =  rem_3_out_gga3(x1, x2, x3)
IF_REM_3_IN_5_GGA3(x1, x2, x3)  =  IF_REM_3_IN_5_GGA3(x1, x2, x3)
REM_3_IN_GGA3(x1, x2, x3)  =  REM_3_IN_GGA2(x1, x2)
GEQ_2_IN_GG2(x1, x2)  =  GEQ_2_IN_GG2(x1, x2)
IF_SUB_3_IN_1_GGA4(x1, x2, x3, x4)  =  IF_SUB_3_IN_1_GGA3(x1, x2, x4)
IF_REM_3_IN_2_GGA4(x1, x2, x3, x4)  =  IF_REM_3_IN_2_GGA3(x1, x2, x4)
NOTZERO_1_IN_G1(x1)  =  NOTZERO_1_IN_G1(x1)
IF_REM_3_IN_4_GGA3(x1, x2, x3)  =  IF_REM_3_IN_4_GGA3(x1, x2, x3)
IF_REM_3_IN_3_GGA5(x1, x2, x3, x4, x5)  =  IF_REM_3_IN_3_GGA3(x1, x2, x5)
IF_REM_3_IN_1_GGA4(x1, x2, x3, x4)  =  IF_REM_3_IN_1_GGA3(x1, x2, x4)
IF_GEQ_2_IN_1_GG3(x1, x2, x3)  =  IF_GEQ_2_IN_1_GG3(x1, x2, x3)
SUB_3_IN_GGA3(x1, x2, x3)  =  SUB_3_IN_GGA2(x1, x2)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph contains 3 SCCs with 9 less nodes.

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

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

GEQ_2_IN_GG2(s_11(X), s_11(Y)) -> GEQ_2_IN_GG2(X, Y)

The TRS R consists of the following rules:

rem_3_in_gga3(X, Y, R) -> if_rem_3_in_1_gga4(X, Y, R, notZero_1_in_g1(Y))
notZero_1_in_g1(s_11(X)) -> notZero_1_out_g1(s_11(X))
if_rem_3_in_1_gga4(X, Y, R, notZero_1_out_g1(Y)) -> if_rem_3_in_2_gga4(X, Y, R, sub_3_in_gga3(X, Y, Z))
sub_3_in_gga3(s_11(X), s_11(Y), Z) -> if_sub_3_in_1_gga4(X, Y, Z, sub_3_in_gga3(X, Y, Z))
sub_3_in_gga3(X, 0_0, X) -> sub_3_out_gga3(X, 0_0, X)
if_sub_3_in_1_gga4(X, Y, Z, sub_3_out_gga3(X, Y, Z)) -> sub_3_out_gga3(s_11(X), s_11(Y), Z)
if_rem_3_in_2_gga4(X, Y, R, sub_3_out_gga3(X, Y, Z)) -> if_rem_3_in_3_gga5(X, Y, R, Z, rem_3_in_gga3(Z, Y, R))
rem_3_in_gga3(X, Y, X) -> if_rem_3_in_4_gga3(X, Y, notZero_1_in_g1(Y))
if_rem_3_in_4_gga3(X, Y, notZero_1_out_g1(Y)) -> if_rem_3_in_5_gga3(X, Y, geq_2_in_gg2(X, Y))
geq_2_in_gg2(s_11(X), s_11(Y)) -> if_geq_2_in_1_gg3(X, Y, geq_2_in_gg2(X, Y))
geq_2_in_gg2(X, 0_0) -> geq_2_out_gg2(X, 0_0)
if_geq_2_in_1_gg3(X, Y, geq_2_out_gg2(X, Y)) -> geq_2_out_gg2(s_11(X), s_11(Y))
if_rem_3_in_5_gga3(X, Y, geq_2_out_gg2(X, Y)) -> rem_3_out_gga3(X, Y, X)
if_rem_3_in_3_gga5(X, Y, R, Z, rem_3_out_gga3(Z, Y, R)) -> rem_3_out_gga3(X, Y, R)

The argument filtering Pi contains the following mapping:
rem_3_in_gga3(x1, x2, x3)  =  rem_3_in_gga2(x1, x2)
s_11(x1)  =  s_11(x1)
0_0  =  0_0
if_rem_3_in_1_gga4(x1, x2, x3, x4)  =  if_rem_3_in_1_gga3(x1, x2, x4)
notZero_1_in_g1(x1)  =  notZero_1_in_g1(x1)
notZero_1_out_g1(x1)  =  notZero_1_out_g1(x1)
if_rem_3_in_2_gga4(x1, x2, x3, x4)  =  if_rem_3_in_2_gga3(x1, x2, x4)
sub_3_in_gga3(x1, x2, x3)  =  sub_3_in_gga2(x1, x2)
if_sub_3_in_1_gga4(x1, x2, x3, x4)  =  if_sub_3_in_1_gga3(x1, x2, x4)
sub_3_out_gga3(x1, x2, x3)  =  sub_3_out_gga3(x1, x2, x3)
if_rem_3_in_3_gga5(x1, x2, x3, x4, x5)  =  if_rem_3_in_3_gga3(x1, x2, x5)
if_rem_3_in_4_gga3(x1, x2, x3)  =  if_rem_3_in_4_gga3(x1, x2, x3)
if_rem_3_in_5_gga3(x1, x2, x3)  =  if_rem_3_in_5_gga3(x1, x2, x3)
geq_2_in_gg2(x1, x2)  =  geq_2_in_gg2(x1, x2)
if_geq_2_in_1_gg3(x1, x2, x3)  =  if_geq_2_in_1_gg3(x1, x2, x3)
geq_2_out_gg2(x1, x2)  =  geq_2_out_gg2(x1, x2)
rem_3_out_gga3(x1, x2, x3)  =  rem_3_out_gga3(x1, x2, x3)
GEQ_2_IN_GG2(x1, x2)  =  GEQ_2_IN_GG2(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
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
              ↳ PiDP
              ↳ PiDP

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

GEQ_2_IN_GG2(s_11(X), s_11(Y)) -> GEQ_2_IN_GG2(X, Y)

R is empty.
Pi is empty.
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
            ↳ AND
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ QDPSizeChangeProof
              ↳ PiDP
              ↳ PiDP

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

GEQ_2_IN_GG2(s_11(X), s_11(Y)) -> GEQ_2_IN_GG2(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 {GEQ_2_IN_GG2}.
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
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
              ↳ PiDP

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

SUB_3_IN_GGA3(s_11(X), s_11(Y), Z) -> SUB_3_IN_GGA3(X, Y, Z)

The TRS R consists of the following rules:

rem_3_in_gga3(X, Y, R) -> if_rem_3_in_1_gga4(X, Y, R, notZero_1_in_g1(Y))
notZero_1_in_g1(s_11(X)) -> notZero_1_out_g1(s_11(X))
if_rem_3_in_1_gga4(X, Y, R, notZero_1_out_g1(Y)) -> if_rem_3_in_2_gga4(X, Y, R, sub_3_in_gga3(X, Y, Z))
sub_3_in_gga3(s_11(X), s_11(Y), Z) -> if_sub_3_in_1_gga4(X, Y, Z, sub_3_in_gga3(X, Y, Z))
sub_3_in_gga3(X, 0_0, X) -> sub_3_out_gga3(X, 0_0, X)
if_sub_3_in_1_gga4(X, Y, Z, sub_3_out_gga3(X, Y, Z)) -> sub_3_out_gga3(s_11(X), s_11(Y), Z)
if_rem_3_in_2_gga4(X, Y, R, sub_3_out_gga3(X, Y, Z)) -> if_rem_3_in_3_gga5(X, Y, R, Z, rem_3_in_gga3(Z, Y, R))
rem_3_in_gga3(X, Y, X) -> if_rem_3_in_4_gga3(X, Y, notZero_1_in_g1(Y))
if_rem_3_in_4_gga3(X, Y, notZero_1_out_g1(Y)) -> if_rem_3_in_5_gga3(X, Y, geq_2_in_gg2(X, Y))
geq_2_in_gg2(s_11(X), s_11(Y)) -> if_geq_2_in_1_gg3(X, Y, geq_2_in_gg2(X, Y))
geq_2_in_gg2(X, 0_0) -> geq_2_out_gg2(X, 0_0)
if_geq_2_in_1_gg3(X, Y, geq_2_out_gg2(X, Y)) -> geq_2_out_gg2(s_11(X), s_11(Y))
if_rem_3_in_5_gga3(X, Y, geq_2_out_gg2(X, Y)) -> rem_3_out_gga3(X, Y, X)
if_rem_3_in_3_gga5(X, Y, R, Z, rem_3_out_gga3(Z, Y, R)) -> rem_3_out_gga3(X, Y, R)

The argument filtering Pi contains the following mapping:
rem_3_in_gga3(x1, x2, x3)  =  rem_3_in_gga2(x1, x2)
s_11(x1)  =  s_11(x1)
0_0  =  0_0
if_rem_3_in_1_gga4(x1, x2, x3, x4)  =  if_rem_3_in_1_gga3(x1, x2, x4)
notZero_1_in_g1(x1)  =  notZero_1_in_g1(x1)
notZero_1_out_g1(x1)  =  notZero_1_out_g1(x1)
if_rem_3_in_2_gga4(x1, x2, x3, x4)  =  if_rem_3_in_2_gga3(x1, x2, x4)
sub_3_in_gga3(x1, x2, x3)  =  sub_3_in_gga2(x1, x2)
if_sub_3_in_1_gga4(x1, x2, x3, x4)  =  if_sub_3_in_1_gga3(x1, x2, x4)
sub_3_out_gga3(x1, x2, x3)  =  sub_3_out_gga3(x1, x2, x3)
if_rem_3_in_3_gga5(x1, x2, x3, x4, x5)  =  if_rem_3_in_3_gga3(x1, x2, x5)
if_rem_3_in_4_gga3(x1, x2, x3)  =  if_rem_3_in_4_gga3(x1, x2, x3)
if_rem_3_in_5_gga3(x1, x2, x3)  =  if_rem_3_in_5_gga3(x1, x2, x3)
geq_2_in_gg2(x1, x2)  =  geq_2_in_gg2(x1, x2)
if_geq_2_in_1_gg3(x1, x2, x3)  =  if_geq_2_in_1_gg3(x1, x2, x3)
geq_2_out_gg2(x1, x2)  =  geq_2_out_gg2(x1, x2)
rem_3_out_gga3(x1, x2, x3)  =  rem_3_out_gga3(x1, x2, x3)
SUB_3_IN_GGA3(x1, x2, x3)  =  SUB_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
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
              ↳ PiDP

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

SUB_3_IN_GGA3(s_11(X), s_11(Y), Z) -> SUB_3_IN_GGA3(X, Y, Z)

R is empty.
The argument filtering Pi contains the following mapping:
s_11(x1)  =  s_11(x1)
SUB_3_IN_GGA3(x1, x2, x3)  =  SUB_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
  ↳ 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:

SUB_3_IN_GGA2(s_11(X), s_11(Y)) -> SUB_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 {SUB_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
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof

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

REM_3_IN_GGA3(X, Y, R) -> IF_REM_3_IN_1_GGA4(X, Y, R, notZero_1_in_g1(Y))
IF_REM_3_IN_2_GGA4(X, Y, R, sub_3_out_gga3(X, Y, Z)) -> REM_3_IN_GGA3(Z, Y, R)
IF_REM_3_IN_1_GGA4(X, Y, R, notZero_1_out_g1(Y)) -> IF_REM_3_IN_2_GGA4(X, Y, R, sub_3_in_gga3(X, Y, Z))

The TRS R consists of the following rules:

rem_3_in_gga3(X, Y, R) -> if_rem_3_in_1_gga4(X, Y, R, notZero_1_in_g1(Y))
notZero_1_in_g1(s_11(X)) -> notZero_1_out_g1(s_11(X))
if_rem_3_in_1_gga4(X, Y, R, notZero_1_out_g1(Y)) -> if_rem_3_in_2_gga4(X, Y, R, sub_3_in_gga3(X, Y, Z))
sub_3_in_gga3(s_11(X), s_11(Y), Z) -> if_sub_3_in_1_gga4(X, Y, Z, sub_3_in_gga3(X, Y, Z))
sub_3_in_gga3(X, 0_0, X) -> sub_3_out_gga3(X, 0_0, X)
if_sub_3_in_1_gga4(X, Y, Z, sub_3_out_gga3(X, Y, Z)) -> sub_3_out_gga3(s_11(X), s_11(Y), Z)
if_rem_3_in_2_gga4(X, Y, R, sub_3_out_gga3(X, Y, Z)) -> if_rem_3_in_3_gga5(X, Y, R, Z, rem_3_in_gga3(Z, Y, R))
rem_3_in_gga3(X, Y, X) -> if_rem_3_in_4_gga3(X, Y, notZero_1_in_g1(Y))
if_rem_3_in_4_gga3(X, Y, notZero_1_out_g1(Y)) -> if_rem_3_in_5_gga3(X, Y, geq_2_in_gg2(X, Y))
geq_2_in_gg2(s_11(X), s_11(Y)) -> if_geq_2_in_1_gg3(X, Y, geq_2_in_gg2(X, Y))
geq_2_in_gg2(X, 0_0) -> geq_2_out_gg2(X, 0_0)
if_geq_2_in_1_gg3(X, Y, geq_2_out_gg2(X, Y)) -> geq_2_out_gg2(s_11(X), s_11(Y))
if_rem_3_in_5_gga3(X, Y, geq_2_out_gg2(X, Y)) -> rem_3_out_gga3(X, Y, X)
if_rem_3_in_3_gga5(X, Y, R, Z, rem_3_out_gga3(Z, Y, R)) -> rem_3_out_gga3(X, Y, R)

The argument filtering Pi contains the following mapping:
rem_3_in_gga3(x1, x2, x3)  =  rem_3_in_gga2(x1, x2)
s_11(x1)  =  s_11(x1)
0_0  =  0_0
if_rem_3_in_1_gga4(x1, x2, x3, x4)  =  if_rem_3_in_1_gga3(x1, x2, x4)
notZero_1_in_g1(x1)  =  notZero_1_in_g1(x1)
notZero_1_out_g1(x1)  =  notZero_1_out_g1(x1)
if_rem_3_in_2_gga4(x1, x2, x3, x4)  =  if_rem_3_in_2_gga3(x1, x2, x4)
sub_3_in_gga3(x1, x2, x3)  =  sub_3_in_gga2(x1, x2)
if_sub_3_in_1_gga4(x1, x2, x3, x4)  =  if_sub_3_in_1_gga3(x1, x2, x4)
sub_3_out_gga3(x1, x2, x3)  =  sub_3_out_gga3(x1, x2, x3)
if_rem_3_in_3_gga5(x1, x2, x3, x4, x5)  =  if_rem_3_in_3_gga3(x1, x2, x5)
if_rem_3_in_4_gga3(x1, x2, x3)  =  if_rem_3_in_4_gga3(x1, x2, x3)
if_rem_3_in_5_gga3(x1, x2, x3)  =  if_rem_3_in_5_gga3(x1, x2, x3)
geq_2_in_gg2(x1, x2)  =  geq_2_in_gg2(x1, x2)
if_geq_2_in_1_gg3(x1, x2, x3)  =  if_geq_2_in_1_gg3(x1, x2, x3)
geq_2_out_gg2(x1, x2)  =  geq_2_out_gg2(x1, x2)
rem_3_out_gga3(x1, x2, x3)  =  rem_3_out_gga3(x1, x2, x3)
REM_3_IN_GGA3(x1, x2, x3)  =  REM_3_IN_GGA2(x1, x2)
IF_REM_3_IN_2_GGA4(x1, x2, x3, x4)  =  IF_REM_3_IN_2_GGA3(x1, x2, x4)
IF_REM_3_IN_1_GGA4(x1, x2, x3, x4)  =  IF_REM_3_IN_1_GGA3(x1, x2, x4)

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

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

REM_3_IN_GGA3(X, Y, R) -> IF_REM_3_IN_1_GGA4(X, Y, R, notZero_1_in_g1(Y))
IF_REM_3_IN_2_GGA4(X, Y, R, sub_3_out_gga3(X, Y, Z)) -> REM_3_IN_GGA3(Z, Y, R)
IF_REM_3_IN_1_GGA4(X, Y, R, notZero_1_out_g1(Y)) -> IF_REM_3_IN_2_GGA4(X, Y, R, sub_3_in_gga3(X, Y, Z))

The TRS R consists of the following rules:

notZero_1_in_g1(s_11(X)) -> notZero_1_out_g1(s_11(X))
sub_3_in_gga3(s_11(X), s_11(Y), Z) -> if_sub_3_in_1_gga4(X, Y, Z, sub_3_in_gga3(X, Y, Z))
sub_3_in_gga3(X, 0_0, X) -> sub_3_out_gga3(X, 0_0, X)
if_sub_3_in_1_gga4(X, Y, Z, sub_3_out_gga3(X, Y, Z)) -> sub_3_out_gga3(s_11(X), s_11(Y), Z)

The argument filtering Pi contains the following mapping:
s_11(x1)  =  s_11(x1)
0_0  =  0_0
notZero_1_in_g1(x1)  =  notZero_1_in_g1(x1)
notZero_1_out_g1(x1)  =  notZero_1_out_g1(x1)
sub_3_in_gga3(x1, x2, x3)  =  sub_3_in_gga2(x1, x2)
if_sub_3_in_1_gga4(x1, x2, x3, x4)  =  if_sub_3_in_1_gga3(x1, x2, x4)
sub_3_out_gga3(x1, x2, x3)  =  sub_3_out_gga3(x1, x2, x3)
REM_3_IN_GGA3(x1, x2, x3)  =  REM_3_IN_GGA2(x1, x2)
IF_REM_3_IN_2_GGA4(x1, x2, x3, x4)  =  IF_REM_3_IN_2_GGA3(x1, x2, x4)
IF_REM_3_IN_1_GGA4(x1, x2, x3, x4)  =  IF_REM_3_IN_1_GGA3(x1, x2, x4)

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

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

REM_3_IN_GGA2(X, Y) -> IF_REM_3_IN_1_GGA3(X, Y, notZero_1_in_g1(Y))
IF_REM_3_IN_2_GGA3(X, Y, sub_3_out_gga3(X, Y, Z)) -> REM_3_IN_GGA2(Z, Y)
IF_REM_3_IN_1_GGA3(X, Y, notZero_1_out_g1(Y)) -> IF_REM_3_IN_2_GGA3(X, Y, sub_3_in_gga2(X, Y))

The TRS R consists of the following rules:

notZero_1_in_g1(s_11(X)) -> notZero_1_out_g1(s_11(X))
sub_3_in_gga2(s_11(X), s_11(Y)) -> if_sub_3_in_1_gga3(X, Y, sub_3_in_gga2(X, Y))
sub_3_in_gga2(X, 0_0) -> sub_3_out_gga3(X, 0_0, X)
if_sub_3_in_1_gga3(X, Y, sub_3_out_gga3(X, Y, Z)) -> sub_3_out_gga3(s_11(X), s_11(Y), Z)

The set Q consists of the following terms:

notZero_1_in_g1(x0)
sub_3_in_gga2(x0, x1)
if_sub_3_in_1_gga3(x0, x1, x2)

We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {IF_REM_3_IN_1_GGA3, REM_3_IN_GGA2, IF_REM_3_IN_2_GGA3}.
By using a polynomial ordering, the following set of Dependency Pairs of this DP problem can be strictly oriented.

IF_REM_3_IN_1_GGA3(X, Y, notZero_1_out_g1(Y)) -> IF_REM_3_IN_2_GGA3(X, Y, sub_3_in_gga2(X, Y))
The remaining Dependency Pairs were at least non-strictly be oriented.

REM_3_IN_GGA2(X, Y) -> IF_REM_3_IN_1_GGA3(X, Y, notZero_1_in_g1(Y))
IF_REM_3_IN_2_GGA3(X, Y, sub_3_out_gga3(X, Y, Z)) -> REM_3_IN_GGA2(Z, Y)
With the implicit AFS we had to orient the following set of usable rules non-strictly.

sub_3_in_gga2(X, 0_0) -> sub_3_out_gga3(X, 0_0, X)
notZero_1_in_g1(s_11(X)) -> notZero_1_out_g1(s_11(X))
if_sub_3_in_1_gga3(X, Y, sub_3_out_gga3(X, Y, Z)) -> sub_3_out_gga3(s_11(X), s_11(Y), Z)
sub_3_in_gga2(s_11(X), s_11(Y)) -> if_sub_3_in_1_gga3(X, Y, sub_3_in_gga2(X, Y))
Used ordering: POLO with Polynomial interpretation:

POL(0_0) = 0   
POL(IF_REM_3_IN_1_GGA3(x1, x2, x3)) = x1 + x3   
POL(sub_3_out_gga3(x1, x2, x3)) = x2 + x3   
POL(REM_3_IN_GGA2(x1, x2)) = x1 + x2   
POL(IF_REM_3_IN_2_GGA3(x1, x2, x3)) = x3   
POL(if_sub_3_in_1_gga3(x1, x2, x3)) = 1 + x3   
POL(sub_3_in_gga2(x1, x2)) = x1   
POL(notZero_1_in_g1(x1)) = x1   
POL(s_11(x1)) = 1 + x1   
POL(notZero_1_out_g1(x1)) = 1   



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

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

REM_3_IN_GGA2(X, Y) -> IF_REM_3_IN_1_GGA3(X, Y, notZero_1_in_g1(Y))
IF_REM_3_IN_2_GGA3(X, Y, sub_3_out_gga3(X, Y, Z)) -> REM_3_IN_GGA2(Z, Y)

The TRS R consists of the following rules:

notZero_1_in_g1(s_11(X)) -> notZero_1_out_g1(s_11(X))
sub_3_in_gga2(s_11(X), s_11(Y)) -> if_sub_3_in_1_gga3(X, Y, sub_3_in_gga2(X, Y))
sub_3_in_gga2(X, 0_0) -> sub_3_out_gga3(X, 0_0, X)
if_sub_3_in_1_gga3(X, Y, sub_3_out_gga3(X, Y, Z)) -> sub_3_out_gga3(s_11(X), s_11(Y), Z)

The set Q consists of the following terms:

notZero_1_in_g1(x0)
sub_3_in_gga2(x0, x1)
if_sub_3_in_1_gga3(x0, x1, x2)

We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {IF_REM_3_IN_1_GGA3, REM_3_IN_GGA2, IF_REM_3_IN_2_GGA3}.
The approximation of the Dependency Graph contains 0 SCCs with 2 less nodes.