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



↳ PROLOG
  ↳ PrologToPiTRSProof

append3(nil0, XS, XS).
append3(cons2(X, XS1), XS2, cons2(X, YS)) :- append3(XS1, XS2, YS).
split3(XS, nil0, XS).
split3(cons2(X, XS), cons2(X, YS1), YS2) :- split3(XS, YS1, YS2).
perm2(nil0, nil0).
perm2(XS, cons2(Y, YS)) :- split3(XS, YS1, cons2(Y, YS2)), append3(YS1, YS2, ZS), perm2(ZS, YS).


With regard to the inferred argument filtering the predicates were used in the following modes:
perm2: (b,f)
split3: (b,f,f)
append3: (b,b,f)
Transforming PROLOG into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:


perm_2_in_ga2(nil_0, nil_0) -> perm_2_out_ga2(nil_0, nil_0)
perm_2_in_ga2(XS, cons_22(Y, YS)) -> if_perm_2_in_1_ga4(XS, Y, YS, split_3_in_gaa3(XS, YS1, cons_22(Y, YS2)))
split_3_in_gaa3(XS, nil_0, XS) -> split_3_out_gaa3(XS, nil_0, XS)
split_3_in_gaa3(cons_22(X, XS), cons_22(X, YS1), YS2) -> if_split_3_in_1_gaa5(X, XS, YS1, YS2, split_3_in_gaa3(XS, YS1, YS2))
if_split_3_in_1_gaa5(X, XS, YS1, YS2, split_3_out_gaa3(XS, YS1, YS2)) -> split_3_out_gaa3(cons_22(X, XS), cons_22(X, YS1), YS2)
if_perm_2_in_1_ga4(XS, Y, YS, split_3_out_gaa3(XS, YS1, cons_22(Y, YS2))) -> if_perm_2_in_2_ga6(XS, Y, YS, YS1, YS2, append_3_in_gga3(YS1, YS2, ZS))
append_3_in_gga3(nil_0, XS, XS) -> append_3_out_gga3(nil_0, XS, XS)
append_3_in_gga3(cons_22(X, XS1), XS2, cons_22(X, YS)) -> if_append_3_in_1_gga5(X, XS1, XS2, YS, append_3_in_gga3(XS1, XS2, YS))
if_append_3_in_1_gga5(X, XS1, XS2, YS, append_3_out_gga3(XS1, XS2, YS)) -> append_3_out_gga3(cons_22(X, XS1), XS2, cons_22(X, YS))
if_perm_2_in_2_ga6(XS, Y, YS, YS1, YS2, append_3_out_gga3(YS1, YS2, ZS)) -> if_perm_2_in_3_ga5(XS, Y, YS, ZS, perm_2_in_ga2(ZS, YS))
if_perm_2_in_3_ga5(XS, Y, YS, ZS, perm_2_out_ga2(ZS, YS)) -> perm_2_out_ga2(XS, cons_22(Y, YS))

The argument filtering Pi contains the following mapping:
perm_2_in_ga2(x1, x2)  =  perm_2_in_ga1(x1)
nil_0  =  nil_0
cons_22(x1, x2)  =  cons_22(x1, x2)
perm_2_out_ga2(x1, x2)  =  perm_2_out_ga1(x2)
if_perm_2_in_1_ga4(x1, x2, x3, x4)  =  if_perm_2_in_1_ga1(x4)
split_3_in_gaa3(x1, x2, x3)  =  split_3_in_gaa1(x1)
split_3_out_gaa3(x1, x2, x3)  =  split_3_out_gaa2(x2, x3)
if_split_3_in_1_gaa5(x1, x2, x3, x4, x5)  =  if_split_3_in_1_gaa2(x1, x5)
if_perm_2_in_2_ga6(x1, x2, x3, x4, x5, x6)  =  if_perm_2_in_2_ga2(x2, x6)
append_3_in_gga3(x1, x2, x3)  =  append_3_in_gga2(x1, x2)
append_3_out_gga3(x1, x2, x3)  =  append_3_out_gga1(x3)
if_append_3_in_1_gga5(x1, x2, x3, x4, x5)  =  if_append_3_in_1_gga2(x1, x5)
if_perm_2_in_3_ga5(x1, x2, x3, x4, x5)  =  if_perm_2_in_3_ga2(x2, 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:

perm_2_in_ga2(nil_0, nil_0) -> perm_2_out_ga2(nil_0, nil_0)
perm_2_in_ga2(XS, cons_22(Y, YS)) -> if_perm_2_in_1_ga4(XS, Y, YS, split_3_in_gaa3(XS, YS1, cons_22(Y, YS2)))
split_3_in_gaa3(XS, nil_0, XS) -> split_3_out_gaa3(XS, nil_0, XS)
split_3_in_gaa3(cons_22(X, XS), cons_22(X, YS1), YS2) -> if_split_3_in_1_gaa5(X, XS, YS1, YS2, split_3_in_gaa3(XS, YS1, YS2))
if_split_3_in_1_gaa5(X, XS, YS1, YS2, split_3_out_gaa3(XS, YS1, YS2)) -> split_3_out_gaa3(cons_22(X, XS), cons_22(X, YS1), YS2)
if_perm_2_in_1_ga4(XS, Y, YS, split_3_out_gaa3(XS, YS1, cons_22(Y, YS2))) -> if_perm_2_in_2_ga6(XS, Y, YS, YS1, YS2, append_3_in_gga3(YS1, YS2, ZS))
append_3_in_gga3(nil_0, XS, XS) -> append_3_out_gga3(nil_0, XS, XS)
append_3_in_gga3(cons_22(X, XS1), XS2, cons_22(X, YS)) -> if_append_3_in_1_gga5(X, XS1, XS2, YS, append_3_in_gga3(XS1, XS2, YS))
if_append_3_in_1_gga5(X, XS1, XS2, YS, append_3_out_gga3(XS1, XS2, YS)) -> append_3_out_gga3(cons_22(X, XS1), XS2, cons_22(X, YS))
if_perm_2_in_2_ga6(XS, Y, YS, YS1, YS2, append_3_out_gga3(YS1, YS2, ZS)) -> if_perm_2_in_3_ga5(XS, Y, YS, ZS, perm_2_in_ga2(ZS, YS))
if_perm_2_in_3_ga5(XS, Y, YS, ZS, perm_2_out_ga2(ZS, YS)) -> perm_2_out_ga2(XS, cons_22(Y, YS))

The argument filtering Pi contains the following mapping:
perm_2_in_ga2(x1, x2)  =  perm_2_in_ga1(x1)
nil_0  =  nil_0
cons_22(x1, x2)  =  cons_22(x1, x2)
perm_2_out_ga2(x1, x2)  =  perm_2_out_ga1(x2)
if_perm_2_in_1_ga4(x1, x2, x3, x4)  =  if_perm_2_in_1_ga1(x4)
split_3_in_gaa3(x1, x2, x3)  =  split_3_in_gaa1(x1)
split_3_out_gaa3(x1, x2, x3)  =  split_3_out_gaa2(x2, x3)
if_split_3_in_1_gaa5(x1, x2, x3, x4, x5)  =  if_split_3_in_1_gaa2(x1, x5)
if_perm_2_in_2_ga6(x1, x2, x3, x4, x5, x6)  =  if_perm_2_in_2_ga2(x2, x6)
append_3_in_gga3(x1, x2, x3)  =  append_3_in_gga2(x1, x2)
append_3_out_gga3(x1, x2, x3)  =  append_3_out_gga1(x3)
if_append_3_in_1_gga5(x1, x2, x3, x4, x5)  =  if_append_3_in_1_gga2(x1, x5)
if_perm_2_in_3_ga5(x1, x2, x3, x4, x5)  =  if_perm_2_in_3_ga2(x2, x5)


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

PERM_2_IN_GA2(XS, cons_22(Y, YS)) -> IF_PERM_2_IN_1_GA4(XS, Y, YS, split_3_in_gaa3(XS, YS1, cons_22(Y, YS2)))
PERM_2_IN_GA2(XS, cons_22(Y, YS)) -> SPLIT_3_IN_GAA3(XS, YS1, cons_22(Y, YS2))
SPLIT_3_IN_GAA3(cons_22(X, XS), cons_22(X, YS1), YS2) -> IF_SPLIT_3_IN_1_GAA5(X, XS, YS1, YS2, split_3_in_gaa3(XS, YS1, YS2))
SPLIT_3_IN_GAA3(cons_22(X, XS), cons_22(X, YS1), YS2) -> SPLIT_3_IN_GAA3(XS, YS1, YS2)
IF_PERM_2_IN_1_GA4(XS, Y, YS, split_3_out_gaa3(XS, YS1, cons_22(Y, YS2))) -> IF_PERM_2_IN_2_GA6(XS, Y, YS, YS1, YS2, append_3_in_gga3(YS1, YS2, ZS))
IF_PERM_2_IN_1_GA4(XS, Y, YS, split_3_out_gaa3(XS, YS1, cons_22(Y, YS2))) -> APPEND_3_IN_GGA3(YS1, YS2, ZS)
APPEND_3_IN_GGA3(cons_22(X, XS1), XS2, cons_22(X, YS)) -> IF_APPEND_3_IN_1_GGA5(X, XS1, XS2, YS, append_3_in_gga3(XS1, XS2, YS))
APPEND_3_IN_GGA3(cons_22(X, XS1), XS2, cons_22(X, YS)) -> APPEND_3_IN_GGA3(XS1, XS2, YS)
IF_PERM_2_IN_2_GA6(XS, Y, YS, YS1, YS2, append_3_out_gga3(YS1, YS2, ZS)) -> IF_PERM_2_IN_3_GA5(XS, Y, YS, ZS, perm_2_in_ga2(ZS, YS))
IF_PERM_2_IN_2_GA6(XS, Y, YS, YS1, YS2, append_3_out_gga3(YS1, YS2, ZS)) -> PERM_2_IN_GA2(ZS, YS)

The TRS R consists of the following rules:

perm_2_in_ga2(nil_0, nil_0) -> perm_2_out_ga2(nil_0, nil_0)
perm_2_in_ga2(XS, cons_22(Y, YS)) -> if_perm_2_in_1_ga4(XS, Y, YS, split_3_in_gaa3(XS, YS1, cons_22(Y, YS2)))
split_3_in_gaa3(XS, nil_0, XS) -> split_3_out_gaa3(XS, nil_0, XS)
split_3_in_gaa3(cons_22(X, XS), cons_22(X, YS1), YS2) -> if_split_3_in_1_gaa5(X, XS, YS1, YS2, split_3_in_gaa3(XS, YS1, YS2))
if_split_3_in_1_gaa5(X, XS, YS1, YS2, split_3_out_gaa3(XS, YS1, YS2)) -> split_3_out_gaa3(cons_22(X, XS), cons_22(X, YS1), YS2)
if_perm_2_in_1_ga4(XS, Y, YS, split_3_out_gaa3(XS, YS1, cons_22(Y, YS2))) -> if_perm_2_in_2_ga6(XS, Y, YS, YS1, YS2, append_3_in_gga3(YS1, YS2, ZS))
append_3_in_gga3(nil_0, XS, XS) -> append_3_out_gga3(nil_0, XS, XS)
append_3_in_gga3(cons_22(X, XS1), XS2, cons_22(X, YS)) -> if_append_3_in_1_gga5(X, XS1, XS2, YS, append_3_in_gga3(XS1, XS2, YS))
if_append_3_in_1_gga5(X, XS1, XS2, YS, append_3_out_gga3(XS1, XS2, YS)) -> append_3_out_gga3(cons_22(X, XS1), XS2, cons_22(X, YS))
if_perm_2_in_2_ga6(XS, Y, YS, YS1, YS2, append_3_out_gga3(YS1, YS2, ZS)) -> if_perm_2_in_3_ga5(XS, Y, YS, ZS, perm_2_in_ga2(ZS, YS))
if_perm_2_in_3_ga5(XS, Y, YS, ZS, perm_2_out_ga2(ZS, YS)) -> perm_2_out_ga2(XS, cons_22(Y, YS))

The argument filtering Pi contains the following mapping:
perm_2_in_ga2(x1, x2)  =  perm_2_in_ga1(x1)
nil_0  =  nil_0
cons_22(x1, x2)  =  cons_22(x1, x2)
perm_2_out_ga2(x1, x2)  =  perm_2_out_ga1(x2)
if_perm_2_in_1_ga4(x1, x2, x3, x4)  =  if_perm_2_in_1_ga1(x4)
split_3_in_gaa3(x1, x2, x3)  =  split_3_in_gaa1(x1)
split_3_out_gaa3(x1, x2, x3)  =  split_3_out_gaa2(x2, x3)
if_split_3_in_1_gaa5(x1, x2, x3, x4, x5)  =  if_split_3_in_1_gaa2(x1, x5)
if_perm_2_in_2_ga6(x1, x2, x3, x4, x5, x6)  =  if_perm_2_in_2_ga2(x2, x6)
append_3_in_gga3(x1, x2, x3)  =  append_3_in_gga2(x1, x2)
append_3_out_gga3(x1, x2, x3)  =  append_3_out_gga1(x3)
if_append_3_in_1_gga5(x1, x2, x3, x4, x5)  =  if_append_3_in_1_gga2(x1, x5)
if_perm_2_in_3_ga5(x1, x2, x3, x4, x5)  =  if_perm_2_in_3_ga2(x2, x5)
IF_SPLIT_3_IN_1_GAA5(x1, x2, x3, x4, x5)  =  IF_SPLIT_3_IN_1_GAA2(x1, x5)
IF_APPEND_3_IN_1_GGA5(x1, x2, x3, x4, x5)  =  IF_APPEND_3_IN_1_GGA2(x1, x5)
SPLIT_3_IN_GAA3(x1, x2, x3)  =  SPLIT_3_IN_GAA1(x1)
IF_PERM_2_IN_3_GA5(x1, x2, x3, x4, x5)  =  IF_PERM_2_IN_3_GA2(x2, x5)
PERM_2_IN_GA2(x1, x2)  =  PERM_2_IN_GA1(x1)
APPEND_3_IN_GGA3(x1, x2, x3)  =  APPEND_3_IN_GGA2(x1, x2)
IF_PERM_2_IN_2_GA6(x1, x2, x3, x4, x5, x6)  =  IF_PERM_2_IN_2_GA2(x2, x6)
IF_PERM_2_IN_1_GA4(x1, x2, x3, x4)  =  IF_PERM_2_IN_1_GA1(x4)

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:

PERM_2_IN_GA2(XS, cons_22(Y, YS)) -> IF_PERM_2_IN_1_GA4(XS, Y, YS, split_3_in_gaa3(XS, YS1, cons_22(Y, YS2)))
PERM_2_IN_GA2(XS, cons_22(Y, YS)) -> SPLIT_3_IN_GAA3(XS, YS1, cons_22(Y, YS2))
SPLIT_3_IN_GAA3(cons_22(X, XS), cons_22(X, YS1), YS2) -> IF_SPLIT_3_IN_1_GAA5(X, XS, YS1, YS2, split_3_in_gaa3(XS, YS1, YS2))
SPLIT_3_IN_GAA3(cons_22(X, XS), cons_22(X, YS1), YS2) -> SPLIT_3_IN_GAA3(XS, YS1, YS2)
IF_PERM_2_IN_1_GA4(XS, Y, YS, split_3_out_gaa3(XS, YS1, cons_22(Y, YS2))) -> IF_PERM_2_IN_2_GA6(XS, Y, YS, YS1, YS2, append_3_in_gga3(YS1, YS2, ZS))
IF_PERM_2_IN_1_GA4(XS, Y, YS, split_3_out_gaa3(XS, YS1, cons_22(Y, YS2))) -> APPEND_3_IN_GGA3(YS1, YS2, ZS)
APPEND_3_IN_GGA3(cons_22(X, XS1), XS2, cons_22(X, YS)) -> IF_APPEND_3_IN_1_GGA5(X, XS1, XS2, YS, append_3_in_gga3(XS1, XS2, YS))
APPEND_3_IN_GGA3(cons_22(X, XS1), XS2, cons_22(X, YS)) -> APPEND_3_IN_GGA3(XS1, XS2, YS)
IF_PERM_2_IN_2_GA6(XS, Y, YS, YS1, YS2, append_3_out_gga3(YS1, YS2, ZS)) -> IF_PERM_2_IN_3_GA5(XS, Y, YS, ZS, perm_2_in_ga2(ZS, YS))
IF_PERM_2_IN_2_GA6(XS, Y, YS, YS1, YS2, append_3_out_gga3(YS1, YS2, ZS)) -> PERM_2_IN_GA2(ZS, YS)

The TRS R consists of the following rules:

perm_2_in_ga2(nil_0, nil_0) -> perm_2_out_ga2(nil_0, nil_0)
perm_2_in_ga2(XS, cons_22(Y, YS)) -> if_perm_2_in_1_ga4(XS, Y, YS, split_3_in_gaa3(XS, YS1, cons_22(Y, YS2)))
split_3_in_gaa3(XS, nil_0, XS) -> split_3_out_gaa3(XS, nil_0, XS)
split_3_in_gaa3(cons_22(X, XS), cons_22(X, YS1), YS2) -> if_split_3_in_1_gaa5(X, XS, YS1, YS2, split_3_in_gaa3(XS, YS1, YS2))
if_split_3_in_1_gaa5(X, XS, YS1, YS2, split_3_out_gaa3(XS, YS1, YS2)) -> split_3_out_gaa3(cons_22(X, XS), cons_22(X, YS1), YS2)
if_perm_2_in_1_ga4(XS, Y, YS, split_3_out_gaa3(XS, YS1, cons_22(Y, YS2))) -> if_perm_2_in_2_ga6(XS, Y, YS, YS1, YS2, append_3_in_gga3(YS1, YS2, ZS))
append_3_in_gga3(nil_0, XS, XS) -> append_3_out_gga3(nil_0, XS, XS)
append_3_in_gga3(cons_22(X, XS1), XS2, cons_22(X, YS)) -> if_append_3_in_1_gga5(X, XS1, XS2, YS, append_3_in_gga3(XS1, XS2, YS))
if_append_3_in_1_gga5(X, XS1, XS2, YS, append_3_out_gga3(XS1, XS2, YS)) -> append_3_out_gga3(cons_22(X, XS1), XS2, cons_22(X, YS))
if_perm_2_in_2_ga6(XS, Y, YS, YS1, YS2, append_3_out_gga3(YS1, YS2, ZS)) -> if_perm_2_in_3_ga5(XS, Y, YS, ZS, perm_2_in_ga2(ZS, YS))
if_perm_2_in_3_ga5(XS, Y, YS, ZS, perm_2_out_ga2(ZS, YS)) -> perm_2_out_ga2(XS, cons_22(Y, YS))

The argument filtering Pi contains the following mapping:
perm_2_in_ga2(x1, x2)  =  perm_2_in_ga1(x1)
nil_0  =  nil_0
cons_22(x1, x2)  =  cons_22(x1, x2)
perm_2_out_ga2(x1, x2)  =  perm_2_out_ga1(x2)
if_perm_2_in_1_ga4(x1, x2, x3, x4)  =  if_perm_2_in_1_ga1(x4)
split_3_in_gaa3(x1, x2, x3)  =  split_3_in_gaa1(x1)
split_3_out_gaa3(x1, x2, x3)  =  split_3_out_gaa2(x2, x3)
if_split_3_in_1_gaa5(x1, x2, x3, x4, x5)  =  if_split_3_in_1_gaa2(x1, x5)
if_perm_2_in_2_ga6(x1, x2, x3, x4, x5, x6)  =  if_perm_2_in_2_ga2(x2, x6)
append_3_in_gga3(x1, x2, x3)  =  append_3_in_gga2(x1, x2)
append_3_out_gga3(x1, x2, x3)  =  append_3_out_gga1(x3)
if_append_3_in_1_gga5(x1, x2, x3, x4, x5)  =  if_append_3_in_1_gga2(x1, x5)
if_perm_2_in_3_ga5(x1, x2, x3, x4, x5)  =  if_perm_2_in_3_ga2(x2, x5)
IF_SPLIT_3_IN_1_GAA5(x1, x2, x3, x4, x5)  =  IF_SPLIT_3_IN_1_GAA2(x1, x5)
IF_APPEND_3_IN_1_GGA5(x1, x2, x3, x4, x5)  =  IF_APPEND_3_IN_1_GGA2(x1, x5)
SPLIT_3_IN_GAA3(x1, x2, x3)  =  SPLIT_3_IN_GAA1(x1)
IF_PERM_2_IN_3_GA5(x1, x2, x3, x4, x5)  =  IF_PERM_2_IN_3_GA2(x2, x5)
PERM_2_IN_GA2(x1, x2)  =  PERM_2_IN_GA1(x1)
APPEND_3_IN_GGA3(x1, x2, x3)  =  APPEND_3_IN_GGA2(x1, x2)
IF_PERM_2_IN_2_GA6(x1, x2, x3, x4, x5, x6)  =  IF_PERM_2_IN_2_GA2(x2, x6)
IF_PERM_2_IN_1_GA4(x1, x2, x3, x4)  =  IF_PERM_2_IN_1_GA1(x4)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph contains 3 SCCs with 5 less nodes.

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

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

APPEND_3_IN_GGA3(cons_22(X, XS1), XS2, cons_22(X, YS)) -> APPEND_3_IN_GGA3(XS1, XS2, YS)

The TRS R consists of the following rules:

perm_2_in_ga2(nil_0, nil_0) -> perm_2_out_ga2(nil_0, nil_0)
perm_2_in_ga2(XS, cons_22(Y, YS)) -> if_perm_2_in_1_ga4(XS, Y, YS, split_3_in_gaa3(XS, YS1, cons_22(Y, YS2)))
split_3_in_gaa3(XS, nil_0, XS) -> split_3_out_gaa3(XS, nil_0, XS)
split_3_in_gaa3(cons_22(X, XS), cons_22(X, YS1), YS2) -> if_split_3_in_1_gaa5(X, XS, YS1, YS2, split_3_in_gaa3(XS, YS1, YS2))
if_split_3_in_1_gaa5(X, XS, YS1, YS2, split_3_out_gaa3(XS, YS1, YS2)) -> split_3_out_gaa3(cons_22(X, XS), cons_22(X, YS1), YS2)
if_perm_2_in_1_ga4(XS, Y, YS, split_3_out_gaa3(XS, YS1, cons_22(Y, YS2))) -> if_perm_2_in_2_ga6(XS, Y, YS, YS1, YS2, append_3_in_gga3(YS1, YS2, ZS))
append_3_in_gga3(nil_0, XS, XS) -> append_3_out_gga3(nil_0, XS, XS)
append_3_in_gga3(cons_22(X, XS1), XS2, cons_22(X, YS)) -> if_append_3_in_1_gga5(X, XS1, XS2, YS, append_3_in_gga3(XS1, XS2, YS))
if_append_3_in_1_gga5(X, XS1, XS2, YS, append_3_out_gga3(XS1, XS2, YS)) -> append_3_out_gga3(cons_22(X, XS1), XS2, cons_22(X, YS))
if_perm_2_in_2_ga6(XS, Y, YS, YS1, YS2, append_3_out_gga3(YS1, YS2, ZS)) -> if_perm_2_in_3_ga5(XS, Y, YS, ZS, perm_2_in_ga2(ZS, YS))
if_perm_2_in_3_ga5(XS, Y, YS, ZS, perm_2_out_ga2(ZS, YS)) -> perm_2_out_ga2(XS, cons_22(Y, YS))

The argument filtering Pi contains the following mapping:
perm_2_in_ga2(x1, x2)  =  perm_2_in_ga1(x1)
nil_0  =  nil_0
cons_22(x1, x2)  =  cons_22(x1, x2)
perm_2_out_ga2(x1, x2)  =  perm_2_out_ga1(x2)
if_perm_2_in_1_ga4(x1, x2, x3, x4)  =  if_perm_2_in_1_ga1(x4)
split_3_in_gaa3(x1, x2, x3)  =  split_3_in_gaa1(x1)
split_3_out_gaa3(x1, x2, x3)  =  split_3_out_gaa2(x2, x3)
if_split_3_in_1_gaa5(x1, x2, x3, x4, x5)  =  if_split_3_in_1_gaa2(x1, x5)
if_perm_2_in_2_ga6(x1, x2, x3, x4, x5, x6)  =  if_perm_2_in_2_ga2(x2, x6)
append_3_in_gga3(x1, x2, x3)  =  append_3_in_gga2(x1, x2)
append_3_out_gga3(x1, x2, x3)  =  append_3_out_gga1(x3)
if_append_3_in_1_gga5(x1, x2, x3, x4, x5)  =  if_append_3_in_1_gga2(x1, x5)
if_perm_2_in_3_ga5(x1, x2, x3, x4, x5)  =  if_perm_2_in_3_ga2(x2, x5)
APPEND_3_IN_GGA3(x1, x2, x3)  =  APPEND_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:

APPEND_3_IN_GGA3(cons_22(X, XS1), XS2, cons_22(X, YS)) -> APPEND_3_IN_GGA3(XS1, XS2, YS)

R is empty.
The argument filtering Pi contains the following mapping:
cons_22(x1, x2)  =  cons_22(x1, x2)
APPEND_3_IN_GGA3(x1, x2, x3)  =  APPEND_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:

APPEND_3_IN_GGA2(cons_22(X, XS1), XS2) -> APPEND_3_IN_GGA2(XS1, XS2)

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

SPLIT_3_IN_GAA3(cons_22(X, XS), cons_22(X, YS1), YS2) -> SPLIT_3_IN_GAA3(XS, YS1, YS2)

The TRS R consists of the following rules:

perm_2_in_ga2(nil_0, nil_0) -> perm_2_out_ga2(nil_0, nil_0)
perm_2_in_ga2(XS, cons_22(Y, YS)) -> if_perm_2_in_1_ga4(XS, Y, YS, split_3_in_gaa3(XS, YS1, cons_22(Y, YS2)))
split_3_in_gaa3(XS, nil_0, XS) -> split_3_out_gaa3(XS, nil_0, XS)
split_3_in_gaa3(cons_22(X, XS), cons_22(X, YS1), YS2) -> if_split_3_in_1_gaa5(X, XS, YS1, YS2, split_3_in_gaa3(XS, YS1, YS2))
if_split_3_in_1_gaa5(X, XS, YS1, YS2, split_3_out_gaa3(XS, YS1, YS2)) -> split_3_out_gaa3(cons_22(X, XS), cons_22(X, YS1), YS2)
if_perm_2_in_1_ga4(XS, Y, YS, split_3_out_gaa3(XS, YS1, cons_22(Y, YS2))) -> if_perm_2_in_2_ga6(XS, Y, YS, YS1, YS2, append_3_in_gga3(YS1, YS2, ZS))
append_3_in_gga3(nil_0, XS, XS) -> append_3_out_gga3(nil_0, XS, XS)
append_3_in_gga3(cons_22(X, XS1), XS2, cons_22(X, YS)) -> if_append_3_in_1_gga5(X, XS1, XS2, YS, append_3_in_gga3(XS1, XS2, YS))
if_append_3_in_1_gga5(X, XS1, XS2, YS, append_3_out_gga3(XS1, XS2, YS)) -> append_3_out_gga3(cons_22(X, XS1), XS2, cons_22(X, YS))
if_perm_2_in_2_ga6(XS, Y, YS, YS1, YS2, append_3_out_gga3(YS1, YS2, ZS)) -> if_perm_2_in_3_ga5(XS, Y, YS, ZS, perm_2_in_ga2(ZS, YS))
if_perm_2_in_3_ga5(XS, Y, YS, ZS, perm_2_out_ga2(ZS, YS)) -> perm_2_out_ga2(XS, cons_22(Y, YS))

The argument filtering Pi contains the following mapping:
perm_2_in_ga2(x1, x2)  =  perm_2_in_ga1(x1)
nil_0  =  nil_0
cons_22(x1, x2)  =  cons_22(x1, x2)
perm_2_out_ga2(x1, x2)  =  perm_2_out_ga1(x2)
if_perm_2_in_1_ga4(x1, x2, x3, x4)  =  if_perm_2_in_1_ga1(x4)
split_3_in_gaa3(x1, x2, x3)  =  split_3_in_gaa1(x1)
split_3_out_gaa3(x1, x2, x3)  =  split_3_out_gaa2(x2, x3)
if_split_3_in_1_gaa5(x1, x2, x3, x4, x5)  =  if_split_3_in_1_gaa2(x1, x5)
if_perm_2_in_2_ga6(x1, x2, x3, x4, x5, x6)  =  if_perm_2_in_2_ga2(x2, x6)
append_3_in_gga3(x1, x2, x3)  =  append_3_in_gga2(x1, x2)
append_3_out_gga3(x1, x2, x3)  =  append_3_out_gga1(x3)
if_append_3_in_1_gga5(x1, x2, x3, x4, x5)  =  if_append_3_in_1_gga2(x1, x5)
if_perm_2_in_3_ga5(x1, x2, x3, x4, x5)  =  if_perm_2_in_3_ga2(x2, x5)
SPLIT_3_IN_GAA3(x1, x2, x3)  =  SPLIT_3_IN_GAA1(x1)

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:

SPLIT_3_IN_GAA3(cons_22(X, XS), cons_22(X, YS1), YS2) -> SPLIT_3_IN_GAA3(XS, YS1, YS2)

R is empty.
The argument filtering Pi contains the following mapping:
cons_22(x1, x2)  =  cons_22(x1, x2)
SPLIT_3_IN_GAA3(x1, x2, x3)  =  SPLIT_3_IN_GAA1(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
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ QDPSizeChangeProof
              ↳ PiDP

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

SPLIT_3_IN_GAA1(cons_22(X, XS)) -> SPLIT_3_IN_GAA1(XS)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {SPLIT_3_IN_GAA1}.
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

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

PERM_2_IN_GA2(XS, cons_22(Y, YS)) -> IF_PERM_2_IN_1_GA4(XS, Y, YS, split_3_in_gaa3(XS, YS1, cons_22(Y, YS2)))
IF_PERM_2_IN_2_GA6(XS, Y, YS, YS1, YS2, append_3_out_gga3(YS1, YS2, ZS)) -> PERM_2_IN_GA2(ZS, YS)
IF_PERM_2_IN_1_GA4(XS, Y, YS, split_3_out_gaa3(XS, YS1, cons_22(Y, YS2))) -> IF_PERM_2_IN_2_GA6(XS, Y, YS, YS1, YS2, append_3_in_gga3(YS1, YS2, ZS))

The TRS R consists of the following rules:

perm_2_in_ga2(nil_0, nil_0) -> perm_2_out_ga2(nil_0, nil_0)
perm_2_in_ga2(XS, cons_22(Y, YS)) -> if_perm_2_in_1_ga4(XS, Y, YS, split_3_in_gaa3(XS, YS1, cons_22(Y, YS2)))
split_3_in_gaa3(XS, nil_0, XS) -> split_3_out_gaa3(XS, nil_0, XS)
split_3_in_gaa3(cons_22(X, XS), cons_22(X, YS1), YS2) -> if_split_3_in_1_gaa5(X, XS, YS1, YS2, split_3_in_gaa3(XS, YS1, YS2))
if_split_3_in_1_gaa5(X, XS, YS1, YS2, split_3_out_gaa3(XS, YS1, YS2)) -> split_3_out_gaa3(cons_22(X, XS), cons_22(X, YS1), YS2)
if_perm_2_in_1_ga4(XS, Y, YS, split_3_out_gaa3(XS, YS1, cons_22(Y, YS2))) -> if_perm_2_in_2_ga6(XS, Y, YS, YS1, YS2, append_3_in_gga3(YS1, YS2, ZS))
append_3_in_gga3(nil_0, XS, XS) -> append_3_out_gga3(nil_0, XS, XS)
append_3_in_gga3(cons_22(X, XS1), XS2, cons_22(X, YS)) -> if_append_3_in_1_gga5(X, XS1, XS2, YS, append_3_in_gga3(XS1, XS2, YS))
if_append_3_in_1_gga5(X, XS1, XS2, YS, append_3_out_gga3(XS1, XS2, YS)) -> append_3_out_gga3(cons_22(X, XS1), XS2, cons_22(X, YS))
if_perm_2_in_2_ga6(XS, Y, YS, YS1, YS2, append_3_out_gga3(YS1, YS2, ZS)) -> if_perm_2_in_3_ga5(XS, Y, YS, ZS, perm_2_in_ga2(ZS, YS))
if_perm_2_in_3_ga5(XS, Y, YS, ZS, perm_2_out_ga2(ZS, YS)) -> perm_2_out_ga2(XS, cons_22(Y, YS))

The argument filtering Pi contains the following mapping:
perm_2_in_ga2(x1, x2)  =  perm_2_in_ga1(x1)
nil_0  =  nil_0
cons_22(x1, x2)  =  cons_22(x1, x2)
perm_2_out_ga2(x1, x2)  =  perm_2_out_ga1(x2)
if_perm_2_in_1_ga4(x1, x2, x3, x4)  =  if_perm_2_in_1_ga1(x4)
split_3_in_gaa3(x1, x2, x3)  =  split_3_in_gaa1(x1)
split_3_out_gaa3(x1, x2, x3)  =  split_3_out_gaa2(x2, x3)
if_split_3_in_1_gaa5(x1, x2, x3, x4, x5)  =  if_split_3_in_1_gaa2(x1, x5)
if_perm_2_in_2_ga6(x1, x2, x3, x4, x5, x6)  =  if_perm_2_in_2_ga2(x2, x6)
append_3_in_gga3(x1, x2, x3)  =  append_3_in_gga2(x1, x2)
append_3_out_gga3(x1, x2, x3)  =  append_3_out_gga1(x3)
if_append_3_in_1_gga5(x1, x2, x3, x4, x5)  =  if_append_3_in_1_gga2(x1, x5)
if_perm_2_in_3_ga5(x1, x2, x3, x4, x5)  =  if_perm_2_in_3_ga2(x2, x5)
PERM_2_IN_GA2(x1, x2)  =  PERM_2_IN_GA1(x1)
IF_PERM_2_IN_2_GA6(x1, x2, x3, x4, x5, x6)  =  IF_PERM_2_IN_2_GA2(x2, x6)
IF_PERM_2_IN_1_GA4(x1, x2, x3, x4)  =  IF_PERM_2_IN_1_GA1(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

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

PERM_2_IN_GA2(XS, cons_22(Y, YS)) -> IF_PERM_2_IN_1_GA4(XS, Y, YS, split_3_in_gaa3(XS, YS1, cons_22(Y, YS2)))
IF_PERM_2_IN_2_GA6(XS, Y, YS, YS1, YS2, append_3_out_gga3(YS1, YS2, ZS)) -> PERM_2_IN_GA2(ZS, YS)
IF_PERM_2_IN_1_GA4(XS, Y, YS, split_3_out_gaa3(XS, YS1, cons_22(Y, YS2))) -> IF_PERM_2_IN_2_GA6(XS, Y, YS, YS1, YS2, append_3_in_gga3(YS1, YS2, ZS))

The TRS R consists of the following rules:

split_3_in_gaa3(XS, nil_0, XS) -> split_3_out_gaa3(XS, nil_0, XS)
split_3_in_gaa3(cons_22(X, XS), cons_22(X, YS1), YS2) -> if_split_3_in_1_gaa5(X, XS, YS1, YS2, split_3_in_gaa3(XS, YS1, YS2))
append_3_in_gga3(nil_0, XS, XS) -> append_3_out_gga3(nil_0, XS, XS)
append_3_in_gga3(cons_22(X, XS1), XS2, cons_22(X, YS)) -> if_append_3_in_1_gga5(X, XS1, XS2, YS, append_3_in_gga3(XS1, XS2, YS))
if_split_3_in_1_gaa5(X, XS, YS1, YS2, split_3_out_gaa3(XS, YS1, YS2)) -> split_3_out_gaa3(cons_22(X, XS), cons_22(X, YS1), YS2)
if_append_3_in_1_gga5(X, XS1, XS2, YS, append_3_out_gga3(XS1, XS2, YS)) -> append_3_out_gga3(cons_22(X, XS1), XS2, cons_22(X, YS))

The argument filtering Pi contains the following mapping:
nil_0  =  nil_0
cons_22(x1, x2)  =  cons_22(x1, x2)
split_3_in_gaa3(x1, x2, x3)  =  split_3_in_gaa1(x1)
split_3_out_gaa3(x1, x2, x3)  =  split_3_out_gaa2(x2, x3)
if_split_3_in_1_gaa5(x1, x2, x3, x4, x5)  =  if_split_3_in_1_gaa2(x1, x5)
append_3_in_gga3(x1, x2, x3)  =  append_3_in_gga2(x1, x2)
append_3_out_gga3(x1, x2, x3)  =  append_3_out_gga1(x3)
if_append_3_in_1_gga5(x1, x2, x3, x4, x5)  =  if_append_3_in_1_gga2(x1, x5)
PERM_2_IN_GA2(x1, x2)  =  PERM_2_IN_GA1(x1)
IF_PERM_2_IN_2_GA6(x1, x2, x3, x4, x5, x6)  =  IF_PERM_2_IN_2_GA2(x2, x6)
IF_PERM_2_IN_1_GA4(x1, x2, x3, x4)  =  IF_PERM_2_IN_1_GA1(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
                        ↳ RuleRemovalProof

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

PERM_2_IN_GA1(XS) -> IF_PERM_2_IN_1_GA1(split_3_in_gaa1(XS))
IF_PERM_2_IN_2_GA2(Y, append_3_out_gga1(ZS)) -> PERM_2_IN_GA1(ZS)
IF_PERM_2_IN_1_GA1(split_3_out_gaa2(YS1, cons_22(Y, YS2))) -> IF_PERM_2_IN_2_GA2(Y, append_3_in_gga2(YS1, YS2))

The TRS R consists of the following rules:

split_3_in_gaa1(XS) -> split_3_out_gaa2(nil_0, XS)
split_3_in_gaa1(cons_22(X, XS)) -> if_split_3_in_1_gaa2(X, split_3_in_gaa1(XS))
append_3_in_gga2(nil_0, XS) -> append_3_out_gga1(XS)
append_3_in_gga2(cons_22(X, XS1), XS2) -> if_append_3_in_1_gga2(X, append_3_in_gga2(XS1, XS2))
if_split_3_in_1_gaa2(X, split_3_out_gaa2(YS1, YS2)) -> split_3_out_gaa2(cons_22(X, YS1), YS2)
if_append_3_in_1_gga2(X, append_3_out_gga1(YS)) -> append_3_out_gga1(cons_22(X, YS))

The set Q consists of the following terms:

split_3_in_gaa1(x0)
append_3_in_gga2(x0, x1)
if_split_3_in_1_gaa2(x0, x1)
if_append_3_in_1_gga2(x0, x1)

We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {IF_PERM_2_IN_1_GA1, PERM_2_IN_GA1, IF_PERM_2_IN_2_GA2}.
By using a polynomial ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.

Strictly oriented rules of the TRS R:

append_3_in_gga2(nil_0, XS) -> append_3_out_gga1(XS)

Used ordering: POLO with Polynomial interpretation:

POL(nil_0) = 2   
POL(append_3_out_gga1(x1)) = 1 + x1   
POL(split_3_in_gaa1(x1)) = 2 + x1   
POL(split_3_out_gaa2(x1, x2)) = x1 + x2   
POL(IF_PERM_2_IN_2_GA2(x1, x2)) = 1 + x1 + x2   
POL(if_append_3_in_1_gga2(x1, x2)) = 1 + x1 + x2   
POL(IF_PERM_2_IN_1_GA1(x1)) = x1   
POL(cons_22(x1, x2)) = 1 + x1 + x2   
POL(if_split_3_in_1_gaa2(x1, x2)) = 1 + x1 + x2   
POL(append_3_in_gga2(x1, x2)) = x1 + x2   
POL(PERM_2_IN_GA1(x1)) = 2 + x1   



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

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

PERM_2_IN_GA1(XS) -> IF_PERM_2_IN_1_GA1(split_3_in_gaa1(XS))
IF_PERM_2_IN_2_GA2(Y, append_3_out_gga1(ZS)) -> PERM_2_IN_GA1(ZS)
IF_PERM_2_IN_1_GA1(split_3_out_gaa2(YS1, cons_22(Y, YS2))) -> IF_PERM_2_IN_2_GA2(Y, append_3_in_gga2(YS1, YS2))

The TRS R consists of the following rules:

split_3_in_gaa1(XS) -> split_3_out_gaa2(nil_0, XS)
split_3_in_gaa1(cons_22(X, XS)) -> if_split_3_in_1_gaa2(X, split_3_in_gaa1(XS))
append_3_in_gga2(cons_22(X, XS1), XS2) -> if_append_3_in_1_gga2(X, append_3_in_gga2(XS1, XS2))
if_split_3_in_1_gaa2(X, split_3_out_gaa2(YS1, YS2)) -> split_3_out_gaa2(cons_22(X, YS1), YS2)
if_append_3_in_1_gga2(X, append_3_out_gga1(YS)) -> append_3_out_gga1(cons_22(X, YS))

The set Q consists of the following terms:

split_3_in_gaa1(x0)
append_3_in_gga2(x0, x1)
if_split_3_in_1_gaa2(x0, x1)
if_append_3_in_1_gga2(x0, x1)

We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {IF_PERM_2_IN_1_GA1, PERM_2_IN_GA1, IF_PERM_2_IN_2_GA2}.
By using a polynomial ordering, the following set of Dependency Pairs of this DP problem can be strictly oriented.

IF_PERM_2_IN_2_GA2(Y, append_3_out_gga1(ZS)) -> PERM_2_IN_GA1(ZS)
The remaining Dependency Pairs were at least non-strictly be oriented.

PERM_2_IN_GA1(XS) -> IF_PERM_2_IN_1_GA1(split_3_in_gaa1(XS))
IF_PERM_2_IN_1_GA1(split_3_out_gaa2(YS1, cons_22(Y, YS2))) -> IF_PERM_2_IN_2_GA2(Y, append_3_in_gga2(YS1, YS2))
With the implicit AFS we had to orient the following set of usable rules non-strictly.

append_3_in_gga2(cons_22(X, XS1), XS2) -> if_append_3_in_1_gga2(X, append_3_in_gga2(XS1, XS2))
if_append_3_in_1_gga2(X, append_3_out_gga1(YS)) -> append_3_out_gga1(cons_22(X, YS))
Used ordering: POLO with Polynomial interpretation:

POL(nil_0) = 0   
POL(append_3_out_gga1(x1)) = 1   
POL(split_3_out_gaa2(x1, x2)) = 0   
POL(split_3_in_gaa1(x1)) = 0   
POL(IF_PERM_2_IN_2_GA2(x1, x2)) = x2   
POL(if_append_3_in_1_gga2(x1, x2)) = x2   
POL(IF_PERM_2_IN_1_GA1(x1)) = 0   
POL(if_split_3_in_1_gaa2(x1, x2)) = 0   
POL(cons_22(x1, x2)) = 0   
POL(append_3_in_gga2(x1, x2)) = 0   
POL(PERM_2_IN_GA1(x1)) = 0   



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

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

PERM_2_IN_GA1(XS) -> IF_PERM_2_IN_1_GA1(split_3_in_gaa1(XS))
IF_PERM_2_IN_1_GA1(split_3_out_gaa2(YS1, cons_22(Y, YS2))) -> IF_PERM_2_IN_2_GA2(Y, append_3_in_gga2(YS1, YS2))

The TRS R consists of the following rules:

split_3_in_gaa1(XS) -> split_3_out_gaa2(nil_0, XS)
split_3_in_gaa1(cons_22(X, XS)) -> if_split_3_in_1_gaa2(X, split_3_in_gaa1(XS))
append_3_in_gga2(cons_22(X, XS1), XS2) -> if_append_3_in_1_gga2(X, append_3_in_gga2(XS1, XS2))
if_split_3_in_1_gaa2(X, split_3_out_gaa2(YS1, YS2)) -> split_3_out_gaa2(cons_22(X, YS1), YS2)
if_append_3_in_1_gga2(X, append_3_out_gga1(YS)) -> append_3_out_gga1(cons_22(X, YS))

The set Q consists of the following terms:

split_3_in_gaa1(x0)
append_3_in_gga2(x0, x1)
if_split_3_in_1_gaa2(x0, x1)
if_append_3_in_1_gga2(x0, x1)

We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {IF_PERM_2_IN_1_GA1, PERM_2_IN_GA1, IF_PERM_2_IN_2_GA2}.
The approximation of the Dependency Graph contains 0 SCCs with 2 less nodes.