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



↳ PROLOG
  ↳ PrologToPiTRSProof

app13(.2(X0, X), Y, .2(X0, Z)) :- app13(X, Y, Z).
app13({}0, Y, Y).
app23(.2(X0, X), Y, .2(X0, Z)) :- app23(X, Y, Z).
app23({}0, Y, Y).
perm2(X, .2(X0, Y)) :- app13(X1, .2(X0, X2), X), app23(X1, X2, Z), perm2(Z, Y).
perm2({}0, {}0).


With regard to the inferred argument filtering the predicates were used in the following modes:
perm2: (b,f)
app13: (f,f,b)
app23: (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(X, ._22(X0, Y)) -> if_perm_2_in_1_ga4(X, X0, Y, app1_3_in_aag3(X1, ._22(X0, X2), X))
app1_3_in_aag3(._22(X0, X), Y, ._22(X0, Z)) -> if_app1_3_in_1_aag5(X0, X, Y, Z, app1_3_in_aag3(X, Y, Z))
app1_3_in_aag3([]_0, Y, Y) -> app1_3_out_aag3([]_0, Y, Y)
if_app1_3_in_1_aag5(X0, X, Y, Z, app1_3_out_aag3(X, Y, Z)) -> app1_3_out_aag3(._22(X0, X), Y, ._22(X0, Z))
if_perm_2_in_1_ga4(X, X0, Y, app1_3_out_aag3(X1, ._22(X0, X2), X)) -> if_perm_2_in_2_ga6(X, X0, Y, X1, X2, app2_3_in_gga3(X1, X2, Z))
app2_3_in_gga3(._22(X0, X), Y, ._22(X0, Z)) -> if_app2_3_in_1_gga5(X0, X, Y, Z, app2_3_in_gga3(X, Y, Z))
app2_3_in_gga3([]_0, Y, Y) -> app2_3_out_gga3([]_0, Y, Y)
if_app2_3_in_1_gga5(X0, X, Y, Z, app2_3_out_gga3(X, Y, Z)) -> app2_3_out_gga3(._22(X0, X), Y, ._22(X0, Z))
if_perm_2_in_2_ga6(X, X0, Y, X1, X2, app2_3_out_gga3(X1, X2, Z)) -> if_perm_2_in_3_ga5(X, X0, Y, Z, perm_2_in_ga2(Z, Y))
perm_2_in_ga2([]_0, []_0) -> perm_2_out_ga2([]_0, []_0)
if_perm_2_in_3_ga5(X, X0, Y, Z, perm_2_out_ga2(Z, Y)) -> perm_2_out_ga2(X, ._22(X0, Y))

The argument filtering Pi contains the following mapping:
perm_2_in_ga2(x1, x2)  =  perm_2_in_ga1(x1)
._22(x1, x2)  =  ._22(x1, x2)
[]_0  =  []_0
if_perm_2_in_1_ga4(x1, x2, x3, x4)  =  if_perm_2_in_1_ga1(x4)
app1_3_in_aag3(x1, x2, x3)  =  app1_3_in_aag1(x3)
if_app1_3_in_1_aag5(x1, x2, x3, x4, x5)  =  if_app1_3_in_1_aag2(x1, x5)
app1_3_out_aag3(x1, x2, x3)  =  app1_3_out_aag2(x1, x2)
if_perm_2_in_2_ga6(x1, x2, x3, x4, x5, x6)  =  if_perm_2_in_2_ga2(x2, x6)
app2_3_in_gga3(x1, x2, x3)  =  app2_3_in_gga2(x1, x2)
if_app2_3_in_1_gga5(x1, x2, x3, x4, x5)  =  if_app2_3_in_1_gga2(x1, x5)
app2_3_out_gga3(x1, x2, x3)  =  app2_3_out_gga1(x3)
if_perm_2_in_3_ga5(x1, x2, x3, x4, x5)  =  if_perm_2_in_3_ga2(x2, x5)
perm_2_out_ga2(x1, x2)  =  perm_2_out_ga1(x2)

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(X, ._22(X0, Y)) -> if_perm_2_in_1_ga4(X, X0, Y, app1_3_in_aag3(X1, ._22(X0, X2), X))
app1_3_in_aag3(._22(X0, X), Y, ._22(X0, Z)) -> if_app1_3_in_1_aag5(X0, X, Y, Z, app1_3_in_aag3(X, Y, Z))
app1_3_in_aag3([]_0, Y, Y) -> app1_3_out_aag3([]_0, Y, Y)
if_app1_3_in_1_aag5(X0, X, Y, Z, app1_3_out_aag3(X, Y, Z)) -> app1_3_out_aag3(._22(X0, X), Y, ._22(X0, Z))
if_perm_2_in_1_ga4(X, X0, Y, app1_3_out_aag3(X1, ._22(X0, X2), X)) -> if_perm_2_in_2_ga6(X, X0, Y, X1, X2, app2_3_in_gga3(X1, X2, Z))
app2_3_in_gga3(._22(X0, X), Y, ._22(X0, Z)) -> if_app2_3_in_1_gga5(X0, X, Y, Z, app2_3_in_gga3(X, Y, Z))
app2_3_in_gga3([]_0, Y, Y) -> app2_3_out_gga3([]_0, Y, Y)
if_app2_3_in_1_gga5(X0, X, Y, Z, app2_3_out_gga3(X, Y, Z)) -> app2_3_out_gga3(._22(X0, X), Y, ._22(X0, Z))
if_perm_2_in_2_ga6(X, X0, Y, X1, X2, app2_3_out_gga3(X1, X2, Z)) -> if_perm_2_in_3_ga5(X, X0, Y, Z, perm_2_in_ga2(Z, Y))
perm_2_in_ga2([]_0, []_0) -> perm_2_out_ga2([]_0, []_0)
if_perm_2_in_3_ga5(X, X0, Y, Z, perm_2_out_ga2(Z, Y)) -> perm_2_out_ga2(X, ._22(X0, Y))

The argument filtering Pi contains the following mapping:
perm_2_in_ga2(x1, x2)  =  perm_2_in_ga1(x1)
._22(x1, x2)  =  ._22(x1, x2)
[]_0  =  []_0
if_perm_2_in_1_ga4(x1, x2, x3, x4)  =  if_perm_2_in_1_ga1(x4)
app1_3_in_aag3(x1, x2, x3)  =  app1_3_in_aag1(x3)
if_app1_3_in_1_aag5(x1, x2, x3, x4, x5)  =  if_app1_3_in_1_aag2(x1, x5)
app1_3_out_aag3(x1, x2, x3)  =  app1_3_out_aag2(x1, x2)
if_perm_2_in_2_ga6(x1, x2, x3, x4, x5, x6)  =  if_perm_2_in_2_ga2(x2, x6)
app2_3_in_gga3(x1, x2, x3)  =  app2_3_in_gga2(x1, x2)
if_app2_3_in_1_gga5(x1, x2, x3, x4, x5)  =  if_app2_3_in_1_gga2(x1, x5)
app2_3_out_gga3(x1, x2, x3)  =  app2_3_out_gga1(x3)
if_perm_2_in_3_ga5(x1, x2, x3, x4, x5)  =  if_perm_2_in_3_ga2(x2, x5)
perm_2_out_ga2(x1, x2)  =  perm_2_out_ga1(x2)


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

PERM_2_IN_GA2(X, ._22(X0, Y)) -> IF_PERM_2_IN_1_GA4(X, X0, Y, app1_3_in_aag3(X1, ._22(X0, X2), X))
PERM_2_IN_GA2(X, ._22(X0, Y)) -> APP1_3_IN_AAG3(X1, ._22(X0, X2), X)
APP1_3_IN_AAG3(._22(X0, X), Y, ._22(X0, Z)) -> IF_APP1_3_IN_1_AAG5(X0, X, Y, Z, app1_3_in_aag3(X, Y, Z))
APP1_3_IN_AAG3(._22(X0, X), Y, ._22(X0, Z)) -> APP1_3_IN_AAG3(X, Y, Z)
IF_PERM_2_IN_1_GA4(X, X0, Y, app1_3_out_aag3(X1, ._22(X0, X2), X)) -> IF_PERM_2_IN_2_GA6(X, X0, Y, X1, X2, app2_3_in_gga3(X1, X2, Z))
IF_PERM_2_IN_1_GA4(X, X0, Y, app1_3_out_aag3(X1, ._22(X0, X2), X)) -> APP2_3_IN_GGA3(X1, X2, Z)
APP2_3_IN_GGA3(._22(X0, X), Y, ._22(X0, Z)) -> IF_APP2_3_IN_1_GGA5(X0, X, Y, Z, app2_3_in_gga3(X, Y, Z))
APP2_3_IN_GGA3(._22(X0, X), Y, ._22(X0, Z)) -> APP2_3_IN_GGA3(X, Y, Z)
IF_PERM_2_IN_2_GA6(X, X0, Y, X1, X2, app2_3_out_gga3(X1, X2, Z)) -> IF_PERM_2_IN_3_GA5(X, X0, Y, Z, perm_2_in_ga2(Z, Y))
IF_PERM_2_IN_2_GA6(X, X0, Y, X1, X2, app2_3_out_gga3(X1, X2, Z)) -> PERM_2_IN_GA2(Z, Y)

The TRS R consists of the following rules:

perm_2_in_ga2(X, ._22(X0, Y)) -> if_perm_2_in_1_ga4(X, X0, Y, app1_3_in_aag3(X1, ._22(X0, X2), X))
app1_3_in_aag3(._22(X0, X), Y, ._22(X0, Z)) -> if_app1_3_in_1_aag5(X0, X, Y, Z, app1_3_in_aag3(X, Y, Z))
app1_3_in_aag3([]_0, Y, Y) -> app1_3_out_aag3([]_0, Y, Y)
if_app1_3_in_1_aag5(X0, X, Y, Z, app1_3_out_aag3(X, Y, Z)) -> app1_3_out_aag3(._22(X0, X), Y, ._22(X0, Z))
if_perm_2_in_1_ga4(X, X0, Y, app1_3_out_aag3(X1, ._22(X0, X2), X)) -> if_perm_2_in_2_ga6(X, X0, Y, X1, X2, app2_3_in_gga3(X1, X2, Z))
app2_3_in_gga3(._22(X0, X), Y, ._22(X0, Z)) -> if_app2_3_in_1_gga5(X0, X, Y, Z, app2_3_in_gga3(X, Y, Z))
app2_3_in_gga3([]_0, Y, Y) -> app2_3_out_gga3([]_0, Y, Y)
if_app2_3_in_1_gga5(X0, X, Y, Z, app2_3_out_gga3(X, Y, Z)) -> app2_3_out_gga3(._22(X0, X), Y, ._22(X0, Z))
if_perm_2_in_2_ga6(X, X0, Y, X1, X2, app2_3_out_gga3(X1, X2, Z)) -> if_perm_2_in_3_ga5(X, X0, Y, Z, perm_2_in_ga2(Z, Y))
perm_2_in_ga2([]_0, []_0) -> perm_2_out_ga2([]_0, []_0)
if_perm_2_in_3_ga5(X, X0, Y, Z, perm_2_out_ga2(Z, Y)) -> perm_2_out_ga2(X, ._22(X0, Y))

The argument filtering Pi contains the following mapping:
perm_2_in_ga2(x1, x2)  =  perm_2_in_ga1(x1)
._22(x1, x2)  =  ._22(x1, x2)
[]_0  =  []_0
if_perm_2_in_1_ga4(x1, x2, x3, x4)  =  if_perm_2_in_1_ga1(x4)
app1_3_in_aag3(x1, x2, x3)  =  app1_3_in_aag1(x3)
if_app1_3_in_1_aag5(x1, x2, x3, x4, x5)  =  if_app1_3_in_1_aag2(x1, x5)
app1_3_out_aag3(x1, x2, x3)  =  app1_3_out_aag2(x1, x2)
if_perm_2_in_2_ga6(x1, x2, x3, x4, x5, x6)  =  if_perm_2_in_2_ga2(x2, x6)
app2_3_in_gga3(x1, x2, x3)  =  app2_3_in_gga2(x1, x2)
if_app2_3_in_1_gga5(x1, x2, x3, x4, x5)  =  if_app2_3_in_1_gga2(x1, x5)
app2_3_out_gga3(x1, x2, x3)  =  app2_3_out_gga1(x3)
if_perm_2_in_3_ga5(x1, x2, x3, x4, x5)  =  if_perm_2_in_3_ga2(x2, x5)
perm_2_out_ga2(x1, x2)  =  perm_2_out_ga1(x2)
APP2_3_IN_GGA3(x1, x2, x3)  =  APP2_3_IN_GGA2(x1, x2)
IF_PERM_2_IN_3_GA5(x1, x2, x3, x4, x5)  =  IF_PERM_2_IN_3_GA2(x2, x5)
APP1_3_IN_AAG3(x1, x2, x3)  =  APP1_3_IN_AAG1(x3)
IF_APP1_3_IN_1_AAG5(x1, x2, x3, x4, x5)  =  IF_APP1_3_IN_1_AAG2(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)
IF_APP2_3_IN_1_GGA5(x1, x2, x3, x4, x5)  =  IF_APP2_3_IN_1_GGA2(x1, x5)

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(X, ._22(X0, Y)) -> IF_PERM_2_IN_1_GA4(X, X0, Y, app1_3_in_aag3(X1, ._22(X0, X2), X))
PERM_2_IN_GA2(X, ._22(X0, Y)) -> APP1_3_IN_AAG3(X1, ._22(X0, X2), X)
APP1_3_IN_AAG3(._22(X0, X), Y, ._22(X0, Z)) -> IF_APP1_3_IN_1_AAG5(X0, X, Y, Z, app1_3_in_aag3(X, Y, Z))
APP1_3_IN_AAG3(._22(X0, X), Y, ._22(X0, Z)) -> APP1_3_IN_AAG3(X, Y, Z)
IF_PERM_2_IN_1_GA4(X, X0, Y, app1_3_out_aag3(X1, ._22(X0, X2), X)) -> IF_PERM_2_IN_2_GA6(X, X0, Y, X1, X2, app2_3_in_gga3(X1, X2, Z))
IF_PERM_2_IN_1_GA4(X, X0, Y, app1_3_out_aag3(X1, ._22(X0, X2), X)) -> APP2_3_IN_GGA3(X1, X2, Z)
APP2_3_IN_GGA3(._22(X0, X), Y, ._22(X0, Z)) -> IF_APP2_3_IN_1_GGA5(X0, X, Y, Z, app2_3_in_gga3(X, Y, Z))
APP2_3_IN_GGA3(._22(X0, X), Y, ._22(X0, Z)) -> APP2_3_IN_GGA3(X, Y, Z)
IF_PERM_2_IN_2_GA6(X, X0, Y, X1, X2, app2_3_out_gga3(X1, X2, Z)) -> IF_PERM_2_IN_3_GA5(X, X0, Y, Z, perm_2_in_ga2(Z, Y))
IF_PERM_2_IN_2_GA6(X, X0, Y, X1, X2, app2_3_out_gga3(X1, X2, Z)) -> PERM_2_IN_GA2(Z, Y)

The TRS R consists of the following rules:

perm_2_in_ga2(X, ._22(X0, Y)) -> if_perm_2_in_1_ga4(X, X0, Y, app1_3_in_aag3(X1, ._22(X0, X2), X))
app1_3_in_aag3(._22(X0, X), Y, ._22(X0, Z)) -> if_app1_3_in_1_aag5(X0, X, Y, Z, app1_3_in_aag3(X, Y, Z))
app1_3_in_aag3([]_0, Y, Y) -> app1_3_out_aag3([]_0, Y, Y)
if_app1_3_in_1_aag5(X0, X, Y, Z, app1_3_out_aag3(X, Y, Z)) -> app1_3_out_aag3(._22(X0, X), Y, ._22(X0, Z))
if_perm_2_in_1_ga4(X, X0, Y, app1_3_out_aag3(X1, ._22(X0, X2), X)) -> if_perm_2_in_2_ga6(X, X0, Y, X1, X2, app2_3_in_gga3(X1, X2, Z))
app2_3_in_gga3(._22(X0, X), Y, ._22(X0, Z)) -> if_app2_3_in_1_gga5(X0, X, Y, Z, app2_3_in_gga3(X, Y, Z))
app2_3_in_gga3([]_0, Y, Y) -> app2_3_out_gga3([]_0, Y, Y)
if_app2_3_in_1_gga5(X0, X, Y, Z, app2_3_out_gga3(X, Y, Z)) -> app2_3_out_gga3(._22(X0, X), Y, ._22(X0, Z))
if_perm_2_in_2_ga6(X, X0, Y, X1, X2, app2_3_out_gga3(X1, X2, Z)) -> if_perm_2_in_3_ga5(X, X0, Y, Z, perm_2_in_ga2(Z, Y))
perm_2_in_ga2([]_0, []_0) -> perm_2_out_ga2([]_0, []_0)
if_perm_2_in_3_ga5(X, X0, Y, Z, perm_2_out_ga2(Z, Y)) -> perm_2_out_ga2(X, ._22(X0, Y))

The argument filtering Pi contains the following mapping:
perm_2_in_ga2(x1, x2)  =  perm_2_in_ga1(x1)
._22(x1, x2)  =  ._22(x1, x2)
[]_0  =  []_0
if_perm_2_in_1_ga4(x1, x2, x3, x4)  =  if_perm_2_in_1_ga1(x4)
app1_3_in_aag3(x1, x2, x3)  =  app1_3_in_aag1(x3)
if_app1_3_in_1_aag5(x1, x2, x3, x4, x5)  =  if_app1_3_in_1_aag2(x1, x5)
app1_3_out_aag3(x1, x2, x3)  =  app1_3_out_aag2(x1, x2)
if_perm_2_in_2_ga6(x1, x2, x3, x4, x5, x6)  =  if_perm_2_in_2_ga2(x2, x6)
app2_3_in_gga3(x1, x2, x3)  =  app2_3_in_gga2(x1, x2)
if_app2_3_in_1_gga5(x1, x2, x3, x4, x5)  =  if_app2_3_in_1_gga2(x1, x5)
app2_3_out_gga3(x1, x2, x3)  =  app2_3_out_gga1(x3)
if_perm_2_in_3_ga5(x1, x2, x3, x4, x5)  =  if_perm_2_in_3_ga2(x2, x5)
perm_2_out_ga2(x1, x2)  =  perm_2_out_ga1(x2)
APP2_3_IN_GGA3(x1, x2, x3)  =  APP2_3_IN_GGA2(x1, x2)
IF_PERM_2_IN_3_GA5(x1, x2, x3, x4, x5)  =  IF_PERM_2_IN_3_GA2(x2, x5)
APP1_3_IN_AAG3(x1, x2, x3)  =  APP1_3_IN_AAG1(x3)
IF_APP1_3_IN_1_AAG5(x1, x2, x3, x4, x5)  =  IF_APP1_3_IN_1_AAG2(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)
IF_APP2_3_IN_1_GGA5(x1, x2, x3, x4, x5)  =  IF_APP2_3_IN_1_GGA2(x1, x5)

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:

APP2_3_IN_GGA3(._22(X0, X), Y, ._22(X0, Z)) -> APP2_3_IN_GGA3(X, Y, Z)

The TRS R consists of the following rules:

perm_2_in_ga2(X, ._22(X0, Y)) -> if_perm_2_in_1_ga4(X, X0, Y, app1_3_in_aag3(X1, ._22(X0, X2), X))
app1_3_in_aag3(._22(X0, X), Y, ._22(X0, Z)) -> if_app1_3_in_1_aag5(X0, X, Y, Z, app1_3_in_aag3(X, Y, Z))
app1_3_in_aag3([]_0, Y, Y) -> app1_3_out_aag3([]_0, Y, Y)
if_app1_3_in_1_aag5(X0, X, Y, Z, app1_3_out_aag3(X, Y, Z)) -> app1_3_out_aag3(._22(X0, X), Y, ._22(X0, Z))
if_perm_2_in_1_ga4(X, X0, Y, app1_3_out_aag3(X1, ._22(X0, X2), X)) -> if_perm_2_in_2_ga6(X, X0, Y, X1, X2, app2_3_in_gga3(X1, X2, Z))
app2_3_in_gga3(._22(X0, X), Y, ._22(X0, Z)) -> if_app2_3_in_1_gga5(X0, X, Y, Z, app2_3_in_gga3(X, Y, Z))
app2_3_in_gga3([]_0, Y, Y) -> app2_3_out_gga3([]_0, Y, Y)
if_app2_3_in_1_gga5(X0, X, Y, Z, app2_3_out_gga3(X, Y, Z)) -> app2_3_out_gga3(._22(X0, X), Y, ._22(X0, Z))
if_perm_2_in_2_ga6(X, X0, Y, X1, X2, app2_3_out_gga3(X1, X2, Z)) -> if_perm_2_in_3_ga5(X, X0, Y, Z, perm_2_in_ga2(Z, Y))
perm_2_in_ga2([]_0, []_0) -> perm_2_out_ga2([]_0, []_0)
if_perm_2_in_3_ga5(X, X0, Y, Z, perm_2_out_ga2(Z, Y)) -> perm_2_out_ga2(X, ._22(X0, Y))

The argument filtering Pi contains the following mapping:
perm_2_in_ga2(x1, x2)  =  perm_2_in_ga1(x1)
._22(x1, x2)  =  ._22(x1, x2)
[]_0  =  []_0
if_perm_2_in_1_ga4(x1, x2, x3, x4)  =  if_perm_2_in_1_ga1(x4)
app1_3_in_aag3(x1, x2, x3)  =  app1_3_in_aag1(x3)
if_app1_3_in_1_aag5(x1, x2, x3, x4, x5)  =  if_app1_3_in_1_aag2(x1, x5)
app1_3_out_aag3(x1, x2, x3)  =  app1_3_out_aag2(x1, x2)
if_perm_2_in_2_ga6(x1, x2, x3, x4, x5, x6)  =  if_perm_2_in_2_ga2(x2, x6)
app2_3_in_gga3(x1, x2, x3)  =  app2_3_in_gga2(x1, x2)
if_app2_3_in_1_gga5(x1, x2, x3, x4, x5)  =  if_app2_3_in_1_gga2(x1, x5)
app2_3_out_gga3(x1, x2, x3)  =  app2_3_out_gga1(x3)
if_perm_2_in_3_ga5(x1, x2, x3, x4, x5)  =  if_perm_2_in_3_ga2(x2, x5)
perm_2_out_ga2(x1, x2)  =  perm_2_out_ga1(x2)
APP2_3_IN_GGA3(x1, x2, x3)  =  APP2_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:

APP2_3_IN_GGA3(._22(X0, X), Y, ._22(X0, Z)) -> APP2_3_IN_GGA3(X, Y, Z)

R is empty.
The argument filtering Pi contains the following mapping:
._22(x1, x2)  =  ._22(x1, x2)
APP2_3_IN_GGA3(x1, x2, x3)  =  APP2_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:

APP2_3_IN_GGA2(._22(X0, X), Y) -> APP2_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 {APP2_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:

APP1_3_IN_AAG3(._22(X0, X), Y, ._22(X0, Z)) -> APP1_3_IN_AAG3(X, Y, Z)

The TRS R consists of the following rules:

perm_2_in_ga2(X, ._22(X0, Y)) -> if_perm_2_in_1_ga4(X, X0, Y, app1_3_in_aag3(X1, ._22(X0, X2), X))
app1_3_in_aag3(._22(X0, X), Y, ._22(X0, Z)) -> if_app1_3_in_1_aag5(X0, X, Y, Z, app1_3_in_aag3(X, Y, Z))
app1_3_in_aag3([]_0, Y, Y) -> app1_3_out_aag3([]_0, Y, Y)
if_app1_3_in_1_aag5(X0, X, Y, Z, app1_3_out_aag3(X, Y, Z)) -> app1_3_out_aag3(._22(X0, X), Y, ._22(X0, Z))
if_perm_2_in_1_ga4(X, X0, Y, app1_3_out_aag3(X1, ._22(X0, X2), X)) -> if_perm_2_in_2_ga6(X, X0, Y, X1, X2, app2_3_in_gga3(X1, X2, Z))
app2_3_in_gga3(._22(X0, X), Y, ._22(X0, Z)) -> if_app2_3_in_1_gga5(X0, X, Y, Z, app2_3_in_gga3(X, Y, Z))
app2_3_in_gga3([]_0, Y, Y) -> app2_3_out_gga3([]_0, Y, Y)
if_app2_3_in_1_gga5(X0, X, Y, Z, app2_3_out_gga3(X, Y, Z)) -> app2_3_out_gga3(._22(X0, X), Y, ._22(X0, Z))
if_perm_2_in_2_ga6(X, X0, Y, X1, X2, app2_3_out_gga3(X1, X2, Z)) -> if_perm_2_in_3_ga5(X, X0, Y, Z, perm_2_in_ga2(Z, Y))
perm_2_in_ga2([]_0, []_0) -> perm_2_out_ga2([]_0, []_0)
if_perm_2_in_3_ga5(X, X0, Y, Z, perm_2_out_ga2(Z, Y)) -> perm_2_out_ga2(X, ._22(X0, Y))

The argument filtering Pi contains the following mapping:
perm_2_in_ga2(x1, x2)  =  perm_2_in_ga1(x1)
._22(x1, x2)  =  ._22(x1, x2)
[]_0  =  []_0
if_perm_2_in_1_ga4(x1, x2, x3, x4)  =  if_perm_2_in_1_ga1(x4)
app1_3_in_aag3(x1, x2, x3)  =  app1_3_in_aag1(x3)
if_app1_3_in_1_aag5(x1, x2, x3, x4, x5)  =  if_app1_3_in_1_aag2(x1, x5)
app1_3_out_aag3(x1, x2, x3)  =  app1_3_out_aag2(x1, x2)
if_perm_2_in_2_ga6(x1, x2, x3, x4, x5, x6)  =  if_perm_2_in_2_ga2(x2, x6)
app2_3_in_gga3(x1, x2, x3)  =  app2_3_in_gga2(x1, x2)
if_app2_3_in_1_gga5(x1, x2, x3, x4, x5)  =  if_app2_3_in_1_gga2(x1, x5)
app2_3_out_gga3(x1, x2, x3)  =  app2_3_out_gga1(x3)
if_perm_2_in_3_ga5(x1, x2, x3, x4, x5)  =  if_perm_2_in_3_ga2(x2, x5)
perm_2_out_ga2(x1, x2)  =  perm_2_out_ga1(x2)
APP1_3_IN_AAG3(x1, x2, x3)  =  APP1_3_IN_AAG1(x3)

We have to consider all (P,R,Pi)-chains
For (infinitary) constructor rewriting we can delete all non-usable rules from R.

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

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

APP1_3_IN_AAG3(._22(X0, X), Y, ._22(X0, Z)) -> APP1_3_IN_AAG3(X, Y, Z)

R is empty.
The argument filtering Pi contains the following mapping:
._22(x1, x2)  =  ._22(x1, x2)
APP1_3_IN_AAG3(x1, x2, x3)  =  APP1_3_IN_AAG1(x3)

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

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

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

APP1_3_IN_AAG1(._22(X0, Z)) -> APP1_3_IN_AAG1(Z)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {APP1_3_IN_AAG1}.
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(X, ._22(X0, Y)) -> IF_PERM_2_IN_1_GA4(X, X0, Y, app1_3_in_aag3(X1, ._22(X0, X2), X))
IF_PERM_2_IN_2_GA6(X, X0, Y, X1, X2, app2_3_out_gga3(X1, X2, Z)) -> PERM_2_IN_GA2(Z, Y)
IF_PERM_2_IN_1_GA4(X, X0, Y, app1_3_out_aag3(X1, ._22(X0, X2), X)) -> IF_PERM_2_IN_2_GA6(X, X0, Y, X1, X2, app2_3_in_gga3(X1, X2, Z))

The TRS R consists of the following rules:

perm_2_in_ga2(X, ._22(X0, Y)) -> if_perm_2_in_1_ga4(X, X0, Y, app1_3_in_aag3(X1, ._22(X0, X2), X))
app1_3_in_aag3(._22(X0, X), Y, ._22(X0, Z)) -> if_app1_3_in_1_aag5(X0, X, Y, Z, app1_3_in_aag3(X, Y, Z))
app1_3_in_aag3([]_0, Y, Y) -> app1_3_out_aag3([]_0, Y, Y)
if_app1_3_in_1_aag5(X0, X, Y, Z, app1_3_out_aag3(X, Y, Z)) -> app1_3_out_aag3(._22(X0, X), Y, ._22(X0, Z))
if_perm_2_in_1_ga4(X, X0, Y, app1_3_out_aag3(X1, ._22(X0, X2), X)) -> if_perm_2_in_2_ga6(X, X0, Y, X1, X2, app2_3_in_gga3(X1, X2, Z))
app2_3_in_gga3(._22(X0, X), Y, ._22(X0, Z)) -> if_app2_3_in_1_gga5(X0, X, Y, Z, app2_3_in_gga3(X, Y, Z))
app2_3_in_gga3([]_0, Y, Y) -> app2_3_out_gga3([]_0, Y, Y)
if_app2_3_in_1_gga5(X0, X, Y, Z, app2_3_out_gga3(X, Y, Z)) -> app2_3_out_gga3(._22(X0, X), Y, ._22(X0, Z))
if_perm_2_in_2_ga6(X, X0, Y, X1, X2, app2_3_out_gga3(X1, X2, Z)) -> if_perm_2_in_3_ga5(X, X0, Y, Z, perm_2_in_ga2(Z, Y))
perm_2_in_ga2([]_0, []_0) -> perm_2_out_ga2([]_0, []_0)
if_perm_2_in_3_ga5(X, X0, Y, Z, perm_2_out_ga2(Z, Y)) -> perm_2_out_ga2(X, ._22(X0, Y))

The argument filtering Pi contains the following mapping:
perm_2_in_ga2(x1, x2)  =  perm_2_in_ga1(x1)
._22(x1, x2)  =  ._22(x1, x2)
[]_0  =  []_0
if_perm_2_in_1_ga4(x1, x2, x3, x4)  =  if_perm_2_in_1_ga1(x4)
app1_3_in_aag3(x1, x2, x3)  =  app1_3_in_aag1(x3)
if_app1_3_in_1_aag5(x1, x2, x3, x4, x5)  =  if_app1_3_in_1_aag2(x1, x5)
app1_3_out_aag3(x1, x2, x3)  =  app1_3_out_aag2(x1, x2)
if_perm_2_in_2_ga6(x1, x2, x3, x4, x5, x6)  =  if_perm_2_in_2_ga2(x2, x6)
app2_3_in_gga3(x1, x2, x3)  =  app2_3_in_gga2(x1, x2)
if_app2_3_in_1_gga5(x1, x2, x3, x4, x5)  =  if_app2_3_in_1_gga2(x1, x5)
app2_3_out_gga3(x1, x2, x3)  =  app2_3_out_gga1(x3)
if_perm_2_in_3_ga5(x1, x2, x3, x4, x5)  =  if_perm_2_in_3_ga2(x2, x5)
perm_2_out_ga2(x1, x2)  =  perm_2_out_ga1(x2)
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(X, ._22(X0, Y)) -> IF_PERM_2_IN_1_GA4(X, X0, Y, app1_3_in_aag3(X1, ._22(X0, X2), X))
IF_PERM_2_IN_2_GA6(X, X0, Y, X1, X2, app2_3_out_gga3(X1, X2, Z)) -> PERM_2_IN_GA2(Z, Y)
IF_PERM_2_IN_1_GA4(X, X0, Y, app1_3_out_aag3(X1, ._22(X0, X2), X)) -> IF_PERM_2_IN_2_GA6(X, X0, Y, X1, X2, app2_3_in_gga3(X1, X2, Z))

The TRS R consists of the following rules:

app1_3_in_aag3(._22(X0, X), Y, ._22(X0, Z)) -> if_app1_3_in_1_aag5(X0, X, Y, Z, app1_3_in_aag3(X, Y, Z))
app1_3_in_aag3([]_0, Y, Y) -> app1_3_out_aag3([]_0, Y, Y)
app2_3_in_gga3(._22(X0, X), Y, ._22(X0, Z)) -> if_app2_3_in_1_gga5(X0, X, Y, Z, app2_3_in_gga3(X, Y, Z))
app2_3_in_gga3([]_0, Y, Y) -> app2_3_out_gga3([]_0, Y, Y)
if_app1_3_in_1_aag5(X0, X, Y, Z, app1_3_out_aag3(X, Y, Z)) -> app1_3_out_aag3(._22(X0, X), Y, ._22(X0, Z))
if_app2_3_in_1_gga5(X0, X, Y, Z, app2_3_out_gga3(X, Y, Z)) -> app2_3_out_gga3(._22(X0, X), Y, ._22(X0, Z))

The argument filtering Pi contains the following mapping:
._22(x1, x2)  =  ._22(x1, x2)
[]_0  =  []_0
app1_3_in_aag3(x1, x2, x3)  =  app1_3_in_aag1(x3)
if_app1_3_in_1_aag5(x1, x2, x3, x4, x5)  =  if_app1_3_in_1_aag2(x1, x5)
app1_3_out_aag3(x1, x2, x3)  =  app1_3_out_aag2(x1, x2)
app2_3_in_gga3(x1, x2, x3)  =  app2_3_in_gga2(x1, x2)
if_app2_3_in_1_gga5(x1, x2, x3, x4, x5)  =  if_app2_3_in_1_gga2(x1, x5)
app2_3_out_gga3(x1, x2, x3)  =  app2_3_out_gga1(x3)
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(X) -> IF_PERM_2_IN_1_GA1(app1_3_in_aag1(X))
IF_PERM_2_IN_2_GA2(X0, app2_3_out_gga1(Z)) -> PERM_2_IN_GA1(Z)
IF_PERM_2_IN_1_GA1(app1_3_out_aag2(X1, ._22(X0, X2))) -> IF_PERM_2_IN_2_GA2(X0, app2_3_in_gga2(X1, X2))

The TRS R consists of the following rules:

app1_3_in_aag1(._22(X0, Z)) -> if_app1_3_in_1_aag2(X0, app1_3_in_aag1(Z))
app1_3_in_aag1(Y) -> app1_3_out_aag2([]_0, Y)
app2_3_in_gga2(._22(X0, X), Y) -> if_app2_3_in_1_gga2(X0, app2_3_in_gga2(X, Y))
app2_3_in_gga2([]_0, Y) -> app2_3_out_gga1(Y)
if_app1_3_in_1_aag2(X0, app1_3_out_aag2(X, Y)) -> app1_3_out_aag2(._22(X0, X), Y)
if_app2_3_in_1_gga2(X0, app2_3_out_gga1(Z)) -> app2_3_out_gga1(._22(X0, Z))

The set Q consists of the following terms:

app1_3_in_aag1(x0)
app2_3_in_gga2(x0, x1)
if_app1_3_in_1_aag2(x0, x1)
if_app2_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:

app2_3_in_gga2([]_0, Y) -> app2_3_out_gga1(Y)

Used ordering: POLO with Polynomial interpretation:

POL(if_app2_3_in_1_gga2(x1, x2)) = 1 + x1 + x2   
POL(._22(x1, x2)) = 1 + x1 + x2   
POL(IF_PERM_2_IN_2_GA2(x1, x2)) = 1 + x1 + x2   
POL(IF_PERM_2_IN_1_GA1(x1)) = x1   
POL(app2_3_in_gga2(x1, x2)) = x1 + x2   
POL(app2_3_out_gga1(x1)) = 1 + x1   
POL(app1_3_in_aag1(x1)) = 2 + x1   
POL(if_app1_3_in_1_aag2(x1, x2)) = 1 + x1 + x2   
POL(app1_3_out_aag2(x1, x2)) = x1 + x2   
POL([]_0) = 2   
POL(PERM_2_IN_GA1(x1)) = 2 + x1   



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

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

PERM_2_IN_GA1(X) -> IF_PERM_2_IN_1_GA1(app1_3_in_aag1(X))
IF_PERM_2_IN_2_GA2(X0, app2_3_out_gga1(Z)) -> PERM_2_IN_GA1(Z)
IF_PERM_2_IN_1_GA1(app1_3_out_aag2(X1, ._22(X0, X2))) -> IF_PERM_2_IN_2_GA2(X0, app2_3_in_gga2(X1, X2))

The TRS R consists of the following rules:

app1_3_in_aag1(._22(X0, Z)) -> if_app1_3_in_1_aag2(X0, app1_3_in_aag1(Z))
app1_3_in_aag1(Y) -> app1_3_out_aag2([]_0, Y)
app2_3_in_gga2(._22(X0, X), Y) -> if_app2_3_in_1_gga2(X0, app2_3_in_gga2(X, Y))
if_app1_3_in_1_aag2(X0, app1_3_out_aag2(X, Y)) -> app1_3_out_aag2(._22(X0, X), Y)
if_app2_3_in_1_gga2(X0, app2_3_out_gga1(Z)) -> app2_3_out_gga1(._22(X0, Z))

The set Q consists of the following terms:

app1_3_in_aag1(x0)
app2_3_in_gga2(x0, x1)
if_app1_3_in_1_aag2(x0, x1)
if_app2_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 dependency pairs:

PERM_2_IN_GA1(X) -> IF_PERM_2_IN_1_GA1(app1_3_in_aag1(X))
IF_PERM_2_IN_2_GA2(X0, app2_3_out_gga1(Z)) -> PERM_2_IN_GA1(Z)


Used ordering: POLO with Polynomial interpretation:

POL(if_app2_3_in_1_gga2(x1, x2)) = x1 + x2   
POL(._22(x1, x2)) = x1 + x2   
POL(IF_PERM_2_IN_2_GA2(x1, x2)) = x1 + x2   
POL(IF_PERM_2_IN_1_GA1(x1)) = x1   
POL(app2_3_in_gga2(x1, x2)) = x1 + x2   
POL(app2_3_out_gga1(x1)) = 2 + x1   
POL(app1_3_in_aag1(x1)) = x1   
POL(if_app1_3_in_1_aag2(x1, x2)) = x1 + x2   
POL(app1_3_out_aag2(x1, x2)) = x1 + x2   
POL([]_0) = 0   
POL(PERM_2_IN_GA1(x1)) = 1 + x1   



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

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

IF_PERM_2_IN_1_GA1(app1_3_out_aag2(X1, ._22(X0, X2))) -> IF_PERM_2_IN_2_GA2(X0, app2_3_in_gga2(X1, X2))

The TRS R consists of the following rules:

app1_3_in_aag1(._22(X0, Z)) -> if_app1_3_in_1_aag2(X0, app1_3_in_aag1(Z))
app1_3_in_aag1(Y) -> app1_3_out_aag2([]_0, Y)
app2_3_in_gga2(._22(X0, X), Y) -> if_app2_3_in_1_gga2(X0, app2_3_in_gga2(X, Y))
if_app1_3_in_1_aag2(X0, app1_3_out_aag2(X, Y)) -> app1_3_out_aag2(._22(X0, X), Y)
if_app2_3_in_1_gga2(X0, app2_3_out_gga1(Z)) -> app2_3_out_gga1(._22(X0, Z))

The set Q consists of the following terms:

app1_3_in_aag1(x0)
app2_3_in_gga2(x0, x1)
if_app1_3_in_1_aag2(x0, x1)
if_app2_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_2_GA2, IF_PERM_2_IN_1_GA1}.
The approximation of the Dependency Graph contains 0 SCCs with 1 less node.