Left Termination proof of /tmp/tmpcm7M9b/quicksort-bf.pl

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

↳ PROLOG

  ↳ PrologToPiTRSProof

qs₂({}₀, {}₀).
qs₂(.₂(X, Xs), Ys) :- part₄(X, Xs, Littles, Bigs), qs₂(Littles, Ls), qs₂(Bigs, Bs), app₃(Ls, .₂(X, Bs), Ys).
part₄(X, .₂(Y, Xs), .₂(Y, Ls), Bs) :- less₂(X, Y), part₄(X, Xs, Ls, Bs).
part₄(X, .₂(Y, Xs), Ls, .₂(Y, Bs)) :- part₄(X, Xs, Ls, Bs).
part₄(underscore, {}₀, {}₀, {}₀).
app₃({}₀, X, X).
app₃(.₂(X, Xs), Ys, .₂(X, Zs)) :- app₃(Xs, Ys, Zs).
less₂(0₀, s₁(underscore1)).
less₂(s₁(X), s₁(Y)) :- less₂(X, Y).

With regard to the inferred argument filtering the predicates were used in the following modes:
qs₂: (b,f)
part₄: (b,b,f,f)
less₂: (b,b)
app₃: (b,b,f)
Transforming PROLOG into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

qs_2_in_ga₂([]_0, []_0) -> qs_2_out_ga₂([]_0, []_0)
qs_2_in_ga₂(._2₂(X, Xs), Ys) -> if_qs_2_in_1_ga₄(X, Xs, Ys, part_4_in_ggaa₄(X, Xs, Littles, Bigs))
part_4_in_ggaa₄(X, ._2₂(Y, Xs), ._2₂(Y, Ls), Bs) -> if_part_4_in_1_ggaa₆(X, Y, Xs, Ls, Bs, less_2_in_gg₂(X, Y))
less_2_in_gg₂(0_0, s_1₁(underscore1)) -> less_2_out_gg₂(0_0, s_1₁(underscore1))
less_2_in_gg₂(s_1₁(X), s_1₁(Y)) -> if_less_2_in_1_gg₃(X, Y, less_2_in_gg₂(X, Y))
if_less_2_in_1_gg₃(X, Y, less_2_out_gg₂(X, Y)) -> less_2_out_gg₂(s_1₁(X), s_1₁(Y))
if_part_4_in_1_ggaa₆(X, Y, Xs, Ls, Bs, less_2_out_gg₂(X, Y)) -> if_part_4_in_2_ggaa₆(X, Y, Xs, Ls, Bs, part_4_in_ggaa₄(X, Xs, Ls, Bs))
part_4_in_ggaa₄(X, ._2₂(Y, Xs), Ls, ._2₂(Y, Bs)) -> if_part_4_in_3_ggaa₆(X, Y, Xs, Ls, Bs, part_4_in_ggaa₄(X, Xs, Ls, Bs))
part_4_in_ggaa₄(underscore, []_0, []_0, []_0) -> part_4_out_ggaa₄(underscore, []_0, []_0, []_0)
if_part_4_in_3_ggaa₆(X, Y, Xs, Ls, Bs, part_4_out_ggaa₄(X, Xs, Ls, Bs)) -> part_4_out_ggaa₄(X, ._2₂(Y, Xs), Ls, ._2₂(Y, Bs))
if_part_4_in_2_ggaa₆(X, Y, Xs, Ls, Bs, part_4_out_ggaa₄(X, Xs, Ls, Bs)) -> part_4_out_ggaa₄(X, ._2₂(Y, Xs), ._2₂(Y, Ls), Bs)
if_qs_2_in_1_ga₄(X, Xs, Ys, part_4_out_ggaa₄(X, Xs, Littles, Bigs)) -> if_qs_2_in_2_ga₆(X, Xs, Ys, Littles, Bigs, qs_2_in_ga₂(Littles, Ls))
if_qs_2_in_2_ga₆(X, Xs, Ys, Littles, Bigs, qs_2_out_ga₂(Littles, Ls)) -> if_qs_2_in_3_ga₆(X, Xs, Ys, Bigs, Ls, qs_2_in_ga₂(Bigs, Bs))
if_qs_2_in_3_ga₆(X, Xs, Ys, Bigs, Ls, qs_2_out_ga₂(Bigs, Bs)) -> if_qs_2_in_4_ga₆(X, Xs, Ys, Ls, Bs, app_3_in_gga₃(Ls, ._2₂(X, Bs), Ys))
app_3_in_gga₃([]_0, X, X) -> app_3_out_gga₃([]_0, X, X)
app_3_in_gga₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_app_3_in_1_gga₅(X, Xs, Ys, Zs, app_3_in_gga₃(Xs, Ys, Zs))
if_app_3_in_1_gga₅(X, Xs, Ys, Zs, app_3_out_gga₃(Xs, Ys, Zs)) -> app_3_out_gga₃(._2₂(X, Xs), Ys, ._2₂(X, Zs))
if_qs_2_in_4_ga₆(X, Xs, Ys, Ls, Bs, app_3_out_gga₃(Ls, ._2₂(X, Bs), Ys)) -> qs_2_out_ga₂(._2₂(X, Xs), Ys)

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:

qs_2_in_ga₂([]_0, []_0) -> qs_2_out_ga₂([]_0, []_0)
qs_2_in_ga₂(._2₂(X, Xs), Ys) -> if_qs_2_in_1_ga₄(X, Xs, Ys, part_4_in_ggaa₄(X, Xs, Littles, Bigs))
part_4_in_ggaa₄(X, ._2₂(Y, Xs), ._2₂(Y, Ls), Bs) -> if_part_4_in_1_ggaa₆(X, Y, Xs, Ls, Bs, less_2_in_gg₂(X, Y))
less_2_in_gg₂(0_0, s_1₁(underscore1)) -> less_2_out_gg₂(0_0, s_1₁(underscore1))
less_2_in_gg₂(s_1₁(X), s_1₁(Y)) -> if_less_2_in_1_gg₃(X, Y, less_2_in_gg₂(X, Y))
if_less_2_in_1_gg₃(X, Y, less_2_out_gg₂(X, Y)) -> less_2_out_gg₂(s_1₁(X), s_1₁(Y))
if_part_4_in_1_ggaa₆(X, Y, Xs, Ls, Bs, less_2_out_gg₂(X, Y)) -> if_part_4_in_2_ggaa₆(X, Y, Xs, Ls, Bs, part_4_in_ggaa₄(X, Xs, Ls, Bs))
part_4_in_ggaa₄(X, ._2₂(Y, Xs), Ls, ._2₂(Y, Bs)) -> if_part_4_in_3_ggaa₆(X, Y, Xs, Ls, Bs, part_4_in_ggaa₄(X, Xs, Ls, Bs))
part_4_in_ggaa₄(underscore, []_0, []_0, []_0) -> part_4_out_ggaa₄(underscore, []_0, []_0, []_0)
if_part_4_in_3_ggaa₆(X, Y, Xs, Ls, Bs, part_4_out_ggaa₄(X, Xs, Ls, Bs)) -> part_4_out_ggaa₄(X, ._2₂(Y, Xs), Ls, ._2₂(Y, Bs))
if_part_4_in_2_ggaa₆(X, Y, Xs, Ls, Bs, part_4_out_ggaa₄(X, Xs, Ls, Bs)) -> part_4_out_ggaa₄(X, ._2₂(Y, Xs), ._2₂(Y, Ls), Bs)
if_qs_2_in_1_ga₄(X, Xs, Ys, part_4_out_ggaa₄(X, Xs, Littles, Bigs)) -> if_qs_2_in_2_ga₆(X, Xs, Ys, Littles, Bigs, qs_2_in_ga₂(Littles, Ls))
if_qs_2_in_2_ga₆(X, Xs, Ys, Littles, Bigs, qs_2_out_ga₂(Littles, Ls)) -> if_qs_2_in_3_ga₆(X, Xs, Ys, Bigs, Ls, qs_2_in_ga₂(Bigs, Bs))
if_qs_2_in_3_ga₆(X, Xs, Ys, Bigs, Ls, qs_2_out_ga₂(Bigs, Bs)) -> if_qs_2_in_4_ga₆(X, Xs, Ys, Ls, Bs, app_3_in_gga₃(Ls, ._2₂(X, Bs), Ys))
app_3_in_gga₃([]_0, X, X) -> app_3_out_gga₃([]_0, X, X)
app_3_in_gga₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_app_3_in_1_gga₅(X, Xs, Ys, Zs, app_3_in_gga₃(Xs, Ys, Zs))
if_app_3_in_1_gga₅(X, Xs, Ys, Zs, app_3_out_gga₃(Xs, Ys, Zs)) -> app_3_out_gga₃(._2₂(X, Xs), Ys, ._2₂(X, Zs))
if_qs_2_in_4_ga₆(X, Xs, Ys, Ls, Bs, app_3_out_gga₃(Ls, ._2₂(X, Bs), Ys)) -> qs_2_out_ga₂(._2₂(X, Xs), Ys)

QS_2_IN_GA₂(._2₂(X, Xs), Ys) -> IF_QS_2_IN_1_GA₄(X, Xs, Ys, part_4_in_ggaa₄(X, Xs, Littles, Bigs))
QS_2_IN_GA₂(._2₂(X, Xs), Ys) -> PART_4_IN_GGAA₄(X, Xs, Littles, Bigs)
PART_4_IN_GGAA₄(X, ._2₂(Y, Xs), ._2₂(Y, Ls), Bs) -> IF_PART_4_IN_1_GGAA₆(X, Y, Xs, Ls, Bs, less_2_in_gg₂(X, Y))
PART_4_IN_GGAA₄(X, ._2₂(Y, Xs), ._2₂(Y, Ls), Bs) -> LESS_2_IN_GG₂(X, Y)
LESS_2_IN_GG₂(s_1₁(X), s_1₁(Y)) -> IF_LESS_2_IN_1_GG₃(X, Y, less_2_in_gg₂(X, Y))
LESS_2_IN_GG₂(s_1₁(X), s_1₁(Y)) -> LESS_2_IN_GG₂(X, Y)
IF_PART_4_IN_1_GGAA₆(X, Y, Xs, Ls, Bs, less_2_out_gg₂(X, Y)) -> IF_PART_4_IN_2_GGAA₆(X, Y, Xs, Ls, Bs, part_4_in_ggaa₄(X, Xs, Ls, Bs))
IF_PART_4_IN_1_GGAA₆(X, Y, Xs, Ls, Bs, less_2_out_gg₂(X, Y)) -> PART_4_IN_GGAA₄(X, Xs, Ls, Bs)
PART_4_IN_GGAA₄(X, ._2₂(Y, Xs), Ls, ._2₂(Y, Bs)) -> IF_PART_4_IN_3_GGAA₆(X, Y, Xs, Ls, Bs, part_4_in_ggaa₄(X, Xs, Ls, Bs))
PART_4_IN_GGAA₄(X, ._2₂(Y, Xs), Ls, ._2₂(Y, Bs)) -> PART_4_IN_GGAA₄(X, Xs, Ls, Bs)
IF_QS_2_IN_1_GA₄(X, Xs, Ys, part_4_out_ggaa₄(X, Xs, Littles, Bigs)) -> IF_QS_2_IN_2_GA₆(X, Xs, Ys, Littles, Bigs, qs_2_in_ga₂(Littles, Ls))
IF_QS_2_IN_1_GA₄(X, Xs, Ys, part_4_out_ggaa₄(X, Xs, Littles, Bigs)) -> QS_2_IN_GA₂(Littles, Ls)
IF_QS_2_IN_2_GA₆(X, Xs, Ys, Littles, Bigs, qs_2_out_ga₂(Littles, Ls)) -> IF_QS_2_IN_3_GA₆(X, Xs, Ys, Bigs, Ls, qs_2_in_ga₂(Bigs, Bs))
IF_QS_2_IN_2_GA₆(X, Xs, Ys, Littles, Bigs, qs_2_out_ga₂(Littles, Ls)) -> QS_2_IN_GA₂(Bigs, Bs)
IF_QS_2_IN_3_GA₆(X, Xs, Ys, Bigs, Ls, qs_2_out_ga₂(Bigs, Bs)) -> IF_QS_2_IN_4_GA₆(X, Xs, Ys, Ls, Bs, app_3_in_gga₃(Ls, ._2₂(X, Bs), Ys))
IF_QS_2_IN_3_GA₆(X, Xs, Ys, Bigs, Ls, qs_2_out_ga₂(Bigs, Bs)) -> APP_3_IN_GGA₃(Ls, ._2₂(X, Bs), Ys)
APP_3_IN_GGA₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> IF_APP_3_IN_1_GGA₅(X, Xs, Ys, Zs, app_3_in_gga₃(Xs, Ys, Zs))
APP_3_IN_GGA₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> APP_3_IN_GGA₃(Xs, Ys, Zs)

The TRS R consists of the following rules:

qs_2_in_ga₂([]_0, []_0) -> qs_2_out_ga₂([]_0, []_0)
qs_2_in_ga₂(._2₂(X, Xs), Ys) -> if_qs_2_in_1_ga₄(X, Xs, Ys, part_4_in_ggaa₄(X, Xs, Littles, Bigs))
part_4_in_ggaa₄(X, ._2₂(Y, Xs), ._2₂(Y, Ls), Bs) -> if_part_4_in_1_ggaa₆(X, Y, Xs, Ls, Bs, less_2_in_gg₂(X, Y))
less_2_in_gg₂(0_0, s_1₁(underscore1)) -> less_2_out_gg₂(0_0, s_1₁(underscore1))
less_2_in_gg₂(s_1₁(X), s_1₁(Y)) -> if_less_2_in_1_gg₃(X, Y, less_2_in_gg₂(X, Y))
if_less_2_in_1_gg₃(X, Y, less_2_out_gg₂(X, Y)) -> less_2_out_gg₂(s_1₁(X), s_1₁(Y))
if_part_4_in_1_ggaa₆(X, Y, Xs, Ls, Bs, less_2_out_gg₂(X, Y)) -> if_part_4_in_2_ggaa₆(X, Y, Xs, Ls, Bs, part_4_in_ggaa₄(X, Xs, Ls, Bs))
part_4_in_ggaa₄(X, ._2₂(Y, Xs), Ls, ._2₂(Y, Bs)) -> if_part_4_in_3_ggaa₆(X, Y, Xs, Ls, Bs, part_4_in_ggaa₄(X, Xs, Ls, Bs))
part_4_in_ggaa₄(underscore, []_0, []_0, []_0) -> part_4_out_ggaa₄(underscore, []_0, []_0, []_0)
if_part_4_in_3_ggaa₆(X, Y, Xs, Ls, Bs, part_4_out_ggaa₄(X, Xs, Ls, Bs)) -> part_4_out_ggaa₄(X, ._2₂(Y, Xs), Ls, ._2₂(Y, Bs))
if_part_4_in_2_ggaa₆(X, Y, Xs, Ls, Bs, part_4_out_ggaa₄(X, Xs, Ls, Bs)) -> part_4_out_ggaa₄(X, ._2₂(Y, Xs), ._2₂(Y, Ls), Bs)
if_qs_2_in_1_ga₄(X, Xs, Ys, part_4_out_ggaa₄(X, Xs, Littles, Bigs)) -> if_qs_2_in_2_ga₆(X, Xs, Ys, Littles, Bigs, qs_2_in_ga₂(Littles, Ls))
if_qs_2_in_2_ga₆(X, Xs, Ys, Littles, Bigs, qs_2_out_ga₂(Littles, Ls)) -> if_qs_2_in_3_ga₆(X, Xs, Ys, Bigs, Ls, qs_2_in_ga₂(Bigs, Bs))
if_qs_2_in_3_ga₆(X, Xs, Ys, Bigs, Ls, qs_2_out_ga₂(Bigs, Bs)) -> if_qs_2_in_4_ga₆(X, Xs, Ys, Ls, Bs, app_3_in_gga₃(Ls, ._2₂(X, Bs), Ys))
app_3_in_gga₃([]_0, X, X) -> app_3_out_gga₃([]_0, X, X)
app_3_in_gga₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_app_3_in_1_gga₅(X, Xs, Ys, Zs, app_3_in_gga₃(Xs, Ys, Zs))
if_app_3_in_1_gga₅(X, Xs, Ys, Zs, app_3_out_gga₃(Xs, Ys, Zs)) -> app_3_out_gga₃(._2₂(X, Xs), Ys, ._2₂(X, Zs))
if_qs_2_in_4_ga₆(X, Xs, Ys, Ls, Bs, app_3_out_gga₃(Ls, ._2₂(X, Bs), Ys)) -> qs_2_out_ga₂(._2₂(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
qs_2_in_ga₂(x1, x2) = qs_2_in_ga₁(x1)
[]_0 = []_0
._2₂(x1, x2) = ._2₂(x1, x2)
0_0 = 0_0
s_1₁(x1) = s_1₁(x1)
qs_2_out_ga₂(x1, x2) = qs_2_out_ga₁(x2)
if_qs_2_in_1_ga₄(x1, x2, x3, x4) = if_qs_2_in_1_ga₂(x1, x4)
part_4_in_ggaa₄(x1, x2, x3, x4) = part_4_in_ggaa₂(x1, x2)
if_part_4_in_1_ggaa₆(x1, x2, x3, x4, x5, x6) = if_part_4_in_1_ggaa₄(x1, x2, x3, x6)
less_2_in_gg₂(x1, x2) = less_2_in_gg₂(x1, x2)
less_2_out_gg₂(x1, x2) = less_2_out_gg
if_less_2_in_1_gg₃(x1, x2, x3) = if_less_2_in_1_gg₁(x3)
if_part_4_in_2_ggaa₆(x1, x2, x3, x4, x5, x6) = if_part_4_in_2_ggaa₂(x2, x6)
if_part_4_in_3_ggaa₆(x1, x2, x3, x4, x5, x6) = if_part_4_in_3_ggaa₂(x2, x6)
part_4_out_ggaa₄(x1, x2, x3, x4) = part_4_out_ggaa₂(x3, x4)
if_qs_2_in_2_ga₆(x1, x2, x3, x4, x5, x6) = if_qs_2_in_2_ga₃(x1, x5, x6)
if_qs_2_in_3_ga₆(x1, x2, x3, x4, x5, x6) = if_qs_2_in_3_ga₃(x1, x5, x6)
if_qs_2_in_4_ga₆(x1, x2, x3, x4, x5, x6) = if_qs_2_in_4_ga₁(x6)
app_3_in_gga₃(x1, x2, x3) = app_3_in_gga₂(x1, x2)
app_3_out_gga₃(x1, x2, x3) = app_3_out_gga₁(x3)
if_app_3_in_1_gga₅(x1, x2, x3, x4, x5) = if_app_3_in_1_gga₂(x1, x5)
PART_4_IN_GGAA₄(x1, x2, x3, x4) = PART_4_IN_GGAA₂(x1, x2)
IF_PART_4_IN_3_GGAA₆(x1, x2, x3, x4, x5, x6) = IF_PART_4_IN_3_GGAA₂(x2, x6)
IF_QS_2_IN_1_GA₄(x1, x2, x3, x4) = IF_QS_2_IN_1_GA₂(x1, x4)
IF_QS_2_IN_3_GA₆(x1, x2, x3, x4, x5, x6) = IF_QS_2_IN_3_GA₃(x1, x5, x6)
APP_3_IN_GGA₃(x1, x2, x3) = APP_3_IN_GGA₂(x1, x2)
IF_PART_4_IN_1_GGAA₆(x1, x2, x3, x4, x5, x6) = IF_PART_4_IN_1_GGAA₄(x1, x2, x3, x6)
IF_QS_2_IN_2_GA₆(x1, x2, x3, x4, x5, x6) = IF_QS_2_IN_2_GA₃(x1, x5, x6)
IF_APP_3_IN_1_GGA₅(x1, x2, x3, x4, x5) = IF_APP_3_IN_1_GGA₂(x1, x5)
QS_2_IN_GA₂(x1, x2) = QS_2_IN_GA₁(x1)
IF_PART_4_IN_2_GGAA₆(x1, x2, x3, x4, x5, x6) = IF_PART_4_IN_2_GGAA₂(x2, x6)
IF_LESS_2_IN_1_GG₃(x1, x2, x3) = IF_LESS_2_IN_1_GG₁(x3)
IF_QS_2_IN_4_GA₆(x1, x2, x3, x4, x5, x6) = IF_QS_2_IN_4_GA₁(x6)
LESS_2_IN_GG₂(x1, x2) = LESS_2_IN_GG₂(x1, x2)

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:

QS_2_IN_GA₂(._2₂(X, Xs), Ys) -> IF_QS_2_IN_1_GA₄(X, Xs, Ys, part_4_in_ggaa₄(X, Xs, Littles, Bigs))
QS_2_IN_GA₂(._2₂(X, Xs), Ys) -> PART_4_IN_GGAA₄(X, Xs, Littles, Bigs)
PART_4_IN_GGAA₄(X, ._2₂(Y, Xs), ._2₂(Y, Ls), Bs) -> IF_PART_4_IN_1_GGAA₆(X, Y, Xs, Ls, Bs, less_2_in_gg₂(X, Y))
PART_4_IN_GGAA₄(X, ._2₂(Y, Xs), ._2₂(Y, Ls), Bs) -> LESS_2_IN_GG₂(X, Y)
LESS_2_IN_GG₂(s_1₁(X), s_1₁(Y)) -> IF_LESS_2_IN_1_GG₃(X, Y, less_2_in_gg₂(X, Y))
LESS_2_IN_GG₂(s_1₁(X), s_1₁(Y)) -> LESS_2_IN_GG₂(X, Y)
IF_PART_4_IN_1_GGAA₆(X, Y, Xs, Ls, Bs, less_2_out_gg₂(X, Y)) -> IF_PART_4_IN_2_GGAA₆(X, Y, Xs, Ls, Bs, part_4_in_ggaa₄(X, Xs, Ls, Bs))
IF_PART_4_IN_1_GGAA₆(X, Y, Xs, Ls, Bs, less_2_out_gg₂(X, Y)) -> PART_4_IN_GGAA₄(X, Xs, Ls, Bs)
PART_4_IN_GGAA₄(X, ._2₂(Y, Xs), Ls, ._2₂(Y, Bs)) -> IF_PART_4_IN_3_GGAA₆(X, Y, Xs, Ls, Bs, part_4_in_ggaa₄(X, Xs, Ls, Bs))
PART_4_IN_GGAA₄(X, ._2₂(Y, Xs), Ls, ._2₂(Y, Bs)) -> PART_4_IN_GGAA₄(X, Xs, Ls, Bs)
IF_QS_2_IN_1_GA₄(X, Xs, Ys, part_4_out_ggaa₄(X, Xs, Littles, Bigs)) -> IF_QS_2_IN_2_GA₆(X, Xs, Ys, Littles, Bigs, qs_2_in_ga₂(Littles, Ls))
IF_QS_2_IN_1_GA₄(X, Xs, Ys, part_4_out_ggaa₄(X, Xs, Littles, Bigs)) -> QS_2_IN_GA₂(Littles, Ls)
IF_QS_2_IN_2_GA₆(X, Xs, Ys, Littles, Bigs, qs_2_out_ga₂(Littles, Ls)) -> IF_QS_2_IN_3_GA₆(X, Xs, Ys, Bigs, Ls, qs_2_in_ga₂(Bigs, Bs))
IF_QS_2_IN_2_GA₆(X, Xs, Ys, Littles, Bigs, qs_2_out_ga₂(Littles, Ls)) -> QS_2_IN_GA₂(Bigs, Bs)
IF_QS_2_IN_3_GA₆(X, Xs, Ys, Bigs, Ls, qs_2_out_ga₂(Bigs, Bs)) -> IF_QS_2_IN_4_GA₆(X, Xs, Ys, Ls, Bs, app_3_in_gga₃(Ls, ._2₂(X, Bs), Ys))
IF_QS_2_IN_3_GA₆(X, Xs, Ys, Bigs, Ls, qs_2_out_ga₂(Bigs, Bs)) -> APP_3_IN_GGA₃(Ls, ._2₂(X, Bs), Ys)
APP_3_IN_GGA₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> IF_APP_3_IN_1_GGA₅(X, Xs, Ys, Zs, app_3_in_gga₃(Xs, Ys, Zs))
APP_3_IN_GGA₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> APP_3_IN_GGA₃(Xs, Ys, Zs)

The TRS R consists of the following rules:

qs_2_in_ga₂([]_0, []_0) -> qs_2_out_ga₂([]_0, []_0)
qs_2_in_ga₂(._2₂(X, Xs), Ys) -> if_qs_2_in_1_ga₄(X, Xs, Ys, part_4_in_ggaa₄(X, Xs, Littles, Bigs))
part_4_in_ggaa₄(X, ._2₂(Y, Xs), ._2₂(Y, Ls), Bs) -> if_part_4_in_1_ggaa₆(X, Y, Xs, Ls, Bs, less_2_in_gg₂(X, Y))
less_2_in_gg₂(0_0, s_1₁(underscore1)) -> less_2_out_gg₂(0_0, s_1₁(underscore1))
less_2_in_gg₂(s_1₁(X), s_1₁(Y)) -> if_less_2_in_1_gg₃(X, Y, less_2_in_gg₂(X, Y))
if_less_2_in_1_gg₃(X, Y, less_2_out_gg₂(X, Y)) -> less_2_out_gg₂(s_1₁(X), s_1₁(Y))
if_part_4_in_1_ggaa₆(X, Y, Xs, Ls, Bs, less_2_out_gg₂(X, Y)) -> if_part_4_in_2_ggaa₆(X, Y, Xs, Ls, Bs, part_4_in_ggaa₄(X, Xs, Ls, Bs))
part_4_in_ggaa₄(X, ._2₂(Y, Xs), Ls, ._2₂(Y, Bs)) -> if_part_4_in_3_ggaa₆(X, Y, Xs, Ls, Bs, part_4_in_ggaa₄(X, Xs, Ls, Bs))
part_4_in_ggaa₄(underscore, []_0, []_0, []_0) -> part_4_out_ggaa₄(underscore, []_0, []_0, []_0)
if_part_4_in_3_ggaa₆(X, Y, Xs, Ls, Bs, part_4_out_ggaa₄(X, Xs, Ls, Bs)) -> part_4_out_ggaa₄(X, ._2₂(Y, Xs), Ls, ._2₂(Y, Bs))
if_part_4_in_2_ggaa₆(X, Y, Xs, Ls, Bs, part_4_out_ggaa₄(X, Xs, Ls, Bs)) -> part_4_out_ggaa₄(X, ._2₂(Y, Xs), ._2₂(Y, Ls), Bs)
if_qs_2_in_1_ga₄(X, Xs, Ys, part_4_out_ggaa₄(X, Xs, Littles, Bigs)) -> if_qs_2_in_2_ga₆(X, Xs, Ys, Littles, Bigs, qs_2_in_ga₂(Littles, Ls))
if_qs_2_in_2_ga₆(X, Xs, Ys, Littles, Bigs, qs_2_out_ga₂(Littles, Ls)) -> if_qs_2_in_3_ga₆(X, Xs, Ys, Bigs, Ls, qs_2_in_ga₂(Bigs, Bs))
if_qs_2_in_3_ga₆(X, Xs, Ys, Bigs, Ls, qs_2_out_ga₂(Bigs, Bs)) -> if_qs_2_in_4_ga₆(X, Xs, Ys, Ls, Bs, app_3_in_gga₃(Ls, ._2₂(X, Bs), Ys))
app_3_in_gga₃([]_0, X, X) -> app_3_out_gga₃([]_0, X, X)
app_3_in_gga₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_app_3_in_1_gga₅(X, Xs, Ys, Zs, app_3_in_gga₃(Xs, Ys, Zs))
if_app_3_in_1_gga₅(X, Xs, Ys, Zs, app_3_out_gga₃(Xs, Ys, Zs)) -> app_3_out_gga₃(._2₂(X, Xs), Ys, ._2₂(X, Zs))
if_qs_2_in_4_ga₆(X, Xs, Ys, Ls, Bs, app_3_out_gga₃(Ls, ._2₂(X, Bs), Ys)) -> qs_2_out_ga₂(._2₂(X, Xs), Ys)

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

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

APP_3_IN_GGA₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> APP_3_IN_GGA₃(Xs, Ys, Zs)

The TRS R consists of the following rules:

qs_2_in_ga₂([]_0, []_0) -> qs_2_out_ga₂([]_0, []_0)
qs_2_in_ga₂(._2₂(X, Xs), Ys) -> if_qs_2_in_1_ga₄(X, Xs, Ys, part_4_in_ggaa₄(X, Xs, Littles, Bigs))
part_4_in_ggaa₄(X, ._2₂(Y, Xs), ._2₂(Y, Ls), Bs) -> if_part_4_in_1_ggaa₆(X, Y, Xs, Ls, Bs, less_2_in_gg₂(X, Y))
less_2_in_gg₂(0_0, s_1₁(underscore1)) -> less_2_out_gg₂(0_0, s_1₁(underscore1))
less_2_in_gg₂(s_1₁(X), s_1₁(Y)) -> if_less_2_in_1_gg₃(X, Y, less_2_in_gg₂(X, Y))
if_less_2_in_1_gg₃(X, Y, less_2_out_gg₂(X, Y)) -> less_2_out_gg₂(s_1₁(X), s_1₁(Y))
if_part_4_in_1_ggaa₆(X, Y, Xs, Ls, Bs, less_2_out_gg₂(X, Y)) -> if_part_4_in_2_ggaa₆(X, Y, Xs, Ls, Bs, part_4_in_ggaa₄(X, Xs, Ls, Bs))
part_4_in_ggaa₄(X, ._2₂(Y, Xs), Ls, ._2₂(Y, Bs)) -> if_part_4_in_3_ggaa₆(X, Y, Xs, Ls, Bs, part_4_in_ggaa₄(X, Xs, Ls, Bs))
part_4_in_ggaa₄(underscore, []_0, []_0, []_0) -> part_4_out_ggaa₄(underscore, []_0, []_0, []_0)
if_part_4_in_3_ggaa₆(X, Y, Xs, Ls, Bs, part_4_out_ggaa₄(X, Xs, Ls, Bs)) -> part_4_out_ggaa₄(X, ._2₂(Y, Xs), Ls, ._2₂(Y, Bs))
if_part_4_in_2_ggaa₆(X, Y, Xs, Ls, Bs, part_4_out_ggaa₄(X, Xs, Ls, Bs)) -> part_4_out_ggaa₄(X, ._2₂(Y, Xs), ._2₂(Y, Ls), Bs)
if_qs_2_in_1_ga₄(X, Xs, Ys, part_4_out_ggaa₄(X, Xs, Littles, Bigs)) -> if_qs_2_in_2_ga₆(X, Xs, Ys, Littles, Bigs, qs_2_in_ga₂(Littles, Ls))
if_qs_2_in_2_ga₆(X, Xs, Ys, Littles, Bigs, qs_2_out_ga₂(Littles, Ls)) -> if_qs_2_in_3_ga₆(X, Xs, Ys, Bigs, Ls, qs_2_in_ga₂(Bigs, Bs))
if_qs_2_in_3_ga₆(X, Xs, Ys, Bigs, Ls, qs_2_out_ga₂(Bigs, Bs)) -> if_qs_2_in_4_ga₆(X, Xs, Ys, Ls, Bs, app_3_in_gga₃(Ls, ._2₂(X, Bs), Ys))
app_3_in_gga₃([]_0, X, X) -> app_3_out_gga₃([]_0, X, X)
app_3_in_gga₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_app_3_in_1_gga₅(X, Xs, Ys, Zs, app_3_in_gga₃(Xs, Ys, Zs))
if_app_3_in_1_gga₅(X, Xs, Ys, Zs, app_3_out_gga₃(Xs, Ys, Zs)) -> app_3_out_gga₃(._2₂(X, Xs), Ys, ._2₂(X, Zs))
if_qs_2_in_4_ga₆(X, Xs, Ys, Ls, Bs, app_3_out_gga₃(Ls, ._2₂(X, Bs), Ys)) -> qs_2_out_ga₂(._2₂(X, Xs), Ys)

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

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

APP_3_IN_GGA₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> APP_3_IN_GGA₃(Xs, Ys, Zs)

R is empty.
The argument filtering Pi contains the following mapping:
._2₂(x1, x2) = ._2₂(x1, x2)
APP_3_IN_GGA₃(x1, x2, x3) = APP_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

              ↳ PiDP

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

APP_3_IN_GGA₂(._2₂(X, Xs), Ys) -> APP_3_IN_GGA₂(Xs, Ys)

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

APP_3_IN_GGA₂(._2₂(X, Xs), Ys) -> APP_3_IN_GGA₂(Xs, Ys)
The graph contains the following edges 1 > 1, 2 >= 2

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

              ↳ PiDP

              ↳ PiDP

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

LESS_2_IN_GG₂(s_1₁(X), s_1₁(Y)) -> LESS_2_IN_GG₂(X, Y)

The TRS R consists of the following rules:

qs_2_in_ga₂([]_0, []_0) -> qs_2_out_ga₂([]_0, []_0)
qs_2_in_ga₂(._2₂(X, Xs), Ys) -> if_qs_2_in_1_ga₄(X, Xs, Ys, part_4_in_ggaa₄(X, Xs, Littles, Bigs))
part_4_in_ggaa₄(X, ._2₂(Y, Xs), ._2₂(Y, Ls), Bs) -> if_part_4_in_1_ggaa₆(X, Y, Xs, Ls, Bs, less_2_in_gg₂(X, Y))
less_2_in_gg₂(0_0, s_1₁(underscore1)) -> less_2_out_gg₂(0_0, s_1₁(underscore1))
less_2_in_gg₂(s_1₁(X), s_1₁(Y)) -> if_less_2_in_1_gg₃(X, Y, less_2_in_gg₂(X, Y))
if_less_2_in_1_gg₃(X, Y, less_2_out_gg₂(X, Y)) -> less_2_out_gg₂(s_1₁(X), s_1₁(Y))
if_part_4_in_1_ggaa₆(X, Y, Xs, Ls, Bs, less_2_out_gg₂(X, Y)) -> if_part_4_in_2_ggaa₆(X, Y, Xs, Ls, Bs, part_4_in_ggaa₄(X, Xs, Ls, Bs))
part_4_in_ggaa₄(X, ._2₂(Y, Xs), Ls, ._2₂(Y, Bs)) -> if_part_4_in_3_ggaa₆(X, Y, Xs, Ls, Bs, part_4_in_ggaa₄(X, Xs, Ls, Bs))
part_4_in_ggaa₄(underscore, []_0, []_0, []_0) -> part_4_out_ggaa₄(underscore, []_0, []_0, []_0)
if_part_4_in_3_ggaa₆(X, Y, Xs, Ls, Bs, part_4_out_ggaa₄(X, Xs, Ls, Bs)) -> part_4_out_ggaa₄(X, ._2₂(Y, Xs), Ls, ._2₂(Y, Bs))
if_part_4_in_2_ggaa₆(X, Y, Xs, Ls, Bs, part_4_out_ggaa₄(X, Xs, Ls, Bs)) -> part_4_out_ggaa₄(X, ._2₂(Y, Xs), ._2₂(Y, Ls), Bs)
if_qs_2_in_1_ga₄(X, Xs, Ys, part_4_out_ggaa₄(X, Xs, Littles, Bigs)) -> if_qs_2_in_2_ga₆(X, Xs, Ys, Littles, Bigs, qs_2_in_ga₂(Littles, Ls))
if_qs_2_in_2_ga₆(X, Xs, Ys, Littles, Bigs, qs_2_out_ga₂(Littles, Ls)) -> if_qs_2_in_3_ga₆(X, Xs, Ys, Bigs, Ls, qs_2_in_ga₂(Bigs, Bs))
if_qs_2_in_3_ga₆(X, Xs, Ys, Bigs, Ls, qs_2_out_ga₂(Bigs, Bs)) -> if_qs_2_in_4_ga₆(X, Xs, Ys, Ls, Bs, app_3_in_gga₃(Ls, ._2₂(X, Bs), Ys))
app_3_in_gga₃([]_0, X, X) -> app_3_out_gga₃([]_0, X, X)
app_3_in_gga₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_app_3_in_1_gga₅(X, Xs, Ys, Zs, app_3_in_gga₃(Xs, Ys, Zs))
if_app_3_in_1_gga₅(X, Xs, Ys, Zs, app_3_out_gga₃(Xs, Ys, Zs)) -> app_3_out_gga₃(._2₂(X, Xs), Ys, ._2₂(X, Zs))
if_qs_2_in_4_ga₆(X, Xs, Ys, Ls, Bs, app_3_out_gga₃(Ls, ._2₂(X, Bs), Ys)) -> qs_2_out_ga₂(._2₂(X, Xs), Ys)

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

              ↳ PiDP

              ↳ PiDP

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

LESS_2_IN_GG₂(s_1₁(X), s_1₁(Y)) -> LESS_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

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ PiDP

              ↳ PiDP

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

LESS_2_IN_GG₂(s_1₁(X), s_1₁(Y)) -> LESS_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 {LESS_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:

LESS_2_IN_GG₂(s_1₁(X), s_1₁(Y)) -> LESS_2_IN_GG₂(X, Y)
The graph contains the following edges 1 > 1, 2 > 2

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

              ↳ PiDP

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

PART_4_IN_GGAA₄(X, ._2₂(Y, Xs), Ls, ._2₂(Y, Bs)) -> PART_4_IN_GGAA₄(X, Xs, Ls, Bs)
PART_4_IN_GGAA₄(X, ._2₂(Y, Xs), ._2₂(Y, Ls), Bs) -> IF_PART_4_IN_1_GGAA₆(X, Y, Xs, Ls, Bs, less_2_in_gg₂(X, Y))
IF_PART_4_IN_1_GGAA₆(X, Y, Xs, Ls, Bs, less_2_out_gg₂(X, Y)) -> PART_4_IN_GGAA₄(X, Xs, Ls, Bs)

The TRS R consists of the following rules:

qs_2_in_ga₂([]_0, []_0) -> qs_2_out_ga₂([]_0, []_0)
qs_2_in_ga₂(._2₂(X, Xs), Ys) -> if_qs_2_in_1_ga₄(X, Xs, Ys, part_4_in_ggaa₄(X, Xs, Littles, Bigs))
part_4_in_ggaa₄(X, ._2₂(Y, Xs), ._2₂(Y, Ls), Bs) -> if_part_4_in_1_ggaa₆(X, Y, Xs, Ls, Bs, less_2_in_gg₂(X, Y))
less_2_in_gg₂(0_0, s_1₁(underscore1)) -> less_2_out_gg₂(0_0, s_1₁(underscore1))
less_2_in_gg₂(s_1₁(X), s_1₁(Y)) -> if_less_2_in_1_gg₃(X, Y, less_2_in_gg₂(X, Y))
if_less_2_in_1_gg₃(X, Y, less_2_out_gg₂(X, Y)) -> less_2_out_gg₂(s_1₁(X), s_1₁(Y))
if_part_4_in_1_ggaa₆(X, Y, Xs, Ls, Bs, less_2_out_gg₂(X, Y)) -> if_part_4_in_2_ggaa₆(X, Y, Xs, Ls, Bs, part_4_in_ggaa₄(X, Xs, Ls, Bs))
part_4_in_ggaa₄(X, ._2₂(Y, Xs), Ls, ._2₂(Y, Bs)) -> if_part_4_in_3_ggaa₆(X, Y, Xs, Ls, Bs, part_4_in_ggaa₄(X, Xs, Ls, Bs))
part_4_in_ggaa₄(underscore, []_0, []_0, []_0) -> part_4_out_ggaa₄(underscore, []_0, []_0, []_0)
if_part_4_in_3_ggaa₆(X, Y, Xs, Ls, Bs, part_4_out_ggaa₄(X, Xs, Ls, Bs)) -> part_4_out_ggaa₄(X, ._2₂(Y, Xs), Ls, ._2₂(Y, Bs))
if_part_4_in_2_ggaa₆(X, Y, Xs, Ls, Bs, part_4_out_ggaa₄(X, Xs, Ls, Bs)) -> part_4_out_ggaa₄(X, ._2₂(Y, Xs), ._2₂(Y, Ls), Bs)
if_qs_2_in_1_ga₄(X, Xs, Ys, part_4_out_ggaa₄(X, Xs, Littles, Bigs)) -> if_qs_2_in_2_ga₆(X, Xs, Ys, Littles, Bigs, qs_2_in_ga₂(Littles, Ls))
if_qs_2_in_2_ga₆(X, Xs, Ys, Littles, Bigs, qs_2_out_ga₂(Littles, Ls)) -> if_qs_2_in_3_ga₆(X, Xs, Ys, Bigs, Ls, qs_2_in_ga₂(Bigs, Bs))
if_qs_2_in_3_ga₆(X, Xs, Ys, Bigs, Ls, qs_2_out_ga₂(Bigs, Bs)) -> if_qs_2_in_4_ga₆(X, Xs, Ys, Ls, Bs, app_3_in_gga₃(Ls, ._2₂(X, Bs), Ys))
app_3_in_gga₃([]_0, X, X) -> app_3_out_gga₃([]_0, X, X)
app_3_in_gga₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_app_3_in_1_gga₅(X, Xs, Ys, Zs, app_3_in_gga₃(Xs, Ys, Zs))
if_app_3_in_1_gga₅(X, Xs, Ys, Zs, app_3_out_gga₃(Xs, Ys, Zs)) -> app_3_out_gga₃(._2₂(X, Xs), Ys, ._2₂(X, Zs))
if_qs_2_in_4_ga₆(X, Xs, Ys, Ls, Bs, app_3_out_gga₃(Ls, ._2₂(X, Bs), Ys)) -> qs_2_out_ga₂(._2₂(X, Xs), Ys)

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

              ↳ PiDP

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

PART_4_IN_GGAA₄(X, ._2₂(Y, Xs), Ls, ._2₂(Y, Bs)) -> PART_4_IN_GGAA₄(X, Xs, Ls, Bs)
PART_4_IN_GGAA₄(X, ._2₂(Y, Xs), ._2₂(Y, Ls), Bs) -> IF_PART_4_IN_1_GGAA₆(X, Y, Xs, Ls, Bs, less_2_in_gg₂(X, Y))
IF_PART_4_IN_1_GGAA₆(X, Y, Xs, Ls, Bs, less_2_out_gg₂(X, Y)) -> PART_4_IN_GGAA₄(X, Xs, Ls, Bs)

The TRS R consists of the following rules:

less_2_in_gg₂(0_0, s_1₁(underscore1)) -> less_2_out_gg₂(0_0, s_1₁(underscore1))
less_2_in_gg₂(s_1₁(X), s_1₁(Y)) -> if_less_2_in_1_gg₃(X, Y, less_2_in_gg₂(X, Y))
if_less_2_in_1_gg₃(X, Y, less_2_out_gg₂(X, Y)) -> less_2_out_gg₂(s_1₁(X), s_1₁(Y))

The argument filtering Pi contains the following mapping:
._2₂(x1, x2) = ._2₂(x1, x2)
0_0 = 0_0
s_1₁(x1) = s_1₁(x1)
less_2_in_gg₂(x1, x2) = less_2_in_gg₂(x1, x2)
less_2_out_gg₂(x1, x2) = less_2_out_gg
if_less_2_in_1_gg₃(x1, x2, x3) = if_less_2_in_1_gg₁(x3)
PART_4_IN_GGAA₄(x1, x2, x3, x4) = PART_4_IN_GGAA₂(x1, x2)
IF_PART_4_IN_1_GGAA₆(x1, x2, x3, x4, x5, x6) = IF_PART_4_IN_1_GGAA₄(x1, x2, x3, x6)

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

                        ↳ QDPSizeChangeProof

              ↳ PiDP

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

PART_4_IN_GGAA₂(X, ._2₂(Y, Xs)) -> PART_4_IN_GGAA₂(X, Xs)
PART_4_IN_GGAA₂(X, ._2₂(Y, Xs)) -> IF_PART_4_IN_1_GGAA₄(X, Y, Xs, less_2_in_gg₂(X, Y))
IF_PART_4_IN_1_GGAA₄(X, Y, Xs, less_2_out_gg) -> PART_4_IN_GGAA₂(X, Xs)

The TRS R consists of the following rules:

less_2_in_gg₂(0_0, s_1₁(underscore1)) -> less_2_out_gg
less_2_in_gg₂(s_1₁(X), s_1₁(Y)) -> if_less_2_in_1_gg₁(less_2_in_gg₂(X, Y))
if_less_2_in_1_gg₁(less_2_out_gg) -> less_2_out_gg

The set Q consists of the following terms:

less_2_in_gg₂(x0, x1)
if_less_2_in_1_gg₁(x0)

We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {PART_4_IN_GGAA₂, IF_PART_4_IN_1_GGAA₄}.
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:

PART_4_IN_GGAA₂(X, ._2₂(Y, Xs)) -> PART_4_IN_GGAA₂(X, Xs)
The graph contains the following edges 1 >= 1, 2 > 2
PART_4_IN_GGAA₂(X, ._2₂(Y, Xs)) -> IF_PART_4_IN_1_GGAA₄(X, Y, Xs, less_2_in_gg₂(X, Y))
The graph contains the following edges 1 >= 1, 2 > 2, 2 > 3
IF_PART_4_IN_1_GGAA₄(X, Y, Xs, less_2_out_gg) -> PART_4_IN_GGAA₂(X, Xs)
The graph contains the following edges 1 >= 1, 3 >= 2

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ PiDPToQDPProof

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

IF_QS_2_IN_2_GA₆(X, Xs, Ys, Littles, Bigs, qs_2_out_ga₂(Littles, Ls)) -> QS_2_IN_GA₂(Bigs, Bs)
IF_QS_2_IN_1_GA₄(X, Xs, Ys, part_4_out_ggaa₄(X, Xs, Littles, Bigs)) -> QS_2_IN_GA₂(Littles, Ls)
QS_2_IN_GA₂(._2₂(X, Xs), Ys) -> IF_QS_2_IN_1_GA₄(X, Xs, Ys, part_4_in_ggaa₄(X, Xs, Littles, Bigs))
IF_QS_2_IN_1_GA₄(X, Xs, Ys, part_4_out_ggaa₄(X, Xs, Littles, Bigs)) -> IF_QS_2_IN_2_GA₆(X, Xs, Ys, Littles, Bigs, qs_2_in_ga₂(Littles, Ls))

The TRS R consists of the following rules:

qs_2_in_ga₂([]_0, []_0) -> qs_2_out_ga₂([]_0, []_0)
qs_2_in_ga₂(._2₂(X, Xs), Ys) -> if_qs_2_in_1_ga₄(X, Xs, Ys, part_4_in_ggaa₄(X, Xs, Littles, Bigs))
part_4_in_ggaa₄(X, ._2₂(Y, Xs), ._2₂(Y, Ls), Bs) -> if_part_4_in_1_ggaa₆(X, Y, Xs, Ls, Bs, less_2_in_gg₂(X, Y))
less_2_in_gg₂(0_0, s_1₁(underscore1)) -> less_2_out_gg₂(0_0, s_1₁(underscore1))
less_2_in_gg₂(s_1₁(X), s_1₁(Y)) -> if_less_2_in_1_gg₃(X, Y, less_2_in_gg₂(X, Y))
if_less_2_in_1_gg₃(X, Y, less_2_out_gg₂(X, Y)) -> less_2_out_gg₂(s_1₁(X), s_1₁(Y))
if_part_4_in_1_ggaa₆(X, Y, Xs, Ls, Bs, less_2_out_gg₂(X, Y)) -> if_part_4_in_2_ggaa₆(X, Y, Xs, Ls, Bs, part_4_in_ggaa₄(X, Xs, Ls, Bs))
part_4_in_ggaa₄(X, ._2₂(Y, Xs), Ls, ._2₂(Y, Bs)) -> if_part_4_in_3_ggaa₆(X, Y, Xs, Ls, Bs, part_4_in_ggaa₄(X, Xs, Ls, Bs))
part_4_in_ggaa₄(underscore, []_0, []_0, []_0) -> part_4_out_ggaa₄(underscore, []_0, []_0, []_0)
if_part_4_in_3_ggaa₆(X, Y, Xs, Ls, Bs, part_4_out_ggaa₄(X, Xs, Ls, Bs)) -> part_4_out_ggaa₄(X, ._2₂(Y, Xs), Ls, ._2₂(Y, Bs))
if_part_4_in_2_ggaa₆(X, Y, Xs, Ls, Bs, part_4_out_ggaa₄(X, Xs, Ls, Bs)) -> part_4_out_ggaa₄(X, ._2₂(Y, Xs), ._2₂(Y, Ls), Bs)
if_qs_2_in_1_ga₄(X, Xs, Ys, part_4_out_ggaa₄(X, Xs, Littles, Bigs)) -> if_qs_2_in_2_ga₆(X, Xs, Ys, Littles, Bigs, qs_2_in_ga₂(Littles, Ls))
if_qs_2_in_2_ga₆(X, Xs, Ys, Littles, Bigs, qs_2_out_ga₂(Littles, Ls)) -> if_qs_2_in_3_ga₆(X, Xs, Ys, Bigs, Ls, qs_2_in_ga₂(Bigs, Bs))
if_qs_2_in_3_ga₆(X, Xs, Ys, Bigs, Ls, qs_2_out_ga₂(Bigs, Bs)) -> if_qs_2_in_4_ga₆(X, Xs, Ys, Ls, Bs, app_3_in_gga₃(Ls, ._2₂(X, Bs), Ys))
app_3_in_gga₃([]_0, X, X) -> app_3_out_gga₃([]_0, X, X)
app_3_in_gga₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_app_3_in_1_gga₅(X, Xs, Ys, Zs, app_3_in_gga₃(Xs, Ys, Zs))
if_app_3_in_1_gga₅(X, Xs, Ys, Zs, app_3_out_gga₃(Xs, Ys, Zs)) -> app_3_out_gga₃(._2₂(X, Xs), Ys, ._2₂(X, Zs))
if_qs_2_in_4_ga₆(X, Xs, Ys, Ls, Bs, app_3_out_gga₃(Ls, ._2₂(X, Bs), Ys)) -> qs_2_out_ga₂(._2₂(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
qs_2_in_ga₂(x1, x2) = qs_2_in_ga₁(x1)
[]_0 = []_0
._2₂(x1, x2) = ._2₂(x1, x2)
0_0 = 0_0
s_1₁(x1) = s_1₁(x1)
qs_2_out_ga₂(x1, x2) = qs_2_out_ga₁(x2)
if_qs_2_in_1_ga₄(x1, x2, x3, x4) = if_qs_2_in_1_ga₂(x1, x4)
part_4_in_ggaa₄(x1, x2, x3, x4) = part_4_in_ggaa₂(x1, x2)
if_part_4_in_1_ggaa₆(x1, x2, x3, x4, x5, x6) = if_part_4_in_1_ggaa₄(x1, x2, x3, x6)
less_2_in_gg₂(x1, x2) = less_2_in_gg₂(x1, x2)
less_2_out_gg₂(x1, x2) = less_2_out_gg
if_less_2_in_1_gg₃(x1, x2, x3) = if_less_2_in_1_gg₁(x3)
if_part_4_in_2_ggaa₆(x1, x2, x3, x4, x5, x6) = if_part_4_in_2_ggaa₂(x2, x6)
if_part_4_in_3_ggaa₆(x1, x2, x3, x4, x5, x6) = if_part_4_in_3_ggaa₂(x2, x6)
part_4_out_ggaa₄(x1, x2, x3, x4) = part_4_out_ggaa₂(x3, x4)
if_qs_2_in_2_ga₆(x1, x2, x3, x4, x5, x6) = if_qs_2_in_2_ga₃(x1, x5, x6)
if_qs_2_in_3_ga₆(x1, x2, x3, x4, x5, x6) = if_qs_2_in_3_ga₃(x1, x5, x6)
if_qs_2_in_4_ga₆(x1, x2, x3, x4, x5, x6) = if_qs_2_in_4_ga₁(x6)
app_3_in_gga₃(x1, x2, x3) = app_3_in_gga₂(x1, x2)
app_3_out_gga₃(x1, x2, x3) = app_3_out_gga₁(x3)
if_app_3_in_1_gga₅(x1, x2, x3, x4, x5) = if_app_3_in_1_gga₂(x1, x5)
IF_QS_2_IN_1_GA₄(x1, x2, x3, x4) = IF_QS_2_IN_1_GA₂(x1, x4)
IF_QS_2_IN_2_GA₆(x1, x2, x3, x4, x5, x6) = IF_QS_2_IN_2_GA₃(x1, x5, x6)
QS_2_IN_GA₂(x1, x2) = QS_2_IN_GA₁(x1)

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

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ PiDPToQDPProof

                  ↳ QDP

                    ↳ QDPPoloProof

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

IF_QS_2_IN_2_GA₃(X, Bigs, qs_2_out_ga₁(Ls)) -> QS_2_IN_GA₁(Bigs)
IF_QS_2_IN_1_GA₂(X, part_4_out_ggaa₂(Littles, Bigs)) -> QS_2_IN_GA₁(Littles)
QS_2_IN_GA₁(._2₂(X, Xs)) -> IF_QS_2_IN_1_GA₂(X, part_4_in_ggaa₂(X, Xs))
IF_QS_2_IN_1_GA₂(X, part_4_out_ggaa₂(Littles, Bigs)) -> IF_QS_2_IN_2_GA₃(X, Bigs, qs_2_in_ga₁(Littles))

The TRS R consists of the following rules:

qs_2_in_ga₁([]_0) -> qs_2_out_ga₁([]_0)
qs_2_in_ga₁(._2₂(X, Xs)) -> if_qs_2_in_1_ga₂(X, part_4_in_ggaa₂(X, Xs))
part_4_in_ggaa₂(X, ._2₂(Y, Xs)) -> if_part_4_in_1_ggaa₄(X, Y, Xs, less_2_in_gg₂(X, Y))
less_2_in_gg₂(0_0, s_1₁(underscore1)) -> less_2_out_gg
less_2_in_gg₂(s_1₁(X), s_1₁(Y)) -> if_less_2_in_1_gg₁(less_2_in_gg₂(X, Y))
if_less_2_in_1_gg₁(less_2_out_gg) -> less_2_out_gg
if_part_4_in_1_ggaa₄(X, Y, Xs, less_2_out_gg) -> if_part_4_in_2_ggaa₂(Y, part_4_in_ggaa₂(X, Xs))
part_4_in_ggaa₂(X, ._2₂(Y, Xs)) -> if_part_4_in_3_ggaa₂(Y, part_4_in_ggaa₂(X, Xs))
part_4_in_ggaa₂(underscore, []_0) -> part_4_out_ggaa₂([]_0, []_0)
if_part_4_in_3_ggaa₂(Y, part_4_out_ggaa₂(Ls, Bs)) -> part_4_out_ggaa₂(Ls, ._2₂(Y, Bs))
if_part_4_in_2_ggaa₂(Y, part_4_out_ggaa₂(Ls, Bs)) -> part_4_out_ggaa₂(._2₂(Y, Ls), Bs)
if_qs_2_in_1_ga₂(X, part_4_out_ggaa₂(Littles, Bigs)) -> if_qs_2_in_2_ga₃(X, Bigs, qs_2_in_ga₁(Littles))
if_qs_2_in_2_ga₃(X, Bigs, qs_2_out_ga₁(Ls)) -> if_qs_2_in_3_ga₃(X, Ls, qs_2_in_ga₁(Bigs))
if_qs_2_in_3_ga₃(X, Ls, qs_2_out_ga₁(Bs)) -> if_qs_2_in_4_ga₁(app_3_in_gga₂(Ls, ._2₂(X, Bs)))
app_3_in_gga₂([]_0, X) -> app_3_out_gga₁(X)
app_3_in_gga₂(._2₂(X, Xs), Ys) -> if_app_3_in_1_gga₂(X, app_3_in_gga₂(Xs, Ys))
if_app_3_in_1_gga₂(X, app_3_out_gga₁(Zs)) -> app_3_out_gga₁(._2₂(X, Zs))
if_qs_2_in_4_ga₁(app_3_out_gga₁(Ys)) -> qs_2_out_ga₁(Ys)

The set Q consists of the following terms:

qs_2_in_ga₁(x0)
part_4_in_ggaa₂(x0, x1)
less_2_in_gg₂(x0, x1)
if_less_2_in_1_gg₁(x0)
if_part_4_in_1_ggaa₄(x0, x1, x2, x3)
if_part_4_in_3_ggaa₂(x0, x1)
if_part_4_in_2_ggaa₂(x0, x1)
if_qs_2_in_1_ga₂(x0, x1)
if_qs_2_in_2_ga₃(x0, x1, x2)
if_qs_2_in_3_ga₃(x0, x1, x2)
app_3_in_gga₂(x0, x1)
if_app_3_in_1_gga₂(x0, x1)
if_qs_2_in_4_ga₁(x0)

We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {QS_2_IN_GA₁, IF_QS_2_IN_2_GA₃, IF_QS_2_IN_1_GA₂}.
By using a polynomial ordering, the following set of Dependency Pairs of this DP problem can be strictly oriented.

QS_2_IN_GA₁(._2₂(X, Xs)) -> IF_QS_2_IN_1_GA₂(X, part_4_in_ggaa₂(X, Xs))

The remaining Dependency Pairs were at least non-strictly be oriented.

IF_QS_2_IN_2_GA₃(X, Bigs, qs_2_out_ga₁(Ls)) -> QS_2_IN_GA₁(Bigs)
IF_QS_2_IN_1_GA₂(X, part_4_out_ggaa₂(Littles, Bigs)) -> QS_2_IN_GA₁(Littles)
IF_QS_2_IN_1_GA₂(X, part_4_out_ggaa₂(Littles, Bigs)) -> IF_QS_2_IN_2_GA₃(X, Bigs, qs_2_in_ga₁(Littles))

With the implicit AFS we had to orient the following set of usable rules non-strictly.

part_4_in_ggaa₂(underscore, []_0) -> part_4_out_ggaa₂([]_0, []_0)
part_4_in_ggaa₂(X, ._2₂(Y, Xs)) -> if_part_4_in_1_ggaa₄(X, Y, Xs, less_2_in_gg₂(X, Y))
part_4_in_ggaa₂(X, ._2₂(Y, Xs)) -> if_part_4_in_3_ggaa₂(Y, part_4_in_ggaa₂(X, Xs))
if_part_4_in_3_ggaa₂(Y, part_4_out_ggaa₂(Ls, Bs)) -> part_4_out_ggaa₂(Ls, ._2₂(Y, Bs))
if_part_4_in_1_ggaa₄(X, Y, Xs, less_2_out_gg) -> if_part_4_in_2_ggaa₂(Y, part_4_in_ggaa₂(X, Xs))
if_part_4_in_2_ggaa₂(Y, part_4_out_ggaa₂(Ls, Bs)) -> part_4_out_ggaa₂(._2₂(Y, Ls), Bs)

Used ordering: POLO with Polynomial interpretation:

POL(if_qs_2_in_4_ga₁(x₁)) = 0
POL(less_2_out_gg) = 0
POL(0_0) = 0
POL(IF_QS_2_IN_1_GA₂(x₁, x₂)) = x₂
POL(qs_2_in_ga₁(x₁)) = 0
POL(if_app_3_in_1_gga₂(x₁, x₂)) = 0
POL(part_4_in_ggaa₂(x₁, x₂)) = x₂
POL(qs_2_out_ga₁(x₁)) = 0
POL(if_part_4_in_1_ggaa₄(x₁, x₂, x₃, x₄)) = 1 + x₃
POL(if_qs_2_in_1_ga₂(x₁, x₂)) = 0
POL([]_0) = 0
POL(if_qs_2_in_2_ga₃(x₁, x₂, x₃)) = 0
POL(part_4_out_ggaa₂(x₁, x₂)) = x₁ + x₂
POL(._2₂(x₁, x₂)) = 1 + x₂
POL(if_less_2_in_1_gg₁(x₁)) = 0
POL(app_3_out_gga₁(x₁)) = 0
POL(IF_QS_2_IN_2_GA₃(x₁, x₂, x₃)) = x₂
POL(if_part_4_in_3_ggaa₂(x₁, x₂)) = 1 + x₂
POL(QS_2_IN_GA₁(x₁)) = x₁
POL(if_part_4_in_2_ggaa₂(x₁, x₂)) = 1 + x₂
POL(s_1₁(x₁)) = 0
POL(if_qs_2_in_3_ga₃(x₁, x₂, x₃)) = 0
POL(app_3_in_gga₂(x₁, x₂)) = 0
POL(less_2_in_gg₂(x₁, x₂)) = 0

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ PiDPToQDPProof

                  ↳ QDP

                    ↳ QDPPoloProof

                      ↳ QDP

                        ↳ DependencyGraphProof

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

IF_QS_2_IN_2_GA₃(X, Bigs, qs_2_out_ga₁(Ls)) -> QS_2_IN_GA₁(Bigs)
IF_QS_2_IN_1_GA₂(X, part_4_out_ggaa₂(Littles, Bigs)) -> QS_2_IN_GA₁(Littles)
IF_QS_2_IN_1_GA₂(X, part_4_out_ggaa₂(Littles, Bigs)) -> IF_QS_2_IN_2_GA₃(X, Bigs, qs_2_in_ga₁(Littles))

The TRS R consists of the following rules:

qs_2_in_ga₁([]_0) -> qs_2_out_ga₁([]_0)
qs_2_in_ga₁(._2₂(X, Xs)) -> if_qs_2_in_1_ga₂(X, part_4_in_ggaa₂(X, Xs))
part_4_in_ggaa₂(X, ._2₂(Y, Xs)) -> if_part_4_in_1_ggaa₄(X, Y, Xs, less_2_in_gg₂(X, Y))
less_2_in_gg₂(0_0, s_1₁(underscore1)) -> less_2_out_gg
less_2_in_gg₂(s_1₁(X), s_1₁(Y)) -> if_less_2_in_1_gg₁(less_2_in_gg₂(X, Y))
if_less_2_in_1_gg₁(less_2_out_gg) -> less_2_out_gg
if_part_4_in_1_ggaa₄(X, Y, Xs, less_2_out_gg) -> if_part_4_in_2_ggaa₂(Y, part_4_in_ggaa₂(X, Xs))
part_4_in_ggaa₂(X, ._2₂(Y, Xs)) -> if_part_4_in_3_ggaa₂(Y, part_4_in_ggaa₂(X, Xs))
part_4_in_ggaa₂(underscore, []_0) -> part_4_out_ggaa₂([]_0, []_0)
if_part_4_in_3_ggaa₂(Y, part_4_out_ggaa₂(Ls, Bs)) -> part_4_out_ggaa₂(Ls, ._2₂(Y, Bs))
if_part_4_in_2_ggaa₂(Y, part_4_out_ggaa₂(Ls, Bs)) -> part_4_out_ggaa₂(._2₂(Y, Ls), Bs)
if_qs_2_in_1_ga₂(X, part_4_out_ggaa₂(Littles, Bigs)) -> if_qs_2_in_2_ga₃(X, Bigs, qs_2_in_ga₁(Littles))
if_qs_2_in_2_ga₃(X, Bigs, qs_2_out_ga₁(Ls)) -> if_qs_2_in_3_ga₃(X, Ls, qs_2_in_ga₁(Bigs))
if_qs_2_in_3_ga₃(X, Ls, qs_2_out_ga₁(Bs)) -> if_qs_2_in_4_ga₁(app_3_in_gga₂(Ls, ._2₂(X, Bs)))
app_3_in_gga₂([]_0, X) -> app_3_out_gga₁(X)
app_3_in_gga₂(._2₂(X, Xs), Ys) -> if_app_3_in_1_gga₂(X, app_3_in_gga₂(Xs, Ys))
if_app_3_in_1_gga₂(X, app_3_out_gga₁(Zs)) -> app_3_out_gga₁(._2₂(X, Zs))
if_qs_2_in_4_ga₁(app_3_out_gga₁(Ys)) -> qs_2_out_ga₁(Ys)

The set Q consists of the following terms:

qs_2_in_ga₁(x0)
part_4_in_ggaa₂(x0, x1)
less_2_in_gg₂(x0, x1)
if_less_2_in_1_gg₁(x0)
if_part_4_in_1_ggaa₄(x0, x1, x2, x3)
if_part_4_in_3_ggaa₂(x0, x1)
if_part_4_in_2_ggaa₂(x0, x1)
if_qs_2_in_1_ga₂(x0, x1)
if_qs_2_in_2_ga₃(x0, x1, x2)
if_qs_2_in_3_ga₃(x0, x1, x2)
app_3_in_gga₂(x0, x1)
if_app_3_in_1_gga₂(x0, x1)
if_qs_2_in_4_ga₁(x0)

We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {QS_2_IN_GA₁, IF_QS_2_IN_2_GA₃, IF_QS_2_IN_1_GA₂}.
The approximation of the Dependency Graph contains 0 SCCs with 3 less nodes.