Left Termination proof of /tmp/tmpEJmfmq/qsort.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(cons₂(X, L), S) :- split₄(L, X, L1, L2), qs₂(L1, S1), qs₂(L2, S2), append₃(S1, cons₂(X, S2), S).
append₃({}₀, L, L).
append₃(cons₂(X, L1), L2, cons₂(X, L3)) :- append₃(L1, L2, L3).
split₄({}₀, X, {}₀, {}₀).
split₄(cons₂(X, L), Y, cons₂(X, L1), L2) :- less₂(X, Y), split₄(L, Y, L1, L2).
split₄(cons₂(X, L), Y, L1, cons₂(X, L2)) :- geq₂(X, Y), split₄(L, Y, L1, L2).
less₂(0₀, s₁(X)).
less₂(s₁(X), s₁(Y)) :- less₂(X, Y).
geq₂(X, X).
geq₂(s₁(X), 0₀).
geq₂(s₁(X), s₁(Y)) :- geq₂(X, Y).

With regard to the inferred argument filtering the predicates were used in the following modes:
qs₂: (b,f)
split₄: (b,b,f,f)
less₂: (b,b)
geq₂: (b,b)
append₃: (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(cons_2₂(X, L), S) -> if_qs_2_in_1_ga₄(X, L, S, split_4_in_ggaa₄(L, X, L1, L2))
split_4_in_ggaa₄([]_0, X, []_0, []_0) -> split_4_out_ggaa₄([]_0, X, []_0, []_0)
split_4_in_ggaa₄(cons_2₂(X, L), Y, cons_2₂(X, L1), L2) -> if_split_4_in_1_ggaa₆(X, L, Y, L1, L2, less_2_in_gg₂(X, Y))
less_2_in_gg₂(0_0, s_1₁(X)) -> less_2_out_gg₂(0_0, s_1₁(X))
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_split_4_in_1_ggaa₆(X, L, Y, L1, L2, less_2_out_gg₂(X, Y)) -> if_split_4_in_2_ggaa₆(X, L, Y, L1, L2, split_4_in_ggaa₄(L, Y, L1, L2))
split_4_in_ggaa₄(cons_2₂(X, L), Y, L1, cons_2₂(X, L2)) -> if_split_4_in_3_ggaa₆(X, L, Y, L1, L2, geq_2_in_gg₂(X, Y))
geq_2_in_gg₂(X, X) -> geq_2_out_gg₂(X, X)
geq_2_in_gg₂(s_1₁(X), 0_0) -> geq_2_out_gg₂(s_1₁(X), 0_0)
geq_2_in_gg₂(s_1₁(X), s_1₁(Y)) -> if_geq_2_in_1_gg₃(X, Y, geq_2_in_gg₂(X, Y))
if_geq_2_in_1_gg₃(X, Y, geq_2_out_gg₂(X, Y)) -> geq_2_out_gg₂(s_1₁(X), s_1₁(Y))
if_split_4_in_3_ggaa₆(X, L, Y, L1, L2, geq_2_out_gg₂(X, Y)) -> if_split_4_in_4_ggaa₆(X, L, Y, L1, L2, split_4_in_ggaa₄(L, Y, L1, L2))
if_split_4_in_4_ggaa₆(X, L, Y, L1, L2, split_4_out_ggaa₄(L, Y, L1, L2)) -> split_4_out_ggaa₄(cons_2₂(X, L), Y, L1, cons_2₂(X, L2))
if_split_4_in_2_ggaa₆(X, L, Y, L1, L2, split_4_out_ggaa₄(L, Y, L1, L2)) -> split_4_out_ggaa₄(cons_2₂(X, L), Y, cons_2₂(X, L1), L2)
if_qs_2_in_1_ga₄(X, L, S, split_4_out_ggaa₄(L, X, L1, L2)) -> if_qs_2_in_2_ga₆(X, L, S, L1, L2, qs_2_in_ga₂(L1, S1))
if_qs_2_in_2_ga₆(X, L, S, L1, L2, qs_2_out_ga₂(L1, S1)) -> if_qs_2_in_3_ga₆(X, L, S, L2, S1, qs_2_in_ga₂(L2, S2))
if_qs_2_in_3_ga₆(X, L, S, L2, S1, qs_2_out_ga₂(L2, S2)) -> if_qs_2_in_4_ga₆(X, L, S, S1, S2, append_3_in_gga₃(S1, cons_2₂(X, S2), S))
append_3_in_gga₃([]_0, L, L) -> append_3_out_gga₃([]_0, L, L)
append_3_in_gga₃(cons_2₂(X, L1), L2, cons_2₂(X, L3)) -> if_append_3_in_1_gga₅(X, L1, L2, L3, append_3_in_gga₃(L1, L2, L3))
if_append_3_in_1_gga₅(X, L1, L2, L3, append_3_out_gga₃(L1, L2, L3)) -> append_3_out_gga₃(cons_2₂(X, L1), L2, cons_2₂(X, L3))
if_qs_2_in_4_ga₆(X, L, S, S1, S2, append_3_out_gga₃(S1, cons_2₂(X, S2), S)) -> qs_2_out_ga₂(cons_2₂(X, L), S)

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(cons_2₂(X, L), S) -> if_qs_2_in_1_ga₄(X, L, S, split_4_in_ggaa₄(L, X, L1, L2))
split_4_in_ggaa₄([]_0, X, []_0, []_0) -> split_4_out_ggaa₄([]_0, X, []_0, []_0)
split_4_in_ggaa₄(cons_2₂(X, L), Y, cons_2₂(X, L1), L2) -> if_split_4_in_1_ggaa₆(X, L, Y, L1, L2, less_2_in_gg₂(X, Y))
less_2_in_gg₂(0_0, s_1₁(X)) -> less_2_out_gg₂(0_0, s_1₁(X))
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_split_4_in_1_ggaa₆(X, L, Y, L1, L2, less_2_out_gg₂(X, Y)) -> if_split_4_in_2_ggaa₆(X, L, Y, L1, L2, split_4_in_ggaa₄(L, Y, L1, L2))
split_4_in_ggaa₄(cons_2₂(X, L), Y, L1, cons_2₂(X, L2)) -> if_split_4_in_3_ggaa₆(X, L, Y, L1, L2, geq_2_in_gg₂(X, Y))
geq_2_in_gg₂(X, X) -> geq_2_out_gg₂(X, X)
geq_2_in_gg₂(s_1₁(X), 0_0) -> geq_2_out_gg₂(s_1₁(X), 0_0)
geq_2_in_gg₂(s_1₁(X), s_1₁(Y)) -> if_geq_2_in_1_gg₃(X, Y, geq_2_in_gg₂(X, Y))
if_geq_2_in_1_gg₃(X, Y, geq_2_out_gg₂(X, Y)) -> geq_2_out_gg₂(s_1₁(X), s_1₁(Y))
if_split_4_in_3_ggaa₆(X, L, Y, L1, L2, geq_2_out_gg₂(X, Y)) -> if_split_4_in_4_ggaa₆(X, L, Y, L1, L2, split_4_in_ggaa₄(L, Y, L1, L2))
if_split_4_in_4_ggaa₆(X, L, Y, L1, L2, split_4_out_ggaa₄(L, Y, L1, L2)) -> split_4_out_ggaa₄(cons_2₂(X, L), Y, L1, cons_2₂(X, L2))
if_split_4_in_2_ggaa₆(X, L, Y, L1, L2, split_4_out_ggaa₄(L, Y, L1, L2)) -> split_4_out_ggaa₄(cons_2₂(X, L), Y, cons_2₂(X, L1), L2)
if_qs_2_in_1_ga₄(X, L, S, split_4_out_ggaa₄(L, X, L1, L2)) -> if_qs_2_in_2_ga₆(X, L, S, L1, L2, qs_2_in_ga₂(L1, S1))
if_qs_2_in_2_ga₆(X, L, S, L1, L2, qs_2_out_ga₂(L1, S1)) -> if_qs_2_in_3_ga₆(X, L, S, L2, S1, qs_2_in_ga₂(L2, S2))
if_qs_2_in_3_ga₆(X, L, S, L2, S1, qs_2_out_ga₂(L2, S2)) -> if_qs_2_in_4_ga₆(X, L, S, S1, S2, append_3_in_gga₃(S1, cons_2₂(X, S2), S))
append_3_in_gga₃([]_0, L, L) -> append_3_out_gga₃([]_0, L, L)
append_3_in_gga₃(cons_2₂(X, L1), L2, cons_2₂(X, L3)) -> if_append_3_in_1_gga₅(X, L1, L2, L3, append_3_in_gga₃(L1, L2, L3))
if_append_3_in_1_gga₅(X, L1, L2, L3, append_3_out_gga₃(L1, L2, L3)) -> append_3_out_gga₃(cons_2₂(X, L1), L2, cons_2₂(X, L3))
if_qs_2_in_4_ga₆(X, L, S, S1, S2, append_3_out_gga₃(S1, cons_2₂(X, S2), S)) -> qs_2_out_ga₂(cons_2₂(X, L), S)

QS_2_IN_GA₂(cons_2₂(X, L), S) -> IF_QS_2_IN_1_GA₄(X, L, S, split_4_in_ggaa₄(L, X, L1, L2))
QS_2_IN_GA₂(cons_2₂(X, L), S) -> SPLIT_4_IN_GGAA₄(L, X, L1, L2)
SPLIT_4_IN_GGAA₄(cons_2₂(X, L), Y, cons_2₂(X, L1), L2) -> IF_SPLIT_4_IN_1_GGAA₆(X, L, Y, L1, L2, less_2_in_gg₂(X, Y))
SPLIT_4_IN_GGAA₄(cons_2₂(X, L), Y, cons_2₂(X, L1), L2) -> 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_SPLIT_4_IN_1_GGAA₆(X, L, Y, L1, L2, less_2_out_gg₂(X, Y)) -> IF_SPLIT_4_IN_2_GGAA₆(X, L, Y, L1, L2, split_4_in_ggaa₄(L, Y, L1, L2))
IF_SPLIT_4_IN_1_GGAA₆(X, L, Y, L1, L2, less_2_out_gg₂(X, Y)) -> SPLIT_4_IN_GGAA₄(L, Y, L1, L2)
SPLIT_4_IN_GGAA₄(cons_2₂(X, L), Y, L1, cons_2₂(X, L2)) -> IF_SPLIT_4_IN_3_GGAA₆(X, L, Y, L1, L2, geq_2_in_gg₂(X, Y))
SPLIT_4_IN_GGAA₄(cons_2₂(X, L), Y, L1, cons_2₂(X, L2)) -> GEQ_2_IN_GG₂(X, Y)
GEQ_2_IN_GG₂(s_1₁(X), s_1₁(Y)) -> IF_GEQ_2_IN_1_GG₃(X, Y, geq_2_in_gg₂(X, Y))
GEQ_2_IN_GG₂(s_1₁(X), s_1₁(Y)) -> GEQ_2_IN_GG₂(X, Y)
IF_SPLIT_4_IN_3_GGAA₆(X, L, Y, L1, L2, geq_2_out_gg₂(X, Y)) -> IF_SPLIT_4_IN_4_GGAA₆(X, L, Y, L1, L2, split_4_in_ggaa₄(L, Y, L1, L2))
IF_SPLIT_4_IN_3_GGAA₆(X, L, Y, L1, L2, geq_2_out_gg₂(X, Y)) -> SPLIT_4_IN_GGAA₄(L, Y, L1, L2)
IF_QS_2_IN_1_GA₄(X, L, S, split_4_out_ggaa₄(L, X, L1, L2)) -> IF_QS_2_IN_2_GA₆(X, L, S, L1, L2, qs_2_in_ga₂(L1, S1))
IF_QS_2_IN_1_GA₄(X, L, S, split_4_out_ggaa₄(L, X, L1, L2)) -> QS_2_IN_GA₂(L1, S1)
IF_QS_2_IN_2_GA₆(X, L, S, L1, L2, qs_2_out_ga₂(L1, S1)) -> IF_QS_2_IN_3_GA₆(X, L, S, L2, S1, qs_2_in_ga₂(L2, S2))
IF_QS_2_IN_2_GA₆(X, L, S, L1, L2, qs_2_out_ga₂(L1, S1)) -> QS_2_IN_GA₂(L2, S2)
IF_QS_2_IN_3_GA₆(X, L, S, L2, S1, qs_2_out_ga₂(L2, S2)) -> IF_QS_2_IN_4_GA₆(X, L, S, S1, S2, append_3_in_gga₃(S1, cons_2₂(X, S2), S))
IF_QS_2_IN_3_GA₆(X, L, S, L2, S1, qs_2_out_ga₂(L2, S2)) -> APPEND_3_IN_GGA₃(S1, cons_2₂(X, S2), S)
APPEND_3_IN_GGA₃(cons_2₂(X, L1), L2, cons_2₂(X, L3)) -> IF_APPEND_3_IN_1_GGA₅(X, L1, L2, L3, append_3_in_gga₃(L1, L2, L3))
APPEND_3_IN_GGA₃(cons_2₂(X, L1), L2, cons_2₂(X, L3)) -> APPEND_3_IN_GGA₃(L1, L2, L3)

The TRS R consists of the following rules:

qs_2_in_ga₂([]_0, []_0) -> qs_2_out_ga₂([]_0, []_0)
qs_2_in_ga₂(cons_2₂(X, L), S) -> if_qs_2_in_1_ga₄(X, L, S, split_4_in_ggaa₄(L, X, L1, L2))
split_4_in_ggaa₄([]_0, X, []_0, []_0) -> split_4_out_ggaa₄([]_0, X, []_0, []_0)
split_4_in_ggaa₄(cons_2₂(X, L), Y, cons_2₂(X, L1), L2) -> if_split_4_in_1_ggaa₆(X, L, Y, L1, L2, less_2_in_gg₂(X, Y))
less_2_in_gg₂(0_0, s_1₁(X)) -> less_2_out_gg₂(0_0, s_1₁(X))
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_split_4_in_1_ggaa₆(X, L, Y, L1, L2, less_2_out_gg₂(X, Y)) -> if_split_4_in_2_ggaa₆(X, L, Y, L1, L2, split_4_in_ggaa₄(L, Y, L1, L2))
split_4_in_ggaa₄(cons_2₂(X, L), Y, L1, cons_2₂(X, L2)) -> if_split_4_in_3_ggaa₆(X, L, Y, L1, L2, geq_2_in_gg₂(X, Y))
geq_2_in_gg₂(X, X) -> geq_2_out_gg₂(X, X)
geq_2_in_gg₂(s_1₁(X), 0_0) -> geq_2_out_gg₂(s_1₁(X), 0_0)
geq_2_in_gg₂(s_1₁(X), s_1₁(Y)) -> if_geq_2_in_1_gg₃(X, Y, geq_2_in_gg₂(X, Y))
if_geq_2_in_1_gg₃(X, Y, geq_2_out_gg₂(X, Y)) -> geq_2_out_gg₂(s_1₁(X), s_1₁(Y))
if_split_4_in_3_ggaa₆(X, L, Y, L1, L2, geq_2_out_gg₂(X, Y)) -> if_split_4_in_4_ggaa₆(X, L, Y, L1, L2, split_4_in_ggaa₄(L, Y, L1, L2))
if_split_4_in_4_ggaa₆(X, L, Y, L1, L2, split_4_out_ggaa₄(L, Y, L1, L2)) -> split_4_out_ggaa₄(cons_2₂(X, L), Y, L1, cons_2₂(X, L2))
if_split_4_in_2_ggaa₆(X, L, Y, L1, L2, split_4_out_ggaa₄(L, Y, L1, L2)) -> split_4_out_ggaa₄(cons_2₂(X, L), Y, cons_2₂(X, L1), L2)
if_qs_2_in_1_ga₄(X, L, S, split_4_out_ggaa₄(L, X, L1, L2)) -> if_qs_2_in_2_ga₆(X, L, S, L1, L2, qs_2_in_ga₂(L1, S1))
if_qs_2_in_2_ga₆(X, L, S, L1, L2, qs_2_out_ga₂(L1, S1)) -> if_qs_2_in_3_ga₆(X, L, S, L2, S1, qs_2_in_ga₂(L2, S2))
if_qs_2_in_3_ga₆(X, L, S, L2, S1, qs_2_out_ga₂(L2, S2)) -> if_qs_2_in_4_ga₆(X, L, S, S1, S2, append_3_in_gga₃(S1, cons_2₂(X, S2), S))
append_3_in_gga₃([]_0, L, L) -> append_3_out_gga₃([]_0, L, L)
append_3_in_gga₃(cons_2₂(X, L1), L2, cons_2₂(X, L3)) -> if_append_3_in_1_gga₅(X, L1, L2, L3, append_3_in_gga₃(L1, L2, L3))
if_append_3_in_1_gga₅(X, L1, L2, L3, append_3_out_gga₃(L1, L2, L3)) -> append_3_out_gga₃(cons_2₂(X, L1), L2, cons_2₂(X, L3))
if_qs_2_in_4_ga₆(X, L, S, S1, S2, append_3_out_gga₃(S1, cons_2₂(X, S2), S)) -> qs_2_out_ga₂(cons_2₂(X, L), S)

The argument filtering Pi contains the following mapping:
qs_2_in_ga₂(x1, x2) = qs_2_in_ga₁(x1)
[]_0 = []_0
cons_2₂(x1, x2) = cons_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)
split_4_in_ggaa₄(x1, x2, x3, x4) = split_4_in_ggaa₂(x1, x2)
split_4_out_ggaa₄(x1, x2, x3, x4) = split_4_out_ggaa₂(x3, x4)
if_split_4_in_1_ggaa₆(x1, x2, x3, x4, x5, x6) = if_split_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_split_4_in_2_ggaa₆(x1, x2, x3, x4, x5, x6) = if_split_4_in_2_ggaa₂(x1, x6)
if_split_4_in_3_ggaa₆(x1, x2, x3, x4, x5, x6) = if_split_4_in_3_ggaa₄(x1, x2, x3, x6)
geq_2_in_gg₂(x1, x2) = geq_2_in_gg₂(x1, x2)
geq_2_out_gg₂(x1, x2) = geq_2_out_gg
if_geq_2_in_1_gg₃(x1, x2, x3) = if_geq_2_in_1_gg₁(x3)
if_split_4_in_4_ggaa₆(x1, x2, x3, x4, x5, x6) = if_split_4_in_4_ggaa₂(x1, x6)
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)
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_SPLIT_4_IN_1_GGAA₆(x1, x2, x3, x4, x5, x6) = IF_SPLIT_4_IN_1_GGAA₄(x1, x2, x3, x6)
SPLIT_4_IN_GGAA₄(x1, x2, x3, x4) = SPLIT_4_IN_GGAA₂(x1, x2)
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)
APPEND_3_IN_GGA₃(x1, x2, x3) = APPEND_3_IN_GGA₂(x1, x2)
IF_GEQ_2_IN_1_GG₃(x1, x2, x3) = IF_GEQ_2_IN_1_GG₁(x3)
IF_SPLIT_4_IN_4_GGAA₆(x1, x2, x3, x4, x5, x6) = IF_SPLIT_4_IN_4_GGAA₂(x1, x6)
IF_APPEND_3_IN_1_GGA₅(x1, x2, x3, x4, x5) = IF_APPEND_3_IN_1_GGA₂(x1, x5)
IF_QS_2_IN_2_GA₆(x1, x2, x3, x4, x5, x6) = IF_QS_2_IN_2_GA₃(x1, x5, x6)
GEQ_2_IN_GG₂(x1, x2) = GEQ_2_IN_GG₂(x1, x2)
QS_2_IN_GA₂(x1, x2) = QS_2_IN_GA₁(x1)
IF_QS_2_IN_4_GA₆(x1, x2, x3, x4, x5, x6) = IF_QS_2_IN_4_GA₁(x6)
IF_LESS_2_IN_1_GG₃(x1, x2, x3) = IF_LESS_2_IN_1_GG₁(x3)
IF_SPLIT_4_IN_3_GGAA₆(x1, x2, x3, x4, x5, x6) = IF_SPLIT_4_IN_3_GGAA₄(x1, x2, x3, x6)
LESS_2_IN_GG₂(x1, x2) = LESS_2_IN_GG₂(x1, x2)
IF_SPLIT_4_IN_2_GGAA₆(x1, x2, x3, x4, x5, x6) = IF_SPLIT_4_IN_2_GGAA₂(x1, x6)

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(cons_2₂(X, L), S) -> IF_QS_2_IN_1_GA₄(X, L, S, split_4_in_ggaa₄(L, X, L1, L2))
QS_2_IN_GA₂(cons_2₂(X, L), S) -> SPLIT_4_IN_GGAA₄(L, X, L1, L2)
SPLIT_4_IN_GGAA₄(cons_2₂(X, L), Y, cons_2₂(X, L1), L2) -> IF_SPLIT_4_IN_1_GGAA₆(X, L, Y, L1, L2, less_2_in_gg₂(X, Y))
SPLIT_4_IN_GGAA₄(cons_2₂(X, L), Y, cons_2₂(X, L1), L2) -> 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_SPLIT_4_IN_1_GGAA₆(X, L, Y, L1, L2, less_2_out_gg₂(X, Y)) -> IF_SPLIT_4_IN_2_GGAA₆(X, L, Y, L1, L2, split_4_in_ggaa₄(L, Y, L1, L2))
IF_SPLIT_4_IN_1_GGAA₆(X, L, Y, L1, L2, less_2_out_gg₂(X, Y)) -> SPLIT_4_IN_GGAA₄(L, Y, L1, L2)
SPLIT_4_IN_GGAA₄(cons_2₂(X, L), Y, L1, cons_2₂(X, L2)) -> IF_SPLIT_4_IN_3_GGAA₆(X, L, Y, L1, L2, geq_2_in_gg₂(X, Y))
SPLIT_4_IN_GGAA₄(cons_2₂(X, L), Y, L1, cons_2₂(X, L2)) -> GEQ_2_IN_GG₂(X, Y)
GEQ_2_IN_GG₂(s_1₁(X), s_1₁(Y)) -> IF_GEQ_2_IN_1_GG₃(X, Y, geq_2_in_gg₂(X, Y))
GEQ_2_IN_GG₂(s_1₁(X), s_1₁(Y)) -> GEQ_2_IN_GG₂(X, Y)
IF_SPLIT_4_IN_3_GGAA₆(X, L, Y, L1, L2, geq_2_out_gg₂(X, Y)) -> IF_SPLIT_4_IN_4_GGAA₆(X, L, Y, L1, L2, split_4_in_ggaa₄(L, Y, L1, L2))
IF_SPLIT_4_IN_3_GGAA₆(X, L, Y, L1, L2, geq_2_out_gg₂(X, Y)) -> SPLIT_4_IN_GGAA₄(L, Y, L1, L2)
IF_QS_2_IN_1_GA₄(X, L, S, split_4_out_ggaa₄(L, X, L1, L2)) -> IF_QS_2_IN_2_GA₆(X, L, S, L1, L2, qs_2_in_ga₂(L1, S1))
IF_QS_2_IN_1_GA₄(X, L, S, split_4_out_ggaa₄(L, X, L1, L2)) -> QS_2_IN_GA₂(L1, S1)
IF_QS_2_IN_2_GA₆(X, L, S, L1, L2, qs_2_out_ga₂(L1, S1)) -> IF_QS_2_IN_3_GA₆(X, L, S, L2, S1, qs_2_in_ga₂(L2, S2))
IF_QS_2_IN_2_GA₆(X, L, S, L1, L2, qs_2_out_ga₂(L1, S1)) -> QS_2_IN_GA₂(L2, S2)
IF_QS_2_IN_3_GA₆(X, L, S, L2, S1, qs_2_out_ga₂(L2, S2)) -> IF_QS_2_IN_4_GA₆(X, L, S, S1, S2, append_3_in_gga₃(S1, cons_2₂(X, S2), S))
IF_QS_2_IN_3_GA₆(X, L, S, L2, S1, qs_2_out_ga₂(L2, S2)) -> APPEND_3_IN_GGA₃(S1, cons_2₂(X, S2), S)
APPEND_3_IN_GGA₃(cons_2₂(X, L1), L2, cons_2₂(X, L3)) -> IF_APPEND_3_IN_1_GGA₅(X, L1, L2, L3, append_3_in_gga₃(L1, L2, L3))
APPEND_3_IN_GGA₃(cons_2₂(X, L1), L2, cons_2₂(X, L3)) -> APPEND_3_IN_GGA₃(L1, L2, L3)

The TRS R consists of the following rules:

qs_2_in_ga₂([]_0, []_0) -> qs_2_out_ga₂([]_0, []_0)
qs_2_in_ga₂(cons_2₂(X, L), S) -> if_qs_2_in_1_ga₄(X, L, S, split_4_in_ggaa₄(L, X, L1, L2))
split_4_in_ggaa₄([]_0, X, []_0, []_0) -> split_4_out_ggaa₄([]_0, X, []_0, []_0)
split_4_in_ggaa₄(cons_2₂(X, L), Y, cons_2₂(X, L1), L2) -> if_split_4_in_1_ggaa₆(X, L, Y, L1, L2, less_2_in_gg₂(X, Y))
less_2_in_gg₂(0_0, s_1₁(X)) -> less_2_out_gg₂(0_0, s_1₁(X))
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_split_4_in_1_ggaa₆(X, L, Y, L1, L2, less_2_out_gg₂(X, Y)) -> if_split_4_in_2_ggaa₆(X, L, Y, L1, L2, split_4_in_ggaa₄(L, Y, L1, L2))
split_4_in_ggaa₄(cons_2₂(X, L), Y, L1, cons_2₂(X, L2)) -> if_split_4_in_3_ggaa₆(X, L, Y, L1, L2, geq_2_in_gg₂(X, Y))
geq_2_in_gg₂(X, X) -> geq_2_out_gg₂(X, X)
geq_2_in_gg₂(s_1₁(X), 0_0) -> geq_2_out_gg₂(s_1₁(X), 0_0)
geq_2_in_gg₂(s_1₁(X), s_1₁(Y)) -> if_geq_2_in_1_gg₃(X, Y, geq_2_in_gg₂(X, Y))
if_geq_2_in_1_gg₃(X, Y, geq_2_out_gg₂(X, Y)) -> geq_2_out_gg₂(s_1₁(X), s_1₁(Y))
if_split_4_in_3_ggaa₆(X, L, Y, L1, L2, geq_2_out_gg₂(X, Y)) -> if_split_4_in_4_ggaa₆(X, L, Y, L1, L2, split_4_in_ggaa₄(L, Y, L1, L2))
if_split_4_in_4_ggaa₆(X, L, Y, L1, L2, split_4_out_ggaa₄(L, Y, L1, L2)) -> split_4_out_ggaa₄(cons_2₂(X, L), Y, L1, cons_2₂(X, L2))
if_split_4_in_2_ggaa₆(X, L, Y, L1, L2, split_4_out_ggaa₄(L, Y, L1, L2)) -> split_4_out_ggaa₄(cons_2₂(X, L), Y, cons_2₂(X, L1), L2)
if_qs_2_in_1_ga₄(X, L, S, split_4_out_ggaa₄(L, X, L1, L2)) -> if_qs_2_in_2_ga₆(X, L, S, L1, L2, qs_2_in_ga₂(L1, S1))
if_qs_2_in_2_ga₆(X, L, S, L1, L2, qs_2_out_ga₂(L1, S1)) -> if_qs_2_in_3_ga₆(X, L, S, L2, S1, qs_2_in_ga₂(L2, S2))
if_qs_2_in_3_ga₆(X, L, S, L2, S1, qs_2_out_ga₂(L2, S2)) -> if_qs_2_in_4_ga₆(X, L, S, S1, S2, append_3_in_gga₃(S1, cons_2₂(X, S2), S))
append_3_in_gga₃([]_0, L, L) -> append_3_out_gga₃([]_0, L, L)
append_3_in_gga₃(cons_2₂(X, L1), L2, cons_2₂(X, L3)) -> if_append_3_in_1_gga₅(X, L1, L2, L3, append_3_in_gga₃(L1, L2, L3))
if_append_3_in_1_gga₅(X, L1, L2, L3, append_3_out_gga₃(L1, L2, L3)) -> append_3_out_gga₃(cons_2₂(X, L1), L2, cons_2₂(X, L3))
if_qs_2_in_4_ga₆(X, L, S, S1, S2, append_3_out_gga₃(S1, cons_2₂(X, S2), S)) -> qs_2_out_ga₂(cons_2₂(X, L), S)

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

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

APPEND_3_IN_GGA₃(cons_2₂(X, L1), L2, cons_2₂(X, L3)) -> APPEND_3_IN_GGA₃(L1, L2, L3)

The TRS R consists of the following rules:

qs_2_in_ga₂([]_0, []_0) -> qs_2_out_ga₂([]_0, []_0)
qs_2_in_ga₂(cons_2₂(X, L), S) -> if_qs_2_in_1_ga₄(X, L, S, split_4_in_ggaa₄(L, X, L1, L2))
split_4_in_ggaa₄([]_0, X, []_0, []_0) -> split_4_out_ggaa₄([]_0, X, []_0, []_0)
split_4_in_ggaa₄(cons_2₂(X, L), Y, cons_2₂(X, L1), L2) -> if_split_4_in_1_ggaa₆(X, L, Y, L1, L2, less_2_in_gg₂(X, Y))
less_2_in_gg₂(0_0, s_1₁(X)) -> less_2_out_gg₂(0_0, s_1₁(X))
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_split_4_in_1_ggaa₆(X, L, Y, L1, L2, less_2_out_gg₂(X, Y)) -> if_split_4_in_2_ggaa₆(X, L, Y, L1, L2, split_4_in_ggaa₄(L, Y, L1, L2))
split_4_in_ggaa₄(cons_2₂(X, L), Y, L1, cons_2₂(X, L2)) -> if_split_4_in_3_ggaa₆(X, L, Y, L1, L2, geq_2_in_gg₂(X, Y))
geq_2_in_gg₂(X, X) -> geq_2_out_gg₂(X, X)
geq_2_in_gg₂(s_1₁(X), 0_0) -> geq_2_out_gg₂(s_1₁(X), 0_0)
geq_2_in_gg₂(s_1₁(X), s_1₁(Y)) -> if_geq_2_in_1_gg₃(X, Y, geq_2_in_gg₂(X, Y))
if_geq_2_in_1_gg₃(X, Y, geq_2_out_gg₂(X, Y)) -> geq_2_out_gg₂(s_1₁(X), s_1₁(Y))
if_split_4_in_3_ggaa₆(X, L, Y, L1, L2, geq_2_out_gg₂(X, Y)) -> if_split_4_in_4_ggaa₆(X, L, Y, L1, L2, split_4_in_ggaa₄(L, Y, L1, L2))
if_split_4_in_4_ggaa₆(X, L, Y, L1, L2, split_4_out_ggaa₄(L, Y, L1, L2)) -> split_4_out_ggaa₄(cons_2₂(X, L), Y, L1, cons_2₂(X, L2))
if_split_4_in_2_ggaa₆(X, L, Y, L1, L2, split_4_out_ggaa₄(L, Y, L1, L2)) -> split_4_out_ggaa₄(cons_2₂(X, L), Y, cons_2₂(X, L1), L2)
if_qs_2_in_1_ga₄(X, L, S, split_4_out_ggaa₄(L, X, L1, L2)) -> if_qs_2_in_2_ga₆(X, L, S, L1, L2, qs_2_in_ga₂(L1, S1))
if_qs_2_in_2_ga₆(X, L, S, L1, L2, qs_2_out_ga₂(L1, S1)) -> if_qs_2_in_3_ga₆(X, L, S, L2, S1, qs_2_in_ga₂(L2, S2))
if_qs_2_in_3_ga₆(X, L, S, L2, S1, qs_2_out_ga₂(L2, S2)) -> if_qs_2_in_4_ga₆(X, L, S, S1, S2, append_3_in_gga₃(S1, cons_2₂(X, S2), S))
append_3_in_gga₃([]_0, L, L) -> append_3_out_gga₃([]_0, L, L)
append_3_in_gga₃(cons_2₂(X, L1), L2, cons_2₂(X, L3)) -> if_append_3_in_1_gga₅(X, L1, L2, L3, append_3_in_gga₃(L1, L2, L3))
if_append_3_in_1_gga₅(X, L1, L2, L3, append_3_out_gga₃(L1, L2, L3)) -> append_3_out_gga₃(cons_2₂(X, L1), L2, cons_2₂(X, L3))
if_qs_2_in_4_ga₆(X, L, S, S1, S2, append_3_out_gga₃(S1, cons_2₂(X, S2), S)) -> qs_2_out_ga₂(cons_2₂(X, L), S)

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

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

APPEND_3_IN_GGA₃(cons_2₂(X, L1), L2, cons_2₂(X, L3)) -> APPEND_3_IN_GGA₃(L1, L2, L3)

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

              ↳ PiDP

              ↳ PiDP

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

APPEND_3_IN_GGA₂(cons_2₂(X, L1), L2) -> APPEND_3_IN_GGA₂(L1, L2)

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:

APPEND_3_IN_GGA₂(cons_2₂(X, L1), L2) -> APPEND_3_IN_GGA₂(L1, L2)
The graph contains the following edges 1 > 1, 2 >= 2

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

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

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

The TRS R consists of the following rules:

qs_2_in_ga₂([]_0, []_0) -> qs_2_out_ga₂([]_0, []_0)
qs_2_in_ga₂(cons_2₂(X, L), S) -> if_qs_2_in_1_ga₄(X, L, S, split_4_in_ggaa₄(L, X, L1, L2))
split_4_in_ggaa₄([]_0, X, []_0, []_0) -> split_4_out_ggaa₄([]_0, X, []_0, []_0)
split_4_in_ggaa₄(cons_2₂(X, L), Y, cons_2₂(X, L1), L2) -> if_split_4_in_1_ggaa₆(X, L, Y, L1, L2, less_2_in_gg₂(X, Y))
less_2_in_gg₂(0_0, s_1₁(X)) -> less_2_out_gg₂(0_0, s_1₁(X))
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_split_4_in_1_ggaa₆(X, L, Y, L1, L2, less_2_out_gg₂(X, Y)) -> if_split_4_in_2_ggaa₆(X, L, Y, L1, L2, split_4_in_ggaa₄(L, Y, L1, L2))
split_4_in_ggaa₄(cons_2₂(X, L), Y, L1, cons_2₂(X, L2)) -> if_split_4_in_3_ggaa₆(X, L, Y, L1, L2, geq_2_in_gg₂(X, Y))
geq_2_in_gg₂(X, X) -> geq_2_out_gg₂(X, X)
geq_2_in_gg₂(s_1₁(X), 0_0) -> geq_2_out_gg₂(s_1₁(X), 0_0)
geq_2_in_gg₂(s_1₁(X), s_1₁(Y)) -> if_geq_2_in_1_gg₃(X, Y, geq_2_in_gg₂(X, Y))
if_geq_2_in_1_gg₃(X, Y, geq_2_out_gg₂(X, Y)) -> geq_2_out_gg₂(s_1₁(X), s_1₁(Y))
if_split_4_in_3_ggaa₆(X, L, Y, L1, L2, geq_2_out_gg₂(X, Y)) -> if_split_4_in_4_ggaa₆(X, L, Y, L1, L2, split_4_in_ggaa₄(L, Y, L1, L2))
if_split_4_in_4_ggaa₆(X, L, Y, L1, L2, split_4_out_ggaa₄(L, Y, L1, L2)) -> split_4_out_ggaa₄(cons_2₂(X, L), Y, L1, cons_2₂(X, L2))
if_split_4_in_2_ggaa₆(X, L, Y, L1, L2, split_4_out_ggaa₄(L, Y, L1, L2)) -> split_4_out_ggaa₄(cons_2₂(X, L), Y, cons_2₂(X, L1), L2)
if_qs_2_in_1_ga₄(X, L, S, split_4_out_ggaa₄(L, X, L1, L2)) -> if_qs_2_in_2_ga₆(X, L, S, L1, L2, qs_2_in_ga₂(L1, S1))
if_qs_2_in_2_ga₆(X, L, S, L1, L2, qs_2_out_ga₂(L1, S1)) -> if_qs_2_in_3_ga₆(X, L, S, L2, S1, qs_2_in_ga₂(L2, S2))
if_qs_2_in_3_ga₆(X, L, S, L2, S1, qs_2_out_ga₂(L2, S2)) -> if_qs_2_in_4_ga₆(X, L, S, S1, S2, append_3_in_gga₃(S1, cons_2₂(X, S2), S))
append_3_in_gga₃([]_0, L, L) -> append_3_out_gga₃([]_0, L, L)
append_3_in_gga₃(cons_2₂(X, L1), L2, cons_2₂(X, L3)) -> if_append_3_in_1_gga₅(X, L1, L2, L3, append_3_in_gga₃(L1, L2, L3))
if_append_3_in_1_gga₅(X, L1, L2, L3, append_3_out_gga₃(L1, L2, L3)) -> append_3_out_gga₃(cons_2₂(X, L1), L2, cons_2₂(X, L3))
if_qs_2_in_4_ga₆(X, L, S, S1, S2, append_3_out_gga₃(S1, cons_2₂(X, S2), S)) -> qs_2_out_ga₂(cons_2₂(X, L), S)

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

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

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

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

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

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

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

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {GEQ_2_IN_GG₂}.
By using the subterm criterion together with the size-change analysis we have proven that there are no infinite chains for this DP problem.

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

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

              ↳ 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(cons_2₂(X, L), S) -> if_qs_2_in_1_ga₄(X, L, S, split_4_in_ggaa₄(L, X, L1, L2))
split_4_in_ggaa₄([]_0, X, []_0, []_0) -> split_4_out_ggaa₄([]_0, X, []_0, []_0)
split_4_in_ggaa₄(cons_2₂(X, L), Y, cons_2₂(X, L1), L2) -> if_split_4_in_1_ggaa₆(X, L, Y, L1, L2, less_2_in_gg₂(X, Y))
less_2_in_gg₂(0_0, s_1₁(X)) -> less_2_out_gg₂(0_0, s_1₁(X))
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_split_4_in_1_ggaa₆(X, L, Y, L1, L2, less_2_out_gg₂(X, Y)) -> if_split_4_in_2_ggaa₆(X, L, Y, L1, L2, split_4_in_ggaa₄(L, Y, L1, L2))
split_4_in_ggaa₄(cons_2₂(X, L), Y, L1, cons_2₂(X, L2)) -> if_split_4_in_3_ggaa₆(X, L, Y, L1, L2, geq_2_in_gg₂(X, Y))
geq_2_in_gg₂(X, X) -> geq_2_out_gg₂(X, X)
geq_2_in_gg₂(s_1₁(X), 0_0) -> geq_2_out_gg₂(s_1₁(X), 0_0)
geq_2_in_gg₂(s_1₁(X), s_1₁(Y)) -> if_geq_2_in_1_gg₃(X, Y, geq_2_in_gg₂(X, Y))
if_geq_2_in_1_gg₃(X, Y, geq_2_out_gg₂(X, Y)) -> geq_2_out_gg₂(s_1₁(X), s_1₁(Y))
if_split_4_in_3_ggaa₆(X, L, Y, L1, L2, geq_2_out_gg₂(X, Y)) -> if_split_4_in_4_ggaa₆(X, L, Y, L1, L2, split_4_in_ggaa₄(L, Y, L1, L2))
if_split_4_in_4_ggaa₆(X, L, Y, L1, L2, split_4_out_ggaa₄(L, Y, L1, L2)) -> split_4_out_ggaa₄(cons_2₂(X, L), Y, L1, cons_2₂(X, L2))
if_split_4_in_2_ggaa₆(X, L, Y, L1, L2, split_4_out_ggaa₄(L, Y, L1, L2)) -> split_4_out_ggaa₄(cons_2₂(X, L), Y, cons_2₂(X, L1), L2)
if_qs_2_in_1_ga₄(X, L, S, split_4_out_ggaa₄(L, X, L1, L2)) -> if_qs_2_in_2_ga₆(X, L, S, L1, L2, qs_2_in_ga₂(L1, S1))
if_qs_2_in_2_ga₆(X, L, S, L1, L2, qs_2_out_ga₂(L1, S1)) -> if_qs_2_in_3_ga₆(X, L, S, L2, S1, qs_2_in_ga₂(L2, S2))
if_qs_2_in_3_ga₆(X, L, S, L2, S1, qs_2_out_ga₂(L2, S2)) -> if_qs_2_in_4_ga₆(X, L, S, S1, S2, append_3_in_gga₃(S1, cons_2₂(X, S2), S))
append_3_in_gga₃([]_0, L, L) -> append_3_out_gga₃([]_0, L, L)
append_3_in_gga₃(cons_2₂(X, L1), L2, cons_2₂(X, L3)) -> if_append_3_in_1_gga₅(X, L1, L2, L3, append_3_in_gga₃(L1, L2, L3))
if_append_3_in_1_gga₅(X, L1, L2, L3, append_3_out_gga₃(L1, L2, L3)) -> append_3_out_gga₃(cons_2₂(X, L1), L2, cons_2₂(X, L3))
if_qs_2_in_4_ga₆(X, L, S, S1, S2, append_3_out_gga₃(S1, cons_2₂(X, S2), S)) -> qs_2_out_ga₂(cons_2₂(X, L), S)

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ 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

              ↳ 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

              ↳ PiDP

                ↳ UsableRulesProof

              ↳ PiDP

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

SPLIT_4_IN_GGAA₄(cons_2₂(X, L), Y, L1, cons_2₂(X, L2)) -> IF_SPLIT_4_IN_3_GGAA₆(X, L, Y, L1, L2, geq_2_in_gg₂(X, Y))
SPLIT_4_IN_GGAA₄(cons_2₂(X, L), Y, cons_2₂(X, L1), L2) -> IF_SPLIT_4_IN_1_GGAA₆(X, L, Y, L1, L2, less_2_in_gg₂(X, Y))
IF_SPLIT_4_IN_3_GGAA₆(X, L, Y, L1, L2, geq_2_out_gg₂(X, Y)) -> SPLIT_4_IN_GGAA₄(L, Y, L1, L2)
IF_SPLIT_4_IN_1_GGAA₆(X, L, Y, L1, L2, less_2_out_gg₂(X, Y)) -> SPLIT_4_IN_GGAA₄(L, Y, L1, L2)

The TRS R consists of the following rules:

qs_2_in_ga₂([]_0, []_0) -> qs_2_out_ga₂([]_0, []_0)
qs_2_in_ga₂(cons_2₂(X, L), S) -> if_qs_2_in_1_ga₄(X, L, S, split_4_in_ggaa₄(L, X, L1, L2))
split_4_in_ggaa₄([]_0, X, []_0, []_0) -> split_4_out_ggaa₄([]_0, X, []_0, []_0)
split_4_in_ggaa₄(cons_2₂(X, L), Y, cons_2₂(X, L1), L2) -> if_split_4_in_1_ggaa₆(X, L, Y, L1, L2, less_2_in_gg₂(X, Y))
less_2_in_gg₂(0_0, s_1₁(X)) -> less_2_out_gg₂(0_0, s_1₁(X))
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_split_4_in_1_ggaa₆(X, L, Y, L1, L2, less_2_out_gg₂(X, Y)) -> if_split_4_in_2_ggaa₆(X, L, Y, L1, L2, split_4_in_ggaa₄(L, Y, L1, L2))
split_4_in_ggaa₄(cons_2₂(X, L), Y, L1, cons_2₂(X, L2)) -> if_split_4_in_3_ggaa₆(X, L, Y, L1, L2, geq_2_in_gg₂(X, Y))
geq_2_in_gg₂(X, X) -> geq_2_out_gg₂(X, X)
geq_2_in_gg₂(s_1₁(X), 0_0) -> geq_2_out_gg₂(s_1₁(X), 0_0)
geq_2_in_gg₂(s_1₁(X), s_1₁(Y)) -> if_geq_2_in_1_gg₃(X, Y, geq_2_in_gg₂(X, Y))
if_geq_2_in_1_gg₃(X, Y, geq_2_out_gg₂(X, Y)) -> geq_2_out_gg₂(s_1₁(X), s_1₁(Y))
if_split_4_in_3_ggaa₆(X, L, Y, L1, L2, geq_2_out_gg₂(X, Y)) -> if_split_4_in_4_ggaa₆(X, L, Y, L1, L2, split_4_in_ggaa₄(L, Y, L1, L2))
if_split_4_in_4_ggaa₆(X, L, Y, L1, L2, split_4_out_ggaa₄(L, Y, L1, L2)) -> split_4_out_ggaa₄(cons_2₂(X, L), Y, L1, cons_2₂(X, L2))
if_split_4_in_2_ggaa₆(X, L, Y, L1, L2, split_4_out_ggaa₄(L, Y, L1, L2)) -> split_4_out_ggaa₄(cons_2₂(X, L), Y, cons_2₂(X, L1), L2)
if_qs_2_in_1_ga₄(X, L, S, split_4_out_ggaa₄(L, X, L1, L2)) -> if_qs_2_in_2_ga₆(X, L, S, L1, L2, qs_2_in_ga₂(L1, S1))
if_qs_2_in_2_ga₆(X, L, S, L1, L2, qs_2_out_ga₂(L1, S1)) -> if_qs_2_in_3_ga₆(X, L, S, L2, S1, qs_2_in_ga₂(L2, S2))
if_qs_2_in_3_ga₆(X, L, S, L2, S1, qs_2_out_ga₂(L2, S2)) -> if_qs_2_in_4_ga₆(X, L, S, S1, S2, append_3_in_gga₃(S1, cons_2₂(X, S2), S))
append_3_in_gga₃([]_0, L, L) -> append_3_out_gga₃([]_0, L, L)
append_3_in_gga₃(cons_2₂(X, L1), L2, cons_2₂(X, L3)) -> if_append_3_in_1_gga₅(X, L1, L2, L3, append_3_in_gga₃(L1, L2, L3))
if_append_3_in_1_gga₅(X, L1, L2, L3, append_3_out_gga₃(L1, L2, L3)) -> append_3_out_gga₃(cons_2₂(X, L1), L2, cons_2₂(X, L3))
if_qs_2_in_4_ga₆(X, L, S, S1, S2, append_3_out_gga₃(S1, cons_2₂(X, S2), S)) -> qs_2_out_ga₂(cons_2₂(X, L), S)

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

              ↳ PiDP

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

SPLIT_4_IN_GGAA₄(cons_2₂(X, L), Y, L1, cons_2₂(X, L2)) -> IF_SPLIT_4_IN_3_GGAA₆(X, L, Y, L1, L2, geq_2_in_gg₂(X, Y))
SPLIT_4_IN_GGAA₄(cons_2₂(X, L), Y, cons_2₂(X, L1), L2) -> IF_SPLIT_4_IN_1_GGAA₆(X, L, Y, L1, L2, less_2_in_gg₂(X, Y))
IF_SPLIT_4_IN_3_GGAA₆(X, L, Y, L1, L2, geq_2_out_gg₂(X, Y)) -> SPLIT_4_IN_GGAA₄(L, Y, L1, L2)
IF_SPLIT_4_IN_1_GGAA₆(X, L, Y, L1, L2, less_2_out_gg₂(X, Y)) -> SPLIT_4_IN_GGAA₄(L, Y, L1, L2)

The TRS R consists of the following rules:

geq_2_in_gg₂(X, X) -> geq_2_out_gg₂(X, X)
geq_2_in_gg₂(s_1₁(X), 0_0) -> geq_2_out_gg₂(s_1₁(X), 0_0)
geq_2_in_gg₂(s_1₁(X), s_1₁(Y)) -> if_geq_2_in_1_gg₃(X, Y, geq_2_in_gg₂(X, Y))
less_2_in_gg₂(0_0, s_1₁(X)) -> less_2_out_gg₂(0_0, s_1₁(X))
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_geq_2_in_1_gg₃(X, Y, geq_2_out_gg₂(X, Y)) -> geq_2_out_gg₂(s_1₁(X), s_1₁(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:
cons_2₂(x1, x2) = cons_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)
geq_2_in_gg₂(x1, x2) = geq_2_in_gg₂(x1, x2)
geq_2_out_gg₂(x1, x2) = geq_2_out_gg
if_geq_2_in_1_gg₃(x1, x2, x3) = if_geq_2_in_1_gg₁(x3)
IF_SPLIT_4_IN_1_GGAA₆(x1, x2, x3, x4, x5, x6) = IF_SPLIT_4_IN_1_GGAA₄(x1, x2, x3, x6)
SPLIT_4_IN_GGAA₄(x1, x2, x3, x4) = SPLIT_4_IN_GGAA₂(x1, x2)
IF_SPLIT_4_IN_3_GGAA₆(x1, x2, x3, x4, x5, x6) = IF_SPLIT_4_IN_3_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

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ PiDP

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

SPLIT_4_IN_GGAA₂(cons_2₂(X, L), Y) -> IF_SPLIT_4_IN_3_GGAA₄(X, L, Y, geq_2_in_gg₂(X, Y))
SPLIT_4_IN_GGAA₂(cons_2₂(X, L), Y) -> IF_SPLIT_4_IN_1_GGAA₄(X, L, Y, less_2_in_gg₂(X, Y))
IF_SPLIT_4_IN_3_GGAA₄(X, L, Y, geq_2_out_gg) -> SPLIT_4_IN_GGAA₂(L, Y)
IF_SPLIT_4_IN_1_GGAA₄(X, L, Y, less_2_out_gg) -> SPLIT_4_IN_GGAA₂(L, Y)

The TRS R consists of the following rules:

geq_2_in_gg₂(X, X) -> geq_2_out_gg
geq_2_in_gg₂(s_1₁(X), 0_0) -> geq_2_out_gg
geq_2_in_gg₂(s_1₁(X), s_1₁(Y)) -> if_geq_2_in_1_gg₁(geq_2_in_gg₂(X, Y))
less_2_in_gg₂(0_0, s_1₁(X)) -> 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_geq_2_in_1_gg₁(geq_2_out_gg) -> geq_2_out_gg
if_less_2_in_1_gg₁(less_2_out_gg) -> less_2_out_gg

The set Q consists of the following terms:

geq_2_in_gg₂(x0, x1)
less_2_in_gg₂(x0, x1)
if_geq_2_in_1_gg₁(x0)
if_less_2_in_1_gg₁(x0)

We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {IF_SPLIT_4_IN_3_GGAA₄, SPLIT_4_IN_GGAA₂, IF_SPLIT_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:

IF_SPLIT_4_IN_3_GGAA₄(X, L, Y, geq_2_out_gg) -> SPLIT_4_IN_GGAA₂(L, Y)
The graph contains the following edges 2 >= 1, 3 >= 2
IF_SPLIT_4_IN_1_GGAA₄(X, L, Y, less_2_out_gg) -> SPLIT_4_IN_GGAA₂(L, Y)
The graph contains the following edges 2 >= 1, 3 >= 2
SPLIT_4_IN_GGAA₂(cons_2₂(X, L), Y) -> IF_SPLIT_4_IN_3_GGAA₄(X, L, Y, geq_2_in_gg₂(X, Y))
The graph contains the following edges 1 > 1, 1 > 2, 2 >= 3
SPLIT_4_IN_GGAA₂(cons_2₂(X, L), Y) -> IF_SPLIT_4_IN_1_GGAA₄(X, L, Y, less_2_in_gg₂(X, Y))
The graph contains the following edges 1 > 1, 1 > 2, 2 >= 3

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ PiDPToQDPProof

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

IF_QS_2_IN_1_GA₄(X, L, S, split_4_out_ggaa₄(L, X, L1, L2)) -> QS_2_IN_GA₂(L1, S1)
IF_QS_2_IN_2_GA₆(X, L, S, L1, L2, qs_2_out_ga₂(L1, S1)) -> QS_2_IN_GA₂(L2, S2)
IF_QS_2_IN_1_GA₄(X, L, S, split_4_out_ggaa₄(L, X, L1, L2)) -> IF_QS_2_IN_2_GA₆(X, L, S, L1, L2, qs_2_in_ga₂(L1, S1))
QS_2_IN_GA₂(cons_2₂(X, L), S) -> IF_QS_2_IN_1_GA₄(X, L, S, split_4_in_ggaa₄(L, X, L1, L2))

The TRS R consists of the following rules:

qs_2_in_ga₂([]_0, []_0) -> qs_2_out_ga₂([]_0, []_0)
qs_2_in_ga₂(cons_2₂(X, L), S) -> if_qs_2_in_1_ga₄(X, L, S, split_4_in_ggaa₄(L, X, L1, L2))
split_4_in_ggaa₄([]_0, X, []_0, []_0) -> split_4_out_ggaa₄([]_0, X, []_0, []_0)
split_4_in_ggaa₄(cons_2₂(X, L), Y, cons_2₂(X, L1), L2) -> if_split_4_in_1_ggaa₆(X, L, Y, L1, L2, less_2_in_gg₂(X, Y))
less_2_in_gg₂(0_0, s_1₁(X)) -> less_2_out_gg₂(0_0, s_1₁(X))
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_split_4_in_1_ggaa₆(X, L, Y, L1, L2, less_2_out_gg₂(X, Y)) -> if_split_4_in_2_ggaa₆(X, L, Y, L1, L2, split_4_in_ggaa₄(L, Y, L1, L2))
split_4_in_ggaa₄(cons_2₂(X, L), Y, L1, cons_2₂(X, L2)) -> if_split_4_in_3_ggaa₆(X, L, Y, L1, L2, geq_2_in_gg₂(X, Y))
geq_2_in_gg₂(X, X) -> geq_2_out_gg₂(X, X)
geq_2_in_gg₂(s_1₁(X), 0_0) -> geq_2_out_gg₂(s_1₁(X), 0_0)
geq_2_in_gg₂(s_1₁(X), s_1₁(Y)) -> if_geq_2_in_1_gg₃(X, Y, geq_2_in_gg₂(X, Y))
if_geq_2_in_1_gg₃(X, Y, geq_2_out_gg₂(X, Y)) -> geq_2_out_gg₂(s_1₁(X), s_1₁(Y))
if_split_4_in_3_ggaa₆(X, L, Y, L1, L2, geq_2_out_gg₂(X, Y)) -> if_split_4_in_4_ggaa₆(X, L, Y, L1, L2, split_4_in_ggaa₄(L, Y, L1, L2))
if_split_4_in_4_ggaa₆(X, L, Y, L1, L2, split_4_out_ggaa₄(L, Y, L1, L2)) -> split_4_out_ggaa₄(cons_2₂(X, L), Y, L1, cons_2₂(X, L2))
if_split_4_in_2_ggaa₆(X, L, Y, L1, L2, split_4_out_ggaa₄(L, Y, L1, L2)) -> split_4_out_ggaa₄(cons_2₂(X, L), Y, cons_2₂(X, L1), L2)
if_qs_2_in_1_ga₄(X, L, S, split_4_out_ggaa₄(L, X, L1, L2)) -> if_qs_2_in_2_ga₆(X, L, S, L1, L2, qs_2_in_ga₂(L1, S1))
if_qs_2_in_2_ga₆(X, L, S, L1, L2, qs_2_out_ga₂(L1, S1)) -> if_qs_2_in_3_ga₆(X, L, S, L2, S1, qs_2_in_ga₂(L2, S2))
if_qs_2_in_3_ga₆(X, L, S, L2, S1, qs_2_out_ga₂(L2, S2)) -> if_qs_2_in_4_ga₆(X, L, S, S1, S2, append_3_in_gga₃(S1, cons_2₂(X, S2), S))
append_3_in_gga₃([]_0, L, L) -> append_3_out_gga₃([]_0, L, L)
append_3_in_gga₃(cons_2₂(X, L1), L2, cons_2₂(X, L3)) -> if_append_3_in_1_gga₅(X, L1, L2, L3, append_3_in_gga₃(L1, L2, L3))
if_append_3_in_1_gga₅(X, L1, L2, L3, append_3_out_gga₃(L1, L2, L3)) -> append_3_out_gga₃(cons_2₂(X, L1), L2, cons_2₂(X, L3))
if_qs_2_in_4_ga₆(X, L, S, S1, S2, append_3_out_gga₃(S1, cons_2₂(X, S2), S)) -> qs_2_out_ga₂(cons_2₂(X, L), S)

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ PiDPToQDPProof

                  ↳ QDP

                    ↳ QDPPoloProof

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

IF_QS_2_IN_1_GA₂(X, split_4_out_ggaa₂(L1, L2)) -> QS_2_IN_GA₁(L1)
IF_QS_2_IN_2_GA₃(X, L2, qs_2_out_ga₁(S1)) -> QS_2_IN_GA₁(L2)
IF_QS_2_IN_1_GA₂(X, split_4_out_ggaa₂(L1, L2)) -> IF_QS_2_IN_2_GA₃(X, L2, qs_2_in_ga₁(L1))
QS_2_IN_GA₁(cons_2₂(X, L)) -> IF_QS_2_IN_1_GA₂(X, split_4_in_ggaa₂(L, X))

The TRS R consists of the following rules:

qs_2_in_ga₁([]_0) -> qs_2_out_ga₁([]_0)
qs_2_in_ga₁(cons_2₂(X, L)) -> if_qs_2_in_1_ga₂(X, split_4_in_ggaa₂(L, X))
split_4_in_ggaa₂([]_0, X) -> split_4_out_ggaa₂([]_0, []_0)
split_4_in_ggaa₂(cons_2₂(X, L), Y) -> if_split_4_in_1_ggaa₄(X, L, Y, less_2_in_gg₂(X, Y))
less_2_in_gg₂(0_0, s_1₁(X)) -> 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_split_4_in_1_ggaa₄(X, L, Y, less_2_out_gg) -> if_split_4_in_2_ggaa₂(X, split_4_in_ggaa₂(L, Y))
split_4_in_ggaa₂(cons_2₂(X, L), Y) -> if_split_4_in_3_ggaa₄(X, L, Y, geq_2_in_gg₂(X, Y))
geq_2_in_gg₂(X, X) -> geq_2_out_gg
geq_2_in_gg₂(s_1₁(X), 0_0) -> geq_2_out_gg
geq_2_in_gg₂(s_1₁(X), s_1₁(Y)) -> if_geq_2_in_1_gg₁(geq_2_in_gg₂(X, Y))
if_geq_2_in_1_gg₁(geq_2_out_gg) -> geq_2_out_gg
if_split_4_in_3_ggaa₄(X, L, Y, geq_2_out_gg) -> if_split_4_in_4_ggaa₂(X, split_4_in_ggaa₂(L, Y))
if_split_4_in_4_ggaa₂(X, split_4_out_ggaa₂(L1, L2)) -> split_4_out_ggaa₂(L1, cons_2₂(X, L2))
if_split_4_in_2_ggaa₂(X, split_4_out_ggaa₂(L1, L2)) -> split_4_out_ggaa₂(cons_2₂(X, L1), L2)
if_qs_2_in_1_ga₂(X, split_4_out_ggaa₂(L1, L2)) -> if_qs_2_in_2_ga₃(X, L2, qs_2_in_ga₁(L1))
if_qs_2_in_2_ga₃(X, L2, qs_2_out_ga₁(S1)) -> if_qs_2_in_3_ga₃(X, S1, qs_2_in_ga₁(L2))
if_qs_2_in_3_ga₃(X, S1, qs_2_out_ga₁(S2)) -> if_qs_2_in_4_ga₁(append_3_in_gga₂(S1, cons_2₂(X, S2)))
append_3_in_gga₂([]_0, L) -> append_3_out_gga₁(L)
append_3_in_gga₂(cons_2₂(X, L1), L2) -> if_append_3_in_1_gga₂(X, append_3_in_gga₂(L1, L2))
if_append_3_in_1_gga₂(X, append_3_out_gga₁(L3)) -> append_3_out_gga₁(cons_2₂(X, L3))
if_qs_2_in_4_ga₁(append_3_out_gga₁(S)) -> qs_2_out_ga₁(S)

The set Q consists of the following terms:

qs_2_in_ga₁(x0)
split_4_in_ggaa₂(x0, x1)
less_2_in_gg₂(x0, x1)
if_less_2_in_1_gg₁(x0)
if_split_4_in_1_ggaa₄(x0, x1, x2, x3)
geq_2_in_gg₂(x0, x1)
if_geq_2_in_1_gg₁(x0)
if_split_4_in_3_ggaa₄(x0, x1, x2, x3)
if_split_4_in_4_ggaa₂(x0, x1)
if_split_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)
append_3_in_gga₂(x0, x1)
if_append_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_1_GA₂, IF_QS_2_IN_2_GA₃}.
By using a polynomial ordering, the following set of Dependency Pairs of this DP problem can be strictly oriented.

QS_2_IN_GA₁(cons_2₂(X, L)) -> IF_QS_2_IN_1_GA₂(X, split_4_in_ggaa₂(L, X))

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

IF_QS_2_IN_1_GA₂(X, split_4_out_ggaa₂(L1, L2)) -> QS_2_IN_GA₁(L1)
IF_QS_2_IN_2_GA₃(X, L2, qs_2_out_ga₁(S1)) -> QS_2_IN_GA₁(L2)
IF_QS_2_IN_1_GA₂(X, split_4_out_ggaa₂(L1, L2)) -> IF_QS_2_IN_2_GA₃(X, L2, qs_2_in_ga₁(L1))

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

split_4_in_ggaa₂(cons_2₂(X, L), Y) -> if_split_4_in_1_ggaa₄(X, L, Y, less_2_in_gg₂(X, Y))
if_split_4_in_4_ggaa₂(X, split_4_out_ggaa₂(L1, L2)) -> split_4_out_ggaa₂(L1, cons_2₂(X, L2))
split_4_in_ggaa₂([]_0, X) -> split_4_out_ggaa₂([]_0, []_0)
if_split_4_in_1_ggaa₄(X, L, Y, less_2_out_gg) -> if_split_4_in_2_ggaa₂(X, split_4_in_ggaa₂(L, Y))
if_split_4_in_2_ggaa₂(X, split_4_out_ggaa₂(L1, L2)) -> split_4_out_ggaa₂(cons_2₂(X, L1), L2)
if_split_4_in_3_ggaa₄(X, L, Y, geq_2_out_gg) -> if_split_4_in_4_ggaa₂(X, split_4_in_ggaa₂(L, Y))
split_4_in_ggaa₂(cons_2₂(X, L), Y) -> if_split_4_in_3_ggaa₄(X, L, Y, geq_2_in_gg₂(X, Y))

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(split_4_in_ggaa₂(x₁, x₂)) = x₁
POL(geq_2_out_gg) = 0
POL(if_append_3_in_1_gga₂(x₁, x₂)) = 0
POL(if_qs_2_in_1_ga₂(x₁, x₂)) = 0
POL(if_qs_2_in_2_ga₃(x₁, x₂, x₃)) = 0
POL(IF_QS_2_IN_2_GA₃(x₁, x₂, x₃)) = x₂
POL(QS_2_IN_GA₁(x₁)) = x₁
POL(if_qs_2_in_3_ga₃(x₁, x₂, x₃)) = 0
POL(cons_2₂(x₁, x₂)) = 1 + x₂
POL(append_3_in_gga₂(x₁, x₂)) = 0
POL(if_split_4_in_1_ggaa₄(x₁, x₂, x₃, x₄)) = 1 + x₂
POL(append_3_out_gga₁(x₁)) = 0
POL(qs_2_in_ga₁(x₁)) = 0
POL(qs_2_out_ga₁(x₁)) = 0
POL([]_0) = 0
POL(split_4_out_ggaa₂(x₁, x₂)) = x₁ + x₂
POL(if_split_4_in_2_ggaa₂(x₁, x₂)) = 1 + x₂
POL(if_geq_2_in_1_gg₁(x₁)) = 0
POL(geq_2_in_gg₂(x₁, x₂)) = 0
POL(if_less_2_in_1_gg₁(x₁)) = 0
POL(s_1₁(x₁)) = 0
POL(if_split_4_in_4_ggaa₂(x₁, x₂)) = 1 + x₂
POL(if_split_4_in_3_ggaa₄(x₁, x₂, x₃, x₄)) = 1 + x₂
POL(less_2_in_gg₂(x₁, x₂)) = 0

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ PiDPToQDPProof

                  ↳ QDP

                    ↳ QDPPoloProof

                      ↳ QDP

                        ↳ DependencyGraphProof

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

IF_QS_2_IN_1_GA₂(X, split_4_out_ggaa₂(L1, L2)) -> QS_2_IN_GA₁(L1)
IF_QS_2_IN_2_GA₃(X, L2, qs_2_out_ga₁(S1)) -> QS_2_IN_GA₁(L2)
IF_QS_2_IN_1_GA₂(X, split_4_out_ggaa₂(L1, L2)) -> IF_QS_2_IN_2_GA₃(X, L2, qs_2_in_ga₁(L1))

The TRS R consists of the following rules:

qs_2_in_ga₁([]_0) -> qs_2_out_ga₁([]_0)
qs_2_in_ga₁(cons_2₂(X, L)) -> if_qs_2_in_1_ga₂(X, split_4_in_ggaa₂(L, X))
split_4_in_ggaa₂([]_0, X) -> split_4_out_ggaa₂([]_0, []_0)
split_4_in_ggaa₂(cons_2₂(X, L), Y) -> if_split_4_in_1_ggaa₄(X, L, Y, less_2_in_gg₂(X, Y))
less_2_in_gg₂(0_0, s_1₁(X)) -> 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_split_4_in_1_ggaa₄(X, L, Y, less_2_out_gg) -> if_split_4_in_2_ggaa₂(X, split_4_in_ggaa₂(L, Y))
split_4_in_ggaa₂(cons_2₂(X, L), Y) -> if_split_4_in_3_ggaa₄(X, L, Y, geq_2_in_gg₂(X, Y))
geq_2_in_gg₂(X, X) -> geq_2_out_gg
geq_2_in_gg₂(s_1₁(X), 0_0) -> geq_2_out_gg
geq_2_in_gg₂(s_1₁(X), s_1₁(Y)) -> if_geq_2_in_1_gg₁(geq_2_in_gg₂(X, Y))
if_geq_2_in_1_gg₁(geq_2_out_gg) -> geq_2_out_gg
if_split_4_in_3_ggaa₄(X, L, Y, geq_2_out_gg) -> if_split_4_in_4_ggaa₂(X, split_4_in_ggaa₂(L, Y))
if_split_4_in_4_ggaa₂(X, split_4_out_ggaa₂(L1, L2)) -> split_4_out_ggaa₂(L1, cons_2₂(X, L2))
if_split_4_in_2_ggaa₂(X, split_4_out_ggaa₂(L1, L2)) -> split_4_out_ggaa₂(cons_2₂(X, L1), L2)
if_qs_2_in_1_ga₂(X, split_4_out_ggaa₂(L1, L2)) -> if_qs_2_in_2_ga₃(X, L2, qs_2_in_ga₁(L1))
if_qs_2_in_2_ga₃(X, L2, qs_2_out_ga₁(S1)) -> if_qs_2_in_3_ga₃(X, S1, qs_2_in_ga₁(L2))
if_qs_2_in_3_ga₃(X, S1, qs_2_out_ga₁(S2)) -> if_qs_2_in_4_ga₁(append_3_in_gga₂(S1, cons_2₂(X, S2)))
append_3_in_gga₂([]_0, L) -> append_3_out_gga₁(L)
append_3_in_gga₂(cons_2₂(X, L1), L2) -> if_append_3_in_1_gga₂(X, append_3_in_gga₂(L1, L2))
if_append_3_in_1_gga₂(X, append_3_out_gga₁(L3)) -> append_3_out_gga₁(cons_2₂(X, L3))
if_qs_2_in_4_ga₁(append_3_out_gga₁(S)) -> qs_2_out_ga₁(S)

The set Q consists of the following terms:

qs_2_in_ga₁(x0)
split_4_in_ggaa₂(x0, x1)
less_2_in_gg₂(x0, x1)
if_less_2_in_1_gg₁(x0)
if_split_4_in_1_ggaa₄(x0, x1, x2, x3)
geq_2_in_gg₂(x0, x1)
if_geq_2_in_1_gg₁(x0)
if_split_4_in_3_ggaa₄(x0, x1, x2, x3)
if_split_4_in_4_ggaa₂(x0, x1)
if_split_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)
append_3_in_gga₂(x0, x1)
if_append_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_1_GA₂, IF_QS_2_IN_2_GA₃}.
The approximation of the Dependency Graph contains 0 SCCs with 3 less nodes.