Left Termination proof of /tmp/tmpndd0j9/slowsort-bf.pl

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

↳ PROLOG

  ↳ PrologToPiTRSProof

ss₂(Xs, Ys) :- perm₂(Xs, Ys), ordered₁(Ys).
perm₂({}₀, {}₀).
perm₂(Xs, .₂(X, Ys)) :- app₃(X1s, .₂(X, X2s), Xs), app₃(X1s, X2s, Zs), perm₂(Zs, Ys).
app₃({}₀, X, X).
app₃(.₂(X, Xs), Ys, .₂(X, Zs)) :- app₃(Xs, Ys, Zs).
ordered₁({}₀).
ordered₁(.₂(underscore, {}₀)).
ordered₁(.₂(X, .₂(Y, Xs))) :- less₂(X, s₁(Y)), ordered₁(.₂(Y, Xs)).
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:
ss₂: (b,f)
perm₂: (b,f)
app₃: (f,f,b) (b,b,f)
ordered₁: (b)
less₂: (b,b)
Transforming PROLOG into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

ss_2_in_ga₂(Xs, Ys) -> if_ss_2_in_1_ga₃(Xs, Ys, perm_2_in_ga₂(Xs, Ys))
perm_2_in_ga₂([]_0, []_0) -> perm_2_out_ga₂([]_0, []_0)
perm_2_in_ga₂(Xs, ._2₂(X, Ys)) -> if_perm_2_in_1_ga₄(Xs, X, Ys, app_3_in_aag₃(X1s, ._2₂(X, X2s), Xs))
app_3_in_aag₃([]_0, X, X) -> app_3_out_aag₃([]_0, X, X)
app_3_in_aag₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_app_3_in_1_aag₅(X, Xs, Ys, Zs, app_3_in_aag₃(Xs, Ys, Zs))
if_app_3_in_1_aag₅(X, Xs, Ys, Zs, app_3_out_aag₃(Xs, Ys, Zs)) -> app_3_out_aag₃(._2₂(X, Xs), Ys, ._2₂(X, Zs))
if_perm_2_in_1_ga₄(Xs, X, Ys, app_3_out_aag₃(X1s, ._2₂(X, X2s), Xs)) -> if_perm_2_in_2_ga₆(Xs, X, Ys, X1s, X2s, app_3_in_gga₃(X1s, X2s, Zs))
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_perm_2_in_2_ga₆(Xs, X, Ys, X1s, X2s, app_3_out_gga₃(X1s, X2s, Zs)) -> if_perm_2_in_3_ga₅(Xs, X, Ys, Zs, perm_2_in_ga₂(Zs, Ys))
if_perm_2_in_3_ga₅(Xs, X, Ys, Zs, perm_2_out_ga₂(Zs, Ys)) -> perm_2_out_ga₂(Xs, ._2₂(X, Ys))
if_ss_2_in_1_ga₃(Xs, Ys, perm_2_out_ga₂(Xs, Ys)) -> if_ss_2_in_2_ga₃(Xs, Ys, ordered_1_in_g₁(Ys))
ordered_1_in_g₁([]_0) -> ordered_1_out_g₁([]_0)
ordered_1_in_g₁(._2₂(underscore, []_0)) -> ordered_1_out_g₁(._2₂(underscore, []_0))
ordered_1_in_g₁(._2₂(X, ._2₂(Y, Xs))) -> if_ordered_1_in_1_g₄(X, Y, Xs, less_2_in_gg₂(X, s_1₁(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_ordered_1_in_1_g₄(X, Y, Xs, less_2_out_gg₂(X, s_1₁(Y))) -> if_ordered_1_in_2_g₄(X, Y, Xs, ordered_1_in_g₁(._2₂(Y, Xs)))
if_ordered_1_in_2_g₄(X, Y, Xs, ordered_1_out_g₁(._2₂(Y, Xs))) -> ordered_1_out_g₁(._2₂(X, ._2₂(Y, Xs)))
if_ss_2_in_2_ga₃(Xs, Ys, ordered_1_out_g₁(Ys)) -> ss_2_out_ga₂(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:

ss_2_in_ga₂(Xs, Ys) -> if_ss_2_in_1_ga₃(Xs, Ys, perm_2_in_ga₂(Xs, Ys))
perm_2_in_ga₂([]_0, []_0) -> perm_2_out_ga₂([]_0, []_0)
perm_2_in_ga₂(Xs, ._2₂(X, Ys)) -> if_perm_2_in_1_ga₄(Xs, X, Ys, app_3_in_aag₃(X1s, ._2₂(X, X2s), Xs))
app_3_in_aag₃([]_0, X, X) -> app_3_out_aag₃([]_0, X, X)
app_3_in_aag₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_app_3_in_1_aag₅(X, Xs, Ys, Zs, app_3_in_aag₃(Xs, Ys, Zs))
if_app_3_in_1_aag₅(X, Xs, Ys, Zs, app_3_out_aag₃(Xs, Ys, Zs)) -> app_3_out_aag₃(._2₂(X, Xs), Ys, ._2₂(X, Zs))
if_perm_2_in_1_ga₄(Xs, X, Ys, app_3_out_aag₃(X1s, ._2₂(X, X2s), Xs)) -> if_perm_2_in_2_ga₆(Xs, X, Ys, X1s, X2s, app_3_in_gga₃(X1s, X2s, Zs))
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_perm_2_in_2_ga₆(Xs, X, Ys, X1s, X2s, app_3_out_gga₃(X1s, X2s, Zs)) -> if_perm_2_in_3_ga₅(Xs, X, Ys, Zs, perm_2_in_ga₂(Zs, Ys))
if_perm_2_in_3_ga₅(Xs, X, Ys, Zs, perm_2_out_ga₂(Zs, Ys)) -> perm_2_out_ga₂(Xs, ._2₂(X, Ys))
if_ss_2_in_1_ga₃(Xs, Ys, perm_2_out_ga₂(Xs, Ys)) -> if_ss_2_in_2_ga₃(Xs, Ys, ordered_1_in_g₁(Ys))
ordered_1_in_g₁([]_0) -> ordered_1_out_g₁([]_0)
ordered_1_in_g₁(._2₂(underscore, []_0)) -> ordered_1_out_g₁(._2₂(underscore, []_0))
ordered_1_in_g₁(._2₂(X, ._2₂(Y, Xs))) -> if_ordered_1_in_1_g₄(X, Y, Xs, less_2_in_gg₂(X, s_1₁(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_ordered_1_in_1_g₄(X, Y, Xs, less_2_out_gg₂(X, s_1₁(Y))) -> if_ordered_1_in_2_g₄(X, Y, Xs, ordered_1_in_g₁(._2₂(Y, Xs)))
if_ordered_1_in_2_g₄(X, Y, Xs, ordered_1_out_g₁(._2₂(Y, Xs))) -> ordered_1_out_g₁(._2₂(X, ._2₂(Y, Xs)))
if_ss_2_in_2_ga₃(Xs, Ys, ordered_1_out_g₁(Ys)) -> ss_2_out_ga₂(Xs, Ys)

SS_2_IN_GA₂(Xs, Ys) -> IF_SS_2_IN_1_GA₃(Xs, Ys, perm_2_in_ga₂(Xs, Ys))
SS_2_IN_GA₂(Xs, Ys) -> PERM_2_IN_GA₂(Xs, Ys)
PERM_2_IN_GA₂(Xs, ._2₂(X, Ys)) -> IF_PERM_2_IN_1_GA₄(Xs, X, Ys, app_3_in_aag₃(X1s, ._2₂(X, X2s), Xs))
PERM_2_IN_GA₂(Xs, ._2₂(X, Ys)) -> APP_3_IN_AAG₃(X1s, ._2₂(X, X2s), Xs)
APP_3_IN_AAG₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> IF_APP_3_IN_1_AAG₅(X, Xs, Ys, Zs, app_3_in_aag₃(Xs, Ys, Zs))
APP_3_IN_AAG₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> APP_3_IN_AAG₃(Xs, Ys, Zs)
IF_PERM_2_IN_1_GA₄(Xs, X, Ys, app_3_out_aag₃(X1s, ._2₂(X, X2s), Xs)) -> IF_PERM_2_IN_2_GA₆(Xs, X, Ys, X1s, X2s, app_3_in_gga₃(X1s, X2s, Zs))
IF_PERM_2_IN_1_GA₄(Xs, X, Ys, app_3_out_aag₃(X1s, ._2₂(X, X2s), Xs)) -> APP_3_IN_GGA₃(X1s, X2s, Zs)
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)
IF_PERM_2_IN_2_GA₆(Xs, X, Ys, X1s, X2s, app_3_out_gga₃(X1s, X2s, Zs)) -> IF_PERM_2_IN_3_GA₅(Xs, X, Ys, Zs, perm_2_in_ga₂(Zs, Ys))
IF_PERM_2_IN_2_GA₆(Xs, X, Ys, X1s, X2s, app_3_out_gga₃(X1s, X2s, Zs)) -> PERM_2_IN_GA₂(Zs, Ys)
IF_SS_2_IN_1_GA₃(Xs, Ys, perm_2_out_ga₂(Xs, Ys)) -> IF_SS_2_IN_2_GA₃(Xs, Ys, ordered_1_in_g₁(Ys))
IF_SS_2_IN_1_GA₃(Xs, Ys, perm_2_out_ga₂(Xs, Ys)) -> ORDERED_1_IN_G₁(Ys)
ORDERED_1_IN_G₁(._2₂(X, ._2₂(Y, Xs))) -> IF_ORDERED_1_IN_1_G₄(X, Y, Xs, less_2_in_gg₂(X, s_1₁(Y)))
ORDERED_1_IN_G₁(._2₂(X, ._2₂(Y, Xs))) -> LESS_2_IN_GG₂(X, s_1₁(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_ORDERED_1_IN_1_G₄(X, Y, Xs, less_2_out_gg₂(X, s_1₁(Y))) -> IF_ORDERED_1_IN_2_G₄(X, Y, Xs, ordered_1_in_g₁(._2₂(Y, Xs)))
IF_ORDERED_1_IN_1_G₄(X, Y, Xs, less_2_out_gg₂(X, s_1₁(Y))) -> ORDERED_1_IN_G₁(._2₂(Y, Xs))

The TRS R consists of the following rules:

ss_2_in_ga₂(Xs, Ys) -> if_ss_2_in_1_ga₃(Xs, Ys, perm_2_in_ga₂(Xs, Ys))
perm_2_in_ga₂([]_0, []_0) -> perm_2_out_ga₂([]_0, []_0)
perm_2_in_ga₂(Xs, ._2₂(X, Ys)) -> if_perm_2_in_1_ga₄(Xs, X, Ys, app_3_in_aag₃(X1s, ._2₂(X, X2s), Xs))
app_3_in_aag₃([]_0, X, X) -> app_3_out_aag₃([]_0, X, X)
app_3_in_aag₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_app_3_in_1_aag₅(X, Xs, Ys, Zs, app_3_in_aag₃(Xs, Ys, Zs))
if_app_3_in_1_aag₅(X, Xs, Ys, Zs, app_3_out_aag₃(Xs, Ys, Zs)) -> app_3_out_aag₃(._2₂(X, Xs), Ys, ._2₂(X, Zs))
if_perm_2_in_1_ga₄(Xs, X, Ys, app_3_out_aag₃(X1s, ._2₂(X, X2s), Xs)) -> if_perm_2_in_2_ga₆(Xs, X, Ys, X1s, X2s, app_3_in_gga₃(X1s, X2s, Zs))
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_perm_2_in_2_ga₆(Xs, X, Ys, X1s, X2s, app_3_out_gga₃(X1s, X2s, Zs)) -> if_perm_2_in_3_ga₅(Xs, X, Ys, Zs, perm_2_in_ga₂(Zs, Ys))
if_perm_2_in_3_ga₅(Xs, X, Ys, Zs, perm_2_out_ga₂(Zs, Ys)) -> perm_2_out_ga₂(Xs, ._2₂(X, Ys))
if_ss_2_in_1_ga₃(Xs, Ys, perm_2_out_ga₂(Xs, Ys)) -> if_ss_2_in_2_ga₃(Xs, Ys, ordered_1_in_g₁(Ys))
ordered_1_in_g₁([]_0) -> ordered_1_out_g₁([]_0)
ordered_1_in_g₁(._2₂(underscore, []_0)) -> ordered_1_out_g₁(._2₂(underscore, []_0))
ordered_1_in_g₁(._2₂(X, ._2₂(Y, Xs))) -> if_ordered_1_in_1_g₄(X, Y, Xs, less_2_in_gg₂(X, s_1₁(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_ordered_1_in_1_g₄(X, Y, Xs, less_2_out_gg₂(X, s_1₁(Y))) -> if_ordered_1_in_2_g₄(X, Y, Xs, ordered_1_in_g₁(._2₂(Y, Xs)))
if_ordered_1_in_2_g₄(X, Y, Xs, ordered_1_out_g₁(._2₂(Y, Xs))) -> ordered_1_out_g₁(._2₂(X, ._2₂(Y, Xs)))
if_ss_2_in_2_ga₃(Xs, Ys, ordered_1_out_g₁(Ys)) -> ss_2_out_ga₂(Xs, Ys)

The argument filtering Pi contains the following mapping:
ss_2_in_ga₂(x1, x2) = ss_2_in_ga₁(x1)
[]_0 = []_0
._2₂(x1, x2) = ._2₂(x1, x2)
s_1₁(x1) = s_1₁(x1)
0_0 = 0_0
if_ss_2_in_1_ga₃(x1, x2, x3) = if_ss_2_in_1_ga₁(x3)
perm_2_in_ga₂(x1, x2) = perm_2_in_ga₁(x1)
perm_2_out_ga₂(x1, x2) = perm_2_out_ga₁(x2)
if_perm_2_in_1_ga₄(x1, x2, x3, x4) = if_perm_2_in_1_ga₁(x4)
app_3_in_aag₃(x1, x2, x3) = app_3_in_aag₁(x3)
app_3_out_aag₃(x1, x2, x3) = app_3_out_aag₂(x1, x2)
if_app_3_in_1_aag₅(x1, x2, x3, x4, x5) = if_app_3_in_1_aag₂(x1, x5)
if_perm_2_in_2_ga₆(x1, x2, x3, x4, x5, x6) = if_perm_2_in_2_ga₂(x2, 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_perm_2_in_3_ga₅(x1, x2, x3, x4, x5) = if_perm_2_in_3_ga₂(x2, x5)
if_ss_2_in_2_ga₃(x1, x2, x3) = if_ss_2_in_2_ga₂(x2, x3)
ordered_1_in_g₁(x1) = ordered_1_in_g₁(x1)
ordered_1_out_g₁(x1) = ordered_1_out_g
if_ordered_1_in_1_g₄(x1, x2, x3, x4) = if_ordered_1_in_1_g₃(x2, x3, x4)
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_ordered_1_in_2_g₄(x1, x2, x3, x4) = if_ordered_1_in_2_g₁(x4)
ss_2_out_ga₂(x1, x2) = ss_2_out_ga₁(x2)
SS_2_IN_GA₂(x1, x2) = SS_2_IN_GA₁(x1)
IF_APP_3_IN_1_AAG₅(x1, x2, x3, x4, x5) = IF_APP_3_IN_1_AAG₂(x1, x5)
APP_3_IN_GGA₃(x1, x2, x3) = APP_3_IN_GGA₂(x1, x2)
PERM_2_IN_GA₂(x1, x2) = PERM_2_IN_GA₁(x1)
IF_SS_2_IN_2_GA₃(x1, x2, x3) = IF_SS_2_IN_2_GA₂(x2, x3)
IF_SS_2_IN_1_GA₃(x1, x2, x3) = IF_SS_2_IN_1_GA₁(x3)
IF_ORDERED_1_IN_2_G₄(x1, x2, x3, x4) = IF_ORDERED_1_IN_2_G₁(x4)
IF_APP_3_IN_1_GGA₅(x1, x2, x3, x4, x5) = IF_APP_3_IN_1_GGA₂(x1, x5)
IF_ORDERED_1_IN_1_G₄(x1, x2, x3, x4) = IF_ORDERED_1_IN_1_G₃(x2, x3, x4)
IF_LESS_2_IN_1_GG₃(x1, x2, x3) = IF_LESS_2_IN_1_GG₁(x3)
IF_PERM_2_IN_3_GA₅(x1, x2, x3, x4, x5) = IF_PERM_2_IN_3_GA₂(x2, x5)
APP_3_IN_AAG₃(x1, x2, x3) = APP_3_IN_AAG₁(x3)
ORDERED_1_IN_G₁(x1) = ORDERED_1_IN_G₁(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)
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:

SS_2_IN_GA₂(Xs, Ys) -> IF_SS_2_IN_1_GA₃(Xs, Ys, perm_2_in_ga₂(Xs, Ys))
SS_2_IN_GA₂(Xs, Ys) -> PERM_2_IN_GA₂(Xs, Ys)
PERM_2_IN_GA₂(Xs, ._2₂(X, Ys)) -> IF_PERM_2_IN_1_GA₄(Xs, X, Ys, app_3_in_aag₃(X1s, ._2₂(X, X2s), Xs))
PERM_2_IN_GA₂(Xs, ._2₂(X, Ys)) -> APP_3_IN_AAG₃(X1s, ._2₂(X, X2s), Xs)
APP_3_IN_AAG₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> IF_APP_3_IN_1_AAG₅(X, Xs, Ys, Zs, app_3_in_aag₃(Xs, Ys, Zs))
APP_3_IN_AAG₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> APP_3_IN_AAG₃(Xs, Ys, Zs)
IF_PERM_2_IN_1_GA₄(Xs, X, Ys, app_3_out_aag₃(X1s, ._2₂(X, X2s), Xs)) -> IF_PERM_2_IN_2_GA₆(Xs, X, Ys, X1s, X2s, app_3_in_gga₃(X1s, X2s, Zs))
IF_PERM_2_IN_1_GA₄(Xs, X, Ys, app_3_out_aag₃(X1s, ._2₂(X, X2s), Xs)) -> APP_3_IN_GGA₃(X1s, X2s, Zs)
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)
IF_PERM_2_IN_2_GA₆(Xs, X, Ys, X1s, X2s, app_3_out_gga₃(X1s, X2s, Zs)) -> IF_PERM_2_IN_3_GA₅(Xs, X, Ys, Zs, perm_2_in_ga₂(Zs, Ys))
IF_PERM_2_IN_2_GA₆(Xs, X, Ys, X1s, X2s, app_3_out_gga₃(X1s, X2s, Zs)) -> PERM_2_IN_GA₂(Zs, Ys)
IF_SS_2_IN_1_GA₃(Xs, Ys, perm_2_out_ga₂(Xs, Ys)) -> IF_SS_2_IN_2_GA₃(Xs, Ys, ordered_1_in_g₁(Ys))
IF_SS_2_IN_1_GA₃(Xs, Ys, perm_2_out_ga₂(Xs, Ys)) -> ORDERED_1_IN_G₁(Ys)
ORDERED_1_IN_G₁(._2₂(X, ._2₂(Y, Xs))) -> IF_ORDERED_1_IN_1_G₄(X, Y, Xs, less_2_in_gg₂(X, s_1₁(Y)))
ORDERED_1_IN_G₁(._2₂(X, ._2₂(Y, Xs))) -> LESS_2_IN_GG₂(X, s_1₁(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_ORDERED_1_IN_1_G₄(X, Y, Xs, less_2_out_gg₂(X, s_1₁(Y))) -> IF_ORDERED_1_IN_2_G₄(X, Y, Xs, ordered_1_in_g₁(._2₂(Y, Xs)))
IF_ORDERED_1_IN_1_G₄(X, Y, Xs, less_2_out_gg₂(X, s_1₁(Y))) -> ORDERED_1_IN_G₁(._2₂(Y, Xs))

The TRS R consists of the following rules:

ss_2_in_ga₂(Xs, Ys) -> if_ss_2_in_1_ga₃(Xs, Ys, perm_2_in_ga₂(Xs, Ys))
perm_2_in_ga₂([]_0, []_0) -> perm_2_out_ga₂([]_0, []_0)
perm_2_in_ga₂(Xs, ._2₂(X, Ys)) -> if_perm_2_in_1_ga₄(Xs, X, Ys, app_3_in_aag₃(X1s, ._2₂(X, X2s), Xs))
app_3_in_aag₃([]_0, X, X) -> app_3_out_aag₃([]_0, X, X)
app_3_in_aag₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_app_3_in_1_aag₅(X, Xs, Ys, Zs, app_3_in_aag₃(Xs, Ys, Zs))
if_app_3_in_1_aag₅(X, Xs, Ys, Zs, app_3_out_aag₃(Xs, Ys, Zs)) -> app_3_out_aag₃(._2₂(X, Xs), Ys, ._2₂(X, Zs))
if_perm_2_in_1_ga₄(Xs, X, Ys, app_3_out_aag₃(X1s, ._2₂(X, X2s), Xs)) -> if_perm_2_in_2_ga₆(Xs, X, Ys, X1s, X2s, app_3_in_gga₃(X1s, X2s, Zs))
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_perm_2_in_2_ga₆(Xs, X, Ys, X1s, X2s, app_3_out_gga₃(X1s, X2s, Zs)) -> if_perm_2_in_3_ga₅(Xs, X, Ys, Zs, perm_2_in_ga₂(Zs, Ys))
if_perm_2_in_3_ga₅(Xs, X, Ys, Zs, perm_2_out_ga₂(Zs, Ys)) -> perm_2_out_ga₂(Xs, ._2₂(X, Ys))
if_ss_2_in_1_ga₃(Xs, Ys, perm_2_out_ga₂(Xs, Ys)) -> if_ss_2_in_2_ga₃(Xs, Ys, ordered_1_in_g₁(Ys))
ordered_1_in_g₁([]_0) -> ordered_1_out_g₁([]_0)
ordered_1_in_g₁(._2₂(underscore, []_0)) -> ordered_1_out_g₁(._2₂(underscore, []_0))
ordered_1_in_g₁(._2₂(X, ._2₂(Y, Xs))) -> if_ordered_1_in_1_g₄(X, Y, Xs, less_2_in_gg₂(X, s_1₁(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_ordered_1_in_1_g₄(X, Y, Xs, less_2_out_gg₂(X, s_1₁(Y))) -> if_ordered_1_in_2_g₄(X, Y, Xs, ordered_1_in_g₁(._2₂(Y, Xs)))
if_ordered_1_in_2_g₄(X, Y, Xs, ordered_1_out_g₁(._2₂(Y, Xs))) -> ordered_1_out_g₁(._2₂(X, ._2₂(Y, Xs)))
if_ss_2_in_2_ga₃(Xs, Ys, ordered_1_out_g₁(Ys)) -> ss_2_out_ga₂(Xs, Ys)

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

              ↳ PiDP

              ↳ PiDP

              ↳ 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:

ss_2_in_ga₂(Xs, Ys) -> if_ss_2_in_1_ga₃(Xs, Ys, perm_2_in_ga₂(Xs, Ys))
perm_2_in_ga₂([]_0, []_0) -> perm_2_out_ga₂([]_0, []_0)
perm_2_in_ga₂(Xs, ._2₂(X, Ys)) -> if_perm_2_in_1_ga₄(Xs, X, Ys, app_3_in_aag₃(X1s, ._2₂(X, X2s), Xs))
app_3_in_aag₃([]_0, X, X) -> app_3_out_aag₃([]_0, X, X)
app_3_in_aag₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_app_3_in_1_aag₅(X, Xs, Ys, Zs, app_3_in_aag₃(Xs, Ys, Zs))
if_app_3_in_1_aag₅(X, Xs, Ys, Zs, app_3_out_aag₃(Xs, Ys, Zs)) -> app_3_out_aag₃(._2₂(X, Xs), Ys, ._2₂(X, Zs))
if_perm_2_in_1_ga₄(Xs, X, Ys, app_3_out_aag₃(X1s, ._2₂(X, X2s), Xs)) -> if_perm_2_in_2_ga₆(Xs, X, Ys, X1s, X2s, app_3_in_gga₃(X1s, X2s, Zs))
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_perm_2_in_2_ga₆(Xs, X, Ys, X1s, X2s, app_3_out_gga₃(X1s, X2s, Zs)) -> if_perm_2_in_3_ga₅(Xs, X, Ys, Zs, perm_2_in_ga₂(Zs, Ys))
if_perm_2_in_3_ga₅(Xs, X, Ys, Zs, perm_2_out_ga₂(Zs, Ys)) -> perm_2_out_ga₂(Xs, ._2₂(X, Ys))
if_ss_2_in_1_ga₃(Xs, Ys, perm_2_out_ga₂(Xs, Ys)) -> if_ss_2_in_2_ga₃(Xs, Ys, ordered_1_in_g₁(Ys))
ordered_1_in_g₁([]_0) -> ordered_1_out_g₁([]_0)
ordered_1_in_g₁(._2₂(underscore, []_0)) -> ordered_1_out_g₁(._2₂(underscore, []_0))
ordered_1_in_g₁(._2₂(X, ._2₂(Y, Xs))) -> if_ordered_1_in_1_g₄(X, Y, Xs, less_2_in_gg₂(X, s_1₁(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_ordered_1_in_1_g₄(X, Y, Xs, less_2_out_gg₂(X, s_1₁(Y))) -> if_ordered_1_in_2_g₄(X, Y, Xs, ordered_1_in_g₁(._2₂(Y, Xs)))
if_ordered_1_in_2_g₄(X, Y, Xs, ordered_1_out_g₁(._2₂(Y, Xs))) -> ordered_1_out_g₁(._2₂(X, ._2₂(Y, Xs)))
if_ss_2_in_2_ga₃(Xs, Ys, ordered_1_out_g₁(Ys)) -> ss_2_out_ga₂(Xs, Ys)

↳ 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:

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

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ PiDP

              ↳ PiDP

              ↳ 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

                ↳ UsableRulesProof

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

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

ORDERED_1_IN_G₁(._2₂(X, ._2₂(Y, Xs))) -> IF_ORDERED_1_IN_1_G₄(X, Y, Xs, less_2_in_gg₂(X, s_1₁(Y)))
IF_ORDERED_1_IN_1_G₄(X, Y, Xs, less_2_out_gg₂(X, s_1₁(Y))) -> ORDERED_1_IN_G₁(._2₂(Y, Xs))

The TRS R consists of the following rules:

ss_2_in_ga₂(Xs, Ys) -> if_ss_2_in_1_ga₃(Xs, Ys, perm_2_in_ga₂(Xs, Ys))
perm_2_in_ga₂([]_0, []_0) -> perm_2_out_ga₂([]_0, []_0)
perm_2_in_ga₂(Xs, ._2₂(X, Ys)) -> if_perm_2_in_1_ga₄(Xs, X, Ys, app_3_in_aag₃(X1s, ._2₂(X, X2s), Xs))
app_3_in_aag₃([]_0, X, X) -> app_3_out_aag₃([]_0, X, X)
app_3_in_aag₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_app_3_in_1_aag₅(X, Xs, Ys, Zs, app_3_in_aag₃(Xs, Ys, Zs))
if_app_3_in_1_aag₅(X, Xs, Ys, Zs, app_3_out_aag₃(Xs, Ys, Zs)) -> app_3_out_aag₃(._2₂(X, Xs), Ys, ._2₂(X, Zs))
if_perm_2_in_1_ga₄(Xs, X, Ys, app_3_out_aag₃(X1s, ._2₂(X, X2s), Xs)) -> if_perm_2_in_2_ga₆(Xs, X, Ys, X1s, X2s, app_3_in_gga₃(X1s, X2s, Zs))
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_perm_2_in_2_ga₆(Xs, X, Ys, X1s, X2s, app_3_out_gga₃(X1s, X2s, Zs)) -> if_perm_2_in_3_ga₅(Xs, X, Ys, Zs, perm_2_in_ga₂(Zs, Ys))
if_perm_2_in_3_ga₅(Xs, X, Ys, Zs, perm_2_out_ga₂(Zs, Ys)) -> perm_2_out_ga₂(Xs, ._2₂(X, Ys))
if_ss_2_in_1_ga₃(Xs, Ys, perm_2_out_ga₂(Xs, Ys)) -> if_ss_2_in_2_ga₃(Xs, Ys, ordered_1_in_g₁(Ys))
ordered_1_in_g₁([]_0) -> ordered_1_out_g₁([]_0)
ordered_1_in_g₁(._2₂(underscore, []_0)) -> ordered_1_out_g₁(._2₂(underscore, []_0))
ordered_1_in_g₁(._2₂(X, ._2₂(Y, Xs))) -> if_ordered_1_in_1_g₄(X, Y, Xs, less_2_in_gg₂(X, s_1₁(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_ordered_1_in_1_g₄(X, Y, Xs, less_2_out_gg₂(X, s_1₁(Y))) -> if_ordered_1_in_2_g₄(X, Y, Xs, ordered_1_in_g₁(._2₂(Y, Xs)))
if_ordered_1_in_2_g₄(X, Y, Xs, ordered_1_out_g₁(._2₂(Y, Xs))) -> ordered_1_out_g₁(._2₂(X, ._2₂(Y, Xs)))
if_ss_2_in_2_ga₃(Xs, Ys, ordered_1_out_g₁(Ys)) -> ss_2_out_ga₂(Xs, Ys)

↳ 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:

ORDERED_1_IN_G₁(._2₂(X, ._2₂(Y, Xs))) -> IF_ORDERED_1_IN_1_G₄(X, Y, Xs, less_2_in_gg₂(X, s_1₁(Y)))
IF_ORDERED_1_IN_1_G₄(X, Y, Xs, less_2_out_gg₂(X, s_1₁(Y))) -> ORDERED_1_IN_G₁(._2₂(Y, Xs))

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)
s_1₁(x1) = s_1₁(x1)
0_0 = 0_0
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_ORDERED_1_IN_1_G₄(x1, x2, x3, x4) = IF_ORDERED_1_IN_1_G₃(x2, x3, x4)
ORDERED_1_IN_G₁(x1) = ORDERED_1_IN_G₁(x1)

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

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ QDPPoloProof

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

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

ORDERED_1_IN_G₁(._2₂(X, ._2₂(Y, Xs))) -> IF_ORDERED_1_IN_1_G₃(Y, Xs, less_2_in_gg₂(X, s_1₁(Y)))
IF_ORDERED_1_IN_1_G₃(Y, Xs, less_2_out_gg) -> ORDERED_1_IN_G₁(._2₂(Y, 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 {IF_ORDERED_1_IN_1_G₃, ORDERED_1_IN_G₁}.
By using a polynomial ordering, the following set of Dependency Pairs of this DP problem can be strictly oriented.

ORDERED_1_IN_G₁(._2₂(X, ._2₂(Y, Xs))) -> IF_ORDERED_1_IN_1_G₃(Y, Xs, less_2_in_gg₂(X, s_1₁(Y)))

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

IF_ORDERED_1_IN_1_G₃(Y, Xs, less_2_out_gg) -> ORDERED_1_IN_G₁(._2₂(Y, Xs))

With the implicit AFS there is no usable rule.

Used ordering: POLO with Polynomial interpretation:

POL(0_0) = 0
POL(less_2_out_gg) = 0
POL(._2₂(x₁, x₂)) = 1 + x₂
POL(ORDERED_1_IN_G₁(x₁)) = x₁
POL(if_less_2_in_1_gg₁(x₁)) = 0
POL(s_1₁(x₁)) = 0
POL(less_2_in_gg₂(x₁, x₂)) = 0
POL(IF_ORDERED_1_IN_1_G₃(x₁, x₂, x₃)) = 1 + x₂

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ QDPPoloProof

                          ↳ QDP

                            ↳ DependencyGraphProof

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

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

IF_ORDERED_1_IN_1_G₃(Y, Xs, less_2_out_gg) -> ORDERED_1_IN_G₁(._2₂(Y, 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 {ORDERED_1_IN_G₁, IF_ORDERED_1_IN_1_G₃}.
The approximation of the Dependency Graph contains 0 SCCs with 1 less node.

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

              ↳ 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:

ss_2_in_ga₂(Xs, Ys) -> if_ss_2_in_1_ga₃(Xs, Ys, perm_2_in_ga₂(Xs, Ys))
perm_2_in_ga₂([]_0, []_0) -> perm_2_out_ga₂([]_0, []_0)
perm_2_in_ga₂(Xs, ._2₂(X, Ys)) -> if_perm_2_in_1_ga₄(Xs, X, Ys, app_3_in_aag₃(X1s, ._2₂(X, X2s), Xs))
app_3_in_aag₃([]_0, X, X) -> app_3_out_aag₃([]_0, X, X)
app_3_in_aag₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_app_3_in_1_aag₅(X, Xs, Ys, Zs, app_3_in_aag₃(Xs, Ys, Zs))
if_app_3_in_1_aag₅(X, Xs, Ys, Zs, app_3_out_aag₃(Xs, Ys, Zs)) -> app_3_out_aag₃(._2₂(X, Xs), Ys, ._2₂(X, Zs))
if_perm_2_in_1_ga₄(Xs, X, Ys, app_3_out_aag₃(X1s, ._2₂(X, X2s), Xs)) -> if_perm_2_in_2_ga₆(Xs, X, Ys, X1s, X2s, app_3_in_gga₃(X1s, X2s, Zs))
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_perm_2_in_2_ga₆(Xs, X, Ys, X1s, X2s, app_3_out_gga₃(X1s, X2s, Zs)) -> if_perm_2_in_3_ga₅(Xs, X, Ys, Zs, perm_2_in_ga₂(Zs, Ys))
if_perm_2_in_3_ga₅(Xs, X, Ys, Zs, perm_2_out_ga₂(Zs, Ys)) -> perm_2_out_ga₂(Xs, ._2₂(X, Ys))
if_ss_2_in_1_ga₃(Xs, Ys, perm_2_out_ga₂(Xs, Ys)) -> if_ss_2_in_2_ga₃(Xs, Ys, ordered_1_in_g₁(Ys))
ordered_1_in_g₁([]_0) -> ordered_1_out_g₁([]_0)
ordered_1_in_g₁(._2₂(underscore, []_0)) -> ordered_1_out_g₁(._2₂(underscore, []_0))
ordered_1_in_g₁(._2₂(X, ._2₂(Y, Xs))) -> if_ordered_1_in_1_g₄(X, Y, Xs, less_2_in_gg₂(X, s_1₁(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_ordered_1_in_1_g₄(X, Y, Xs, less_2_out_gg₂(X, s_1₁(Y))) -> if_ordered_1_in_2_g₄(X, Y, Xs, ordered_1_in_g₁(._2₂(Y, Xs)))
if_ordered_1_in_2_g₄(X, Y, Xs, ordered_1_out_g₁(._2₂(Y, Xs))) -> ordered_1_out_g₁(._2₂(X, ._2₂(Y, Xs)))
if_ss_2_in_2_ga₃(Xs, Ys, ordered_1_out_g₁(Ys)) -> ss_2_out_ga₂(Xs, Ys)

↳ 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:

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

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ 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

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

              ↳ PiDP

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

APP_3_IN_AAG₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> APP_3_IN_AAG₃(Xs, Ys, Zs)

The TRS R consists of the following rules:

ss_2_in_ga₂(Xs, Ys) -> if_ss_2_in_1_ga₃(Xs, Ys, perm_2_in_ga₂(Xs, Ys))
perm_2_in_ga₂([]_0, []_0) -> perm_2_out_ga₂([]_0, []_0)
perm_2_in_ga₂(Xs, ._2₂(X, Ys)) -> if_perm_2_in_1_ga₄(Xs, X, Ys, app_3_in_aag₃(X1s, ._2₂(X, X2s), Xs))
app_3_in_aag₃([]_0, X, X) -> app_3_out_aag₃([]_0, X, X)
app_3_in_aag₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_app_3_in_1_aag₅(X, Xs, Ys, Zs, app_3_in_aag₃(Xs, Ys, Zs))
if_app_3_in_1_aag₅(X, Xs, Ys, Zs, app_3_out_aag₃(Xs, Ys, Zs)) -> app_3_out_aag₃(._2₂(X, Xs), Ys, ._2₂(X, Zs))
if_perm_2_in_1_ga₄(Xs, X, Ys, app_3_out_aag₃(X1s, ._2₂(X, X2s), Xs)) -> if_perm_2_in_2_ga₆(Xs, X, Ys, X1s, X2s, app_3_in_gga₃(X1s, X2s, Zs))
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_perm_2_in_2_ga₆(Xs, X, Ys, X1s, X2s, app_3_out_gga₃(X1s, X2s, Zs)) -> if_perm_2_in_3_ga₅(Xs, X, Ys, Zs, perm_2_in_ga₂(Zs, Ys))
if_perm_2_in_3_ga₅(Xs, X, Ys, Zs, perm_2_out_ga₂(Zs, Ys)) -> perm_2_out_ga₂(Xs, ._2₂(X, Ys))
if_ss_2_in_1_ga₃(Xs, Ys, perm_2_out_ga₂(Xs, Ys)) -> if_ss_2_in_2_ga₃(Xs, Ys, ordered_1_in_g₁(Ys))
ordered_1_in_g₁([]_0) -> ordered_1_out_g₁([]_0)
ordered_1_in_g₁(._2₂(underscore, []_0)) -> ordered_1_out_g₁(._2₂(underscore, []_0))
ordered_1_in_g₁(._2₂(X, ._2₂(Y, Xs))) -> if_ordered_1_in_1_g₄(X, Y, Xs, less_2_in_gg₂(X, s_1₁(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_ordered_1_in_1_g₄(X, Y, Xs, less_2_out_gg₂(X, s_1₁(Y))) -> if_ordered_1_in_2_g₄(X, Y, Xs, ordered_1_in_g₁(._2₂(Y, Xs)))
if_ordered_1_in_2_g₄(X, Y, Xs, ordered_1_out_g₁(._2₂(Y, Xs))) -> ordered_1_out_g₁(._2₂(X, ._2₂(Y, Xs)))
if_ss_2_in_2_ga₃(Xs, Ys, ordered_1_out_g₁(Ys)) -> ss_2_out_ga₂(Xs, Ys)

↳ 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:

APP_3_IN_AAG₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> APP_3_IN_AAG₃(Xs, Ys, Zs)

R is empty.
The argument filtering Pi contains the following mapping:
._2₂(x1, x2) = ._2₂(x1, x2)
APP_3_IN_AAG₃(x1, x2, x3) = APP_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

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ PiDP

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

APP_3_IN_AAG₁(._2₂(X, Zs)) -> APP_3_IN_AAG₁(Zs)

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_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:

APP_3_IN_AAG₁(._2₂(X, Zs)) -> APP_3_IN_AAG₁(Zs)
The graph contains the following edges 1 > 1

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

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

IF_PERM_2_IN_2_GA₆(Xs, X, Ys, X1s, X2s, app_3_out_gga₃(X1s, X2s, Zs)) -> PERM_2_IN_GA₂(Zs, Ys)
PERM_2_IN_GA₂(Xs, ._2₂(X, Ys)) -> IF_PERM_2_IN_1_GA₄(Xs, X, Ys, app_3_in_aag₃(X1s, ._2₂(X, X2s), Xs))
IF_PERM_2_IN_1_GA₄(Xs, X, Ys, app_3_out_aag₃(X1s, ._2₂(X, X2s), Xs)) -> IF_PERM_2_IN_2_GA₆(Xs, X, Ys, X1s, X2s, app_3_in_gga₃(X1s, X2s, Zs))

The TRS R consists of the following rules:

ss_2_in_ga₂(Xs, Ys) -> if_ss_2_in_1_ga₃(Xs, Ys, perm_2_in_ga₂(Xs, Ys))
perm_2_in_ga₂([]_0, []_0) -> perm_2_out_ga₂([]_0, []_0)
perm_2_in_ga₂(Xs, ._2₂(X, Ys)) -> if_perm_2_in_1_ga₄(Xs, X, Ys, app_3_in_aag₃(X1s, ._2₂(X, X2s), Xs))
app_3_in_aag₃([]_0, X, X) -> app_3_out_aag₃([]_0, X, X)
app_3_in_aag₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_app_3_in_1_aag₅(X, Xs, Ys, Zs, app_3_in_aag₃(Xs, Ys, Zs))
if_app_3_in_1_aag₅(X, Xs, Ys, Zs, app_3_out_aag₃(Xs, Ys, Zs)) -> app_3_out_aag₃(._2₂(X, Xs), Ys, ._2₂(X, Zs))
if_perm_2_in_1_ga₄(Xs, X, Ys, app_3_out_aag₃(X1s, ._2₂(X, X2s), Xs)) -> if_perm_2_in_2_ga₆(Xs, X, Ys, X1s, X2s, app_3_in_gga₃(X1s, X2s, Zs))
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_perm_2_in_2_ga₆(Xs, X, Ys, X1s, X2s, app_3_out_gga₃(X1s, X2s, Zs)) -> if_perm_2_in_3_ga₅(Xs, X, Ys, Zs, perm_2_in_ga₂(Zs, Ys))
if_perm_2_in_3_ga₅(Xs, X, Ys, Zs, perm_2_out_ga₂(Zs, Ys)) -> perm_2_out_ga₂(Xs, ._2₂(X, Ys))
if_ss_2_in_1_ga₃(Xs, Ys, perm_2_out_ga₂(Xs, Ys)) -> if_ss_2_in_2_ga₃(Xs, Ys, ordered_1_in_g₁(Ys))
ordered_1_in_g₁([]_0) -> ordered_1_out_g₁([]_0)
ordered_1_in_g₁(._2₂(underscore, []_0)) -> ordered_1_out_g₁(._2₂(underscore, []_0))
ordered_1_in_g₁(._2₂(X, ._2₂(Y, Xs))) -> if_ordered_1_in_1_g₄(X, Y, Xs, less_2_in_gg₂(X, s_1₁(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_ordered_1_in_1_g₄(X, Y, Xs, less_2_out_gg₂(X, s_1₁(Y))) -> if_ordered_1_in_2_g₄(X, Y, Xs, ordered_1_in_g₁(._2₂(Y, Xs)))
if_ordered_1_in_2_g₄(X, Y, Xs, ordered_1_out_g₁(._2₂(Y, Xs))) -> ordered_1_out_g₁(._2₂(X, ._2₂(Y, Xs)))
if_ss_2_in_2_ga₃(Xs, Ys, ordered_1_out_g₁(Ys)) -> ss_2_out_ga₂(Xs, Ys)

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

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

IF_PERM_2_IN_2_GA₆(Xs, X, Ys, X1s, X2s, app_3_out_gga₃(X1s, X2s, Zs)) -> PERM_2_IN_GA₂(Zs, Ys)
PERM_2_IN_GA₂(Xs, ._2₂(X, Ys)) -> IF_PERM_2_IN_1_GA₄(Xs, X, Ys, app_3_in_aag₃(X1s, ._2₂(X, X2s), Xs))
IF_PERM_2_IN_1_GA₄(Xs, X, Ys, app_3_out_aag₃(X1s, ._2₂(X, X2s), Xs)) -> IF_PERM_2_IN_2_GA₆(Xs, X, Ys, X1s, X2s, app_3_in_gga₃(X1s, X2s, Zs))

The TRS R consists of the following rules:

app_3_in_aag₃([]_0, X, X) -> app_3_out_aag₃([]_0, X, X)
app_3_in_aag₃(._2₂(X, Xs), Ys, ._2₂(X, Zs)) -> if_app_3_in_1_aag₅(X, Xs, Ys, Zs, app_3_in_aag₃(Xs, Ys, Zs))
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_aag₅(X, Xs, Ys, Zs, app_3_out_aag₃(Xs, Ys, Zs)) -> app_3_out_aag₃(._2₂(X, Xs), Ys, ._2₂(X, 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))

The argument filtering Pi contains the following mapping:
[]_0 = []_0
._2₂(x1, x2) = ._2₂(x1, x2)
app_3_in_aag₃(x1, x2, x3) = app_3_in_aag₁(x3)
app_3_out_aag₃(x1, x2, x3) = app_3_out_aag₂(x1, x2)
if_app_3_in_1_aag₅(x1, x2, x3, x4, x5) = if_app_3_in_1_aag₂(x1, x5)
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)
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

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ QDPPoloProof

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

IF_PERM_2_IN_2_GA₂(X, app_3_out_gga₁(Zs)) -> PERM_2_IN_GA₁(Zs)
PERM_2_IN_GA₁(Xs) -> IF_PERM_2_IN_1_GA₁(app_3_in_aag₁(Xs))
IF_PERM_2_IN_1_GA₁(app_3_out_aag₂(X1s, ._2₂(X, X2s))) -> IF_PERM_2_IN_2_GA₂(X, app_3_in_gga₂(X1s, X2s))

The TRS R consists of the following rules:

app_3_in_aag₁(X) -> app_3_out_aag₂([]_0, X)
app_3_in_aag₁(._2₂(X, Zs)) -> if_app_3_in_1_aag₂(X, app_3_in_aag₁(Zs))
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_aag₂(X, app_3_out_aag₂(Xs, Ys)) -> app_3_out_aag₂(._2₂(X, Xs), Ys)
if_app_3_in_1_gga₂(X, app_3_out_gga₁(Zs)) -> app_3_out_gga₁(._2₂(X, Zs))

The set Q consists of the following terms:

app_3_in_aag₁(x0)
app_3_in_gga₂(x0, x1)
if_app_3_in_1_aag₂(x0, x1)
if_app_3_in_1_gga₂(x0, x1)

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

IF_PERM_2_IN_1_GA₁(app_3_out_aag₂(X1s, ._2₂(X, X2s))) -> IF_PERM_2_IN_2_GA₂(X, app_3_in_gga₂(X1s, X2s))

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

IF_PERM_2_IN_2_GA₂(X, app_3_out_gga₁(Zs)) -> PERM_2_IN_GA₁(Zs)
PERM_2_IN_GA₁(Xs) -> IF_PERM_2_IN_1_GA₁(app_3_in_aag₁(Xs))

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

app_3_in_aag₁(._2₂(X, Zs)) -> if_app_3_in_1_aag₂(X, app_3_in_aag₁(Zs))
app_3_in_aag₁(X) -> app_3_out_aag₂([]_0, X)
if_app_3_in_1_gga₂(X, app_3_out_gga₁(Zs)) -> app_3_out_gga₁(._2₂(X, Zs))
app_3_in_gga₂(._2₂(X, Xs), Ys) -> if_app_3_in_1_gga₂(X, app_3_in_gga₂(Xs, Ys))
app_3_in_gga₂([]_0, X) -> app_3_out_gga₁(X)
if_app_3_in_1_aag₂(X, app_3_out_aag₂(Xs, Ys)) -> app_3_out_aag₂(._2₂(X, Xs), Ys)

Used ordering: POLO with Polynomial interpretation:

POL(._2₂(x₁, x₂)) = 1 + x₂
POL(if_app_3_in_1_gga₂(x₁, x₂)) = 1 + x₂
POL(IF_PERM_2_IN_2_GA₂(x₁, x₂)) = x₂
POL(IF_PERM_2_IN_1_GA₁(x₁)) = x₁
POL(app_3_out_gga₁(x₁)) = x₁
POL(app_3_in_aag₁(x₁)) = x₁
POL(app_3_out_aag₂(x₁, x₂)) = x₁ + x₂
POL(if_app_3_in_1_aag₂(x₁, x₂)) = 1 + x₂
POL([]_0) = 0
POL(app_3_in_gga₂(x₁, x₂)) = x₁ + x₂
POL(PERM_2_IN_GA₁(x₁)) = x₁

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ QDPPoloProof

                          ↳ QDP

                            ↳ DependencyGraphProof

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

IF_PERM_2_IN_2_GA₂(X, app_3_out_gga₁(Zs)) -> PERM_2_IN_GA₁(Zs)
PERM_2_IN_GA₁(Xs) -> IF_PERM_2_IN_1_GA₁(app_3_in_aag₁(Xs))

The TRS R consists of the following rules:

app_3_in_aag₁(X) -> app_3_out_aag₂([]_0, X)
app_3_in_aag₁(._2₂(X, Zs)) -> if_app_3_in_1_aag₂(X, app_3_in_aag₁(Zs))
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_aag₂(X, app_3_out_aag₂(Xs, Ys)) -> app_3_out_aag₂(._2₂(X, Xs), Ys)
if_app_3_in_1_gga₂(X, app_3_out_gga₁(Zs)) -> app_3_out_gga₁(._2₂(X, Zs))

The set Q consists of the following terms:

app_3_in_aag₁(x0)
app_3_in_gga₂(x0, x1)
if_app_3_in_1_aag₂(x0, x1)
if_app_3_in_1_gga₂(x0, x1)

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