Left Termination proof of /tmp/tmpfG1JrH/pl7.6.2c.pl

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

↳ PROLOG

  ↳ PrologToPiTRSProof

reach₄(X, Y, E, L) :- member₂(.₂(X, .₂(Y, {}₀)), E).
reach₄(X, Z, E, L) :- member1₂(.₂(X, .₂(Y, {}₀)), E), member₂(Y, L), delete₃(Y, L, V1), reach₄(Y, Z, E, V1).
member₂(H, .₂(H, L)).
member₂(X, .₂(H, L)) :- member₂(X, L).
member1₂(H, .₂(H, L)).
member1₂(X, .₂(H, L)) :- member1₂(X, L).
delete₃(X, .₂(X, Y), Y).
delete₃(X, .₂(H, T1), .₂(H, T2)) :- delete₃(X, T1, T2).

With regard to the inferred argument filtering the predicates were used in the following modes:
reach₄: (b,b,b,b)
member₂: (b,b)
member1₂: (f,b)
delete₃: (b,b,f)
Transforming PROLOG into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

reach_4_in_gggg₄(X, Y, E, L) -> if_reach_4_in_1_gggg₅(X, Y, E, L, member_2_in_gg₂(._2₂(X, ._2₂(Y, []_0)), E))
member_2_in_gg₂(H, ._2₂(H, L)) -> member_2_out_gg₂(H, ._2₂(H, L))
member_2_in_gg₂(X, ._2₂(H, L)) -> if_member_2_in_1_gg₄(X, H, L, member_2_in_gg₂(X, L))
if_member_2_in_1_gg₄(X, H, L, member_2_out_gg₂(X, L)) -> member_2_out_gg₂(X, ._2₂(H, L))
if_reach_4_in_1_gggg₅(X, Y, E, L, member_2_out_gg₂(._2₂(X, ._2₂(Y, []_0)), E)) -> reach_4_out_gggg₄(X, Y, E, L)
reach_4_in_gggg₄(X, Z, E, L) -> if_reach_4_in_2_gggg₅(X, Z, E, L, member1_2_in_ag₂(._2₂(X, ._2₂(Y, []_0)), E))
member1_2_in_ag₂(H, ._2₂(H, L)) -> member1_2_out_ag₂(H, ._2₂(H, L))
member1_2_in_ag₂(X, ._2₂(H, L)) -> if_member1_2_in_1_ag₄(X, H, L, member1_2_in_ag₂(X, L))
if_member1_2_in_1_ag₄(X, H, L, member1_2_out_ag₂(X, L)) -> member1_2_out_ag₂(X, ._2₂(H, L))
if_reach_4_in_2_gggg₅(X, Z, E, L, member1_2_out_ag₂(._2₂(X, ._2₂(Y, []_0)), E)) -> if_reach_4_in_3_gggg₆(X, Z, E, L, Y, member_2_in_gg₂(Y, L))
if_reach_4_in_3_gggg₆(X, Z, E, L, Y, member_2_out_gg₂(Y, L)) -> if_reach_4_in_4_gggg₆(X, Z, E, L, Y, delete_3_in_gga₃(Y, L, V1))
delete_3_in_gga₃(X, ._2₂(X, Y), Y) -> delete_3_out_gga₃(X, ._2₂(X, Y), Y)
delete_3_in_gga₃(X, ._2₂(H, T1), ._2₂(H, T2)) -> if_delete_3_in_1_gga₅(X, H, T1, T2, delete_3_in_gga₃(X, T1, T2))
if_delete_3_in_1_gga₅(X, H, T1, T2, delete_3_out_gga₃(X, T1, T2)) -> delete_3_out_gga₃(X, ._2₂(H, T1), ._2₂(H, T2))
if_reach_4_in_4_gggg₆(X, Z, E, L, Y, delete_3_out_gga₃(Y, L, V1)) -> if_reach_4_in_5_gggg₇(X, Z, E, L, Y, V1, reach_4_in_gggg₄(Y, Z, E, V1))
if_reach_4_in_5_gggg₇(X, Z, E, L, Y, V1, reach_4_out_gggg₄(Y, Z, E, V1)) -> reach_4_out_gggg₄(X, Z, E, L)

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:

reach_4_in_gggg₄(X, Y, E, L) -> if_reach_4_in_1_gggg₅(X, Y, E, L, member_2_in_gg₂(._2₂(X, ._2₂(Y, []_0)), E))
member_2_in_gg₂(H, ._2₂(H, L)) -> member_2_out_gg₂(H, ._2₂(H, L))
member_2_in_gg₂(X, ._2₂(H, L)) -> if_member_2_in_1_gg₄(X, H, L, member_2_in_gg₂(X, L))
if_member_2_in_1_gg₄(X, H, L, member_2_out_gg₂(X, L)) -> member_2_out_gg₂(X, ._2₂(H, L))
if_reach_4_in_1_gggg₅(X, Y, E, L, member_2_out_gg₂(._2₂(X, ._2₂(Y, []_0)), E)) -> reach_4_out_gggg₄(X, Y, E, L)
reach_4_in_gggg₄(X, Z, E, L) -> if_reach_4_in_2_gggg₅(X, Z, E, L, member1_2_in_ag₂(._2₂(X, ._2₂(Y, []_0)), E))
member1_2_in_ag₂(H, ._2₂(H, L)) -> member1_2_out_ag₂(H, ._2₂(H, L))
member1_2_in_ag₂(X, ._2₂(H, L)) -> if_member1_2_in_1_ag₄(X, H, L, member1_2_in_ag₂(X, L))
if_member1_2_in_1_ag₄(X, H, L, member1_2_out_ag₂(X, L)) -> member1_2_out_ag₂(X, ._2₂(H, L))
if_reach_4_in_2_gggg₅(X, Z, E, L, member1_2_out_ag₂(._2₂(X, ._2₂(Y, []_0)), E)) -> if_reach_4_in_3_gggg₆(X, Z, E, L, Y, member_2_in_gg₂(Y, L))
if_reach_4_in_3_gggg₆(X, Z, E, L, Y, member_2_out_gg₂(Y, L)) -> if_reach_4_in_4_gggg₆(X, Z, E, L, Y, delete_3_in_gga₃(Y, L, V1))
delete_3_in_gga₃(X, ._2₂(X, Y), Y) -> delete_3_out_gga₃(X, ._2₂(X, Y), Y)
delete_3_in_gga₃(X, ._2₂(H, T1), ._2₂(H, T2)) -> if_delete_3_in_1_gga₅(X, H, T1, T2, delete_3_in_gga₃(X, T1, T2))
if_delete_3_in_1_gga₅(X, H, T1, T2, delete_3_out_gga₃(X, T1, T2)) -> delete_3_out_gga₃(X, ._2₂(H, T1), ._2₂(H, T2))
if_reach_4_in_4_gggg₆(X, Z, E, L, Y, delete_3_out_gga₃(Y, L, V1)) -> if_reach_4_in_5_gggg₇(X, Z, E, L, Y, V1, reach_4_in_gggg₄(Y, Z, E, V1))
if_reach_4_in_5_gggg₇(X, Z, E, L, Y, V1, reach_4_out_gggg₄(Y, Z, E, V1)) -> reach_4_out_gggg₄(X, Z, E, L)

REACH_4_IN_GGGG₄(X, Y, E, L) -> IF_REACH_4_IN_1_GGGG₅(X, Y, E, L, member_2_in_gg₂(._2₂(X, ._2₂(Y, []_0)), E))
REACH_4_IN_GGGG₄(X, Y, E, L) -> MEMBER_2_IN_GG₂(._2₂(X, ._2₂(Y, []_0)), E)
MEMBER_2_IN_GG₂(X, ._2₂(H, L)) -> IF_MEMBER_2_IN_1_GG₄(X, H, L, member_2_in_gg₂(X, L))
MEMBER_2_IN_GG₂(X, ._2₂(H, L)) -> MEMBER_2_IN_GG₂(X, L)
REACH_4_IN_GGGG₄(X, Z, E, L) -> IF_REACH_4_IN_2_GGGG₅(X, Z, E, L, member1_2_in_ag₂(._2₂(X, ._2₂(Y, []_0)), E))
REACH_4_IN_GGGG₄(X, Z, E, L) -> MEMBER1_2_IN_AG₂(._2₂(X, ._2₂(Y, []_0)), E)
MEMBER1_2_IN_AG₂(X, ._2₂(H, L)) -> IF_MEMBER1_2_IN_1_AG₄(X, H, L, member1_2_in_ag₂(X, L))
MEMBER1_2_IN_AG₂(X, ._2₂(H, L)) -> MEMBER1_2_IN_AG₂(X, L)
IF_REACH_4_IN_2_GGGG₅(X, Z, E, L, member1_2_out_ag₂(._2₂(X, ._2₂(Y, []_0)), E)) -> IF_REACH_4_IN_3_GGGG₆(X, Z, E, L, Y, member_2_in_gg₂(Y, L))
IF_REACH_4_IN_2_GGGG₅(X, Z, E, L, member1_2_out_ag₂(._2₂(X, ._2₂(Y, []_0)), E)) -> MEMBER_2_IN_GG₂(Y, L)
IF_REACH_4_IN_3_GGGG₆(X, Z, E, L, Y, member_2_out_gg₂(Y, L)) -> IF_REACH_4_IN_4_GGGG₆(X, Z, E, L, Y, delete_3_in_gga₃(Y, L, V1))
IF_REACH_4_IN_3_GGGG₆(X, Z, E, L, Y, member_2_out_gg₂(Y, L)) -> DELETE_3_IN_GGA₃(Y, L, V1)
DELETE_3_IN_GGA₃(X, ._2₂(H, T1), ._2₂(H, T2)) -> IF_DELETE_3_IN_1_GGA₅(X, H, T1, T2, delete_3_in_gga₃(X, T1, T2))
DELETE_3_IN_GGA₃(X, ._2₂(H, T1), ._2₂(H, T2)) -> DELETE_3_IN_GGA₃(X, T1, T2)
IF_REACH_4_IN_4_GGGG₆(X, Z, E, L, Y, delete_3_out_gga₃(Y, L, V1)) -> IF_REACH_4_IN_5_GGGG₇(X, Z, E, L, Y, V1, reach_4_in_gggg₄(Y, Z, E, V1))
IF_REACH_4_IN_4_GGGG₆(X, Z, E, L, Y, delete_3_out_gga₃(Y, L, V1)) -> REACH_4_IN_GGGG₄(Y, Z, E, V1)

The TRS R consists of the following rules:

reach_4_in_gggg₄(X, Y, E, L) -> if_reach_4_in_1_gggg₅(X, Y, E, L, member_2_in_gg₂(._2₂(X, ._2₂(Y, []_0)), E))
member_2_in_gg₂(H, ._2₂(H, L)) -> member_2_out_gg₂(H, ._2₂(H, L))
member_2_in_gg₂(X, ._2₂(H, L)) -> if_member_2_in_1_gg₄(X, H, L, member_2_in_gg₂(X, L))
if_member_2_in_1_gg₄(X, H, L, member_2_out_gg₂(X, L)) -> member_2_out_gg₂(X, ._2₂(H, L))
if_reach_4_in_1_gggg₅(X, Y, E, L, member_2_out_gg₂(._2₂(X, ._2₂(Y, []_0)), E)) -> reach_4_out_gggg₄(X, Y, E, L)
reach_4_in_gggg₄(X, Z, E, L) -> if_reach_4_in_2_gggg₅(X, Z, E, L, member1_2_in_ag₂(._2₂(X, ._2₂(Y, []_0)), E))
member1_2_in_ag₂(H, ._2₂(H, L)) -> member1_2_out_ag₂(H, ._2₂(H, L))
member1_2_in_ag₂(X, ._2₂(H, L)) -> if_member1_2_in_1_ag₄(X, H, L, member1_2_in_ag₂(X, L))
if_member1_2_in_1_ag₄(X, H, L, member1_2_out_ag₂(X, L)) -> member1_2_out_ag₂(X, ._2₂(H, L))
if_reach_4_in_2_gggg₅(X, Z, E, L, member1_2_out_ag₂(._2₂(X, ._2₂(Y, []_0)), E)) -> if_reach_4_in_3_gggg₆(X, Z, E, L, Y, member_2_in_gg₂(Y, L))
if_reach_4_in_3_gggg₆(X, Z, E, L, Y, member_2_out_gg₂(Y, L)) -> if_reach_4_in_4_gggg₆(X, Z, E, L, Y, delete_3_in_gga₃(Y, L, V1))
delete_3_in_gga₃(X, ._2₂(X, Y), Y) -> delete_3_out_gga₃(X, ._2₂(X, Y), Y)
delete_3_in_gga₃(X, ._2₂(H, T1), ._2₂(H, T2)) -> if_delete_3_in_1_gga₅(X, H, T1, T2, delete_3_in_gga₃(X, T1, T2))
if_delete_3_in_1_gga₅(X, H, T1, T2, delete_3_out_gga₃(X, T1, T2)) -> delete_3_out_gga₃(X, ._2₂(H, T1), ._2₂(H, T2))
if_reach_4_in_4_gggg₆(X, Z, E, L, Y, delete_3_out_gga₃(Y, L, V1)) -> if_reach_4_in_5_gggg₇(X, Z, E, L, Y, V1, reach_4_in_gggg₄(Y, Z, E, V1))
if_reach_4_in_5_gggg₇(X, Z, E, L, Y, V1, reach_4_out_gggg₄(Y, Z, E, V1)) -> reach_4_out_gggg₄(X, Z, E, L)

The argument filtering Pi contains the following mapping:
reach_4_in_gggg₄(x1, x2, x3, x4) = reach_4_in_gggg₄(x1, x2, x3, x4)
._2₂(x1, x2) = ._2₂(x1, x2)
[]_0 = []_0
if_reach_4_in_1_gggg₅(x1, x2, x3, x4, x5) = if_reach_4_in_1_gggg₁(x5)
member_2_in_gg₂(x1, x2) = member_2_in_gg₂(x1, x2)
member_2_out_gg₂(x1, x2) = member_2_out_gg
if_member_2_in_1_gg₄(x1, x2, x3, x4) = if_member_2_in_1_gg₁(x4)
reach_4_out_gggg₄(x1, x2, x3, x4) = reach_4_out_gggg
if_reach_4_in_2_gggg₅(x1, x2, x3, x4, x5) = if_reach_4_in_2_gggg₄(x2, x3, x4, x5)
member1_2_in_ag₂(x1, x2) = member1_2_in_ag₁(x2)
member1_2_out_ag₂(x1, x2) = member1_2_out_ag₁(x1)
if_member1_2_in_1_ag₄(x1, x2, x3, x4) = if_member1_2_in_1_ag₁(x4)
if_reach_4_in_3_gggg₆(x1, x2, x3, x4, x5, x6) = if_reach_4_in_3_gggg₅(x2, x3, x4, x5, x6)
if_reach_4_in_4_gggg₆(x1, x2, x3, x4, x5, x6) = if_reach_4_in_4_gggg₄(x2, x3, x5, x6)
delete_3_in_gga₃(x1, x2, x3) = delete_3_in_gga₂(x1, x2)
delete_3_out_gga₃(x1, x2, x3) = delete_3_out_gga₁(x3)
if_delete_3_in_1_gga₅(x1, x2, x3, x4, x5) = if_delete_3_in_1_gga₂(x2, x5)
if_reach_4_in_5_gggg₇(x1, x2, x3, x4, x5, x6, x7) = if_reach_4_in_5_gggg₁(x7)
DELETE_3_IN_GGA₃(x1, x2, x3) = DELETE_3_IN_GGA₂(x1, x2)
IF_MEMBER_2_IN_1_GG₄(x1, x2, x3, x4) = IF_MEMBER_2_IN_1_GG₁(x4)
IF_REACH_4_IN_4_GGGG₆(x1, x2, x3, x4, x5, x6) = IF_REACH_4_IN_4_GGGG₄(x2, x3, x5, x6)
REACH_4_IN_GGGG₄(x1, x2, x3, x4) = REACH_4_IN_GGGG₄(x1, x2, x3, x4)
IF_DELETE_3_IN_1_GGA₅(x1, x2, x3, x4, x5) = IF_DELETE_3_IN_1_GGA₂(x2, x5)
IF_REACH_4_IN_1_GGGG₅(x1, x2, x3, x4, x5) = IF_REACH_4_IN_1_GGGG₁(x5)
MEMBER_2_IN_GG₂(x1, x2) = MEMBER_2_IN_GG₂(x1, x2)
MEMBER1_2_IN_AG₂(x1, x2) = MEMBER1_2_IN_AG₁(x2)
IF_REACH_4_IN_5_GGGG₇(x1, x2, x3, x4, x5, x6, x7) = IF_REACH_4_IN_5_GGGG₁(x7)
IF_MEMBER1_2_IN_1_AG₄(x1, x2, x3, x4) = IF_MEMBER1_2_IN_1_AG₁(x4)
IF_REACH_4_IN_2_GGGG₅(x1, x2, x3, x4, x5) = IF_REACH_4_IN_2_GGGG₄(x2, x3, x4, x5)
IF_REACH_4_IN_3_GGGG₆(x1, x2, x3, x4, x5, x6) = IF_REACH_4_IN_3_GGGG₅(x2, x3, x4, x5, 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:

REACH_4_IN_GGGG₄(X, Y, E, L) -> IF_REACH_4_IN_1_GGGG₅(X, Y, E, L, member_2_in_gg₂(._2₂(X, ._2₂(Y, []_0)), E))
REACH_4_IN_GGGG₄(X, Y, E, L) -> MEMBER_2_IN_GG₂(._2₂(X, ._2₂(Y, []_0)), E)
MEMBER_2_IN_GG₂(X, ._2₂(H, L)) -> IF_MEMBER_2_IN_1_GG₄(X, H, L, member_2_in_gg₂(X, L))
MEMBER_2_IN_GG₂(X, ._2₂(H, L)) -> MEMBER_2_IN_GG₂(X, L)
REACH_4_IN_GGGG₄(X, Z, E, L) -> IF_REACH_4_IN_2_GGGG₅(X, Z, E, L, member1_2_in_ag₂(._2₂(X, ._2₂(Y, []_0)), E))
REACH_4_IN_GGGG₄(X, Z, E, L) -> MEMBER1_2_IN_AG₂(._2₂(X, ._2₂(Y, []_0)), E)
MEMBER1_2_IN_AG₂(X, ._2₂(H, L)) -> IF_MEMBER1_2_IN_1_AG₄(X, H, L, member1_2_in_ag₂(X, L))
MEMBER1_2_IN_AG₂(X, ._2₂(H, L)) -> MEMBER1_2_IN_AG₂(X, L)
IF_REACH_4_IN_2_GGGG₅(X, Z, E, L, member1_2_out_ag₂(._2₂(X, ._2₂(Y, []_0)), E)) -> IF_REACH_4_IN_3_GGGG₆(X, Z, E, L, Y, member_2_in_gg₂(Y, L))
IF_REACH_4_IN_2_GGGG₅(X, Z, E, L, member1_2_out_ag₂(._2₂(X, ._2₂(Y, []_0)), E)) -> MEMBER_2_IN_GG₂(Y, L)
IF_REACH_4_IN_3_GGGG₆(X, Z, E, L, Y, member_2_out_gg₂(Y, L)) -> IF_REACH_4_IN_4_GGGG₆(X, Z, E, L, Y, delete_3_in_gga₃(Y, L, V1))
IF_REACH_4_IN_3_GGGG₆(X, Z, E, L, Y, member_2_out_gg₂(Y, L)) -> DELETE_3_IN_GGA₃(Y, L, V1)
DELETE_3_IN_GGA₃(X, ._2₂(H, T1), ._2₂(H, T2)) -> IF_DELETE_3_IN_1_GGA₅(X, H, T1, T2, delete_3_in_gga₃(X, T1, T2))
DELETE_3_IN_GGA₃(X, ._2₂(H, T1), ._2₂(H, T2)) -> DELETE_3_IN_GGA₃(X, T1, T2)
IF_REACH_4_IN_4_GGGG₆(X, Z, E, L, Y, delete_3_out_gga₃(Y, L, V1)) -> IF_REACH_4_IN_5_GGGG₇(X, Z, E, L, Y, V1, reach_4_in_gggg₄(Y, Z, E, V1))
IF_REACH_4_IN_4_GGGG₆(X, Z, E, L, Y, delete_3_out_gga₃(Y, L, V1)) -> REACH_4_IN_GGGG₄(Y, Z, E, V1)

The TRS R consists of the following rules:

reach_4_in_gggg₄(X, Y, E, L) -> if_reach_4_in_1_gggg₅(X, Y, E, L, member_2_in_gg₂(._2₂(X, ._2₂(Y, []_0)), E))
member_2_in_gg₂(H, ._2₂(H, L)) -> member_2_out_gg₂(H, ._2₂(H, L))
member_2_in_gg₂(X, ._2₂(H, L)) -> if_member_2_in_1_gg₄(X, H, L, member_2_in_gg₂(X, L))
if_member_2_in_1_gg₄(X, H, L, member_2_out_gg₂(X, L)) -> member_2_out_gg₂(X, ._2₂(H, L))
if_reach_4_in_1_gggg₅(X, Y, E, L, member_2_out_gg₂(._2₂(X, ._2₂(Y, []_0)), E)) -> reach_4_out_gggg₄(X, Y, E, L)
reach_4_in_gggg₄(X, Z, E, L) -> if_reach_4_in_2_gggg₅(X, Z, E, L, member1_2_in_ag₂(._2₂(X, ._2₂(Y, []_0)), E))
member1_2_in_ag₂(H, ._2₂(H, L)) -> member1_2_out_ag₂(H, ._2₂(H, L))
member1_2_in_ag₂(X, ._2₂(H, L)) -> if_member1_2_in_1_ag₄(X, H, L, member1_2_in_ag₂(X, L))
if_member1_2_in_1_ag₄(X, H, L, member1_2_out_ag₂(X, L)) -> member1_2_out_ag₂(X, ._2₂(H, L))
if_reach_4_in_2_gggg₅(X, Z, E, L, member1_2_out_ag₂(._2₂(X, ._2₂(Y, []_0)), E)) -> if_reach_4_in_3_gggg₆(X, Z, E, L, Y, member_2_in_gg₂(Y, L))
if_reach_4_in_3_gggg₆(X, Z, E, L, Y, member_2_out_gg₂(Y, L)) -> if_reach_4_in_4_gggg₆(X, Z, E, L, Y, delete_3_in_gga₃(Y, L, V1))
delete_3_in_gga₃(X, ._2₂(X, Y), Y) -> delete_3_out_gga₃(X, ._2₂(X, Y), Y)
delete_3_in_gga₃(X, ._2₂(H, T1), ._2₂(H, T2)) -> if_delete_3_in_1_gga₅(X, H, T1, T2, delete_3_in_gga₃(X, T1, T2))
if_delete_3_in_1_gga₅(X, H, T1, T2, delete_3_out_gga₃(X, T1, T2)) -> delete_3_out_gga₃(X, ._2₂(H, T1), ._2₂(H, T2))
if_reach_4_in_4_gggg₆(X, Z, E, L, Y, delete_3_out_gga₃(Y, L, V1)) -> if_reach_4_in_5_gggg₇(X, Z, E, L, Y, V1, reach_4_in_gggg₄(Y, Z, E, V1))
if_reach_4_in_5_gggg₇(X, Z, E, L, Y, V1, reach_4_out_gggg₄(Y, Z, E, V1)) -> reach_4_out_gggg₄(X, Z, E, L)

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

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

DELETE_3_IN_GGA₃(X, ._2₂(H, T1), ._2₂(H, T2)) -> DELETE_3_IN_GGA₃(X, T1, T2)

The TRS R consists of the following rules:

reach_4_in_gggg₄(X, Y, E, L) -> if_reach_4_in_1_gggg₅(X, Y, E, L, member_2_in_gg₂(._2₂(X, ._2₂(Y, []_0)), E))
member_2_in_gg₂(H, ._2₂(H, L)) -> member_2_out_gg₂(H, ._2₂(H, L))
member_2_in_gg₂(X, ._2₂(H, L)) -> if_member_2_in_1_gg₄(X, H, L, member_2_in_gg₂(X, L))
if_member_2_in_1_gg₄(X, H, L, member_2_out_gg₂(X, L)) -> member_2_out_gg₂(X, ._2₂(H, L))
if_reach_4_in_1_gggg₅(X, Y, E, L, member_2_out_gg₂(._2₂(X, ._2₂(Y, []_0)), E)) -> reach_4_out_gggg₄(X, Y, E, L)
reach_4_in_gggg₄(X, Z, E, L) -> if_reach_4_in_2_gggg₅(X, Z, E, L, member1_2_in_ag₂(._2₂(X, ._2₂(Y, []_0)), E))
member1_2_in_ag₂(H, ._2₂(H, L)) -> member1_2_out_ag₂(H, ._2₂(H, L))
member1_2_in_ag₂(X, ._2₂(H, L)) -> if_member1_2_in_1_ag₄(X, H, L, member1_2_in_ag₂(X, L))
if_member1_2_in_1_ag₄(X, H, L, member1_2_out_ag₂(X, L)) -> member1_2_out_ag₂(X, ._2₂(H, L))
if_reach_4_in_2_gggg₅(X, Z, E, L, member1_2_out_ag₂(._2₂(X, ._2₂(Y, []_0)), E)) -> if_reach_4_in_3_gggg₆(X, Z, E, L, Y, member_2_in_gg₂(Y, L))
if_reach_4_in_3_gggg₆(X, Z, E, L, Y, member_2_out_gg₂(Y, L)) -> if_reach_4_in_4_gggg₆(X, Z, E, L, Y, delete_3_in_gga₃(Y, L, V1))
delete_3_in_gga₃(X, ._2₂(X, Y), Y) -> delete_3_out_gga₃(X, ._2₂(X, Y), Y)
delete_3_in_gga₃(X, ._2₂(H, T1), ._2₂(H, T2)) -> if_delete_3_in_1_gga₅(X, H, T1, T2, delete_3_in_gga₃(X, T1, T2))
if_delete_3_in_1_gga₅(X, H, T1, T2, delete_3_out_gga₃(X, T1, T2)) -> delete_3_out_gga₃(X, ._2₂(H, T1), ._2₂(H, T2))
if_reach_4_in_4_gggg₆(X, Z, E, L, Y, delete_3_out_gga₃(Y, L, V1)) -> if_reach_4_in_5_gggg₇(X, Z, E, L, Y, V1, reach_4_in_gggg₄(Y, Z, E, V1))
if_reach_4_in_5_gggg₇(X, Z, E, L, Y, V1, reach_4_out_gggg₄(Y, Z, E, V1)) -> reach_4_out_gggg₄(X, Z, E, L)

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

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

DELETE_3_IN_GGA₃(X, ._2₂(H, T1), ._2₂(H, T2)) -> DELETE_3_IN_GGA₃(X, T1, T2)

R is empty.
The argument filtering Pi contains the following mapping:
._2₂(x1, x2) = ._2₂(x1, x2)
DELETE_3_IN_GGA₃(x1, x2, x3) = DELETE_3_IN_GGA₂(x1, x2)

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

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

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

DELETE_3_IN_GGA₂(X, ._2₂(H, T1)) -> DELETE_3_IN_GGA₂(X, T1)

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

DELETE_3_IN_GGA₂(X, ._2₂(H, T1)) -> DELETE_3_IN_GGA₂(X, T1)
The graph contains the following edges 1 >= 1, 2 > 2

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

              ↳ PiDP

              ↳ PiDP

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

MEMBER1_2_IN_AG₂(X, ._2₂(H, L)) -> MEMBER1_2_IN_AG₂(X, L)

The TRS R consists of the following rules:

reach_4_in_gggg₄(X, Y, E, L) -> if_reach_4_in_1_gggg₅(X, Y, E, L, member_2_in_gg₂(._2₂(X, ._2₂(Y, []_0)), E))
member_2_in_gg₂(H, ._2₂(H, L)) -> member_2_out_gg₂(H, ._2₂(H, L))
member_2_in_gg₂(X, ._2₂(H, L)) -> if_member_2_in_1_gg₄(X, H, L, member_2_in_gg₂(X, L))
if_member_2_in_1_gg₄(X, H, L, member_2_out_gg₂(X, L)) -> member_2_out_gg₂(X, ._2₂(H, L))
if_reach_4_in_1_gggg₅(X, Y, E, L, member_2_out_gg₂(._2₂(X, ._2₂(Y, []_0)), E)) -> reach_4_out_gggg₄(X, Y, E, L)
reach_4_in_gggg₄(X, Z, E, L) -> if_reach_4_in_2_gggg₅(X, Z, E, L, member1_2_in_ag₂(._2₂(X, ._2₂(Y, []_0)), E))
member1_2_in_ag₂(H, ._2₂(H, L)) -> member1_2_out_ag₂(H, ._2₂(H, L))
member1_2_in_ag₂(X, ._2₂(H, L)) -> if_member1_2_in_1_ag₄(X, H, L, member1_2_in_ag₂(X, L))
if_member1_2_in_1_ag₄(X, H, L, member1_2_out_ag₂(X, L)) -> member1_2_out_ag₂(X, ._2₂(H, L))
if_reach_4_in_2_gggg₅(X, Z, E, L, member1_2_out_ag₂(._2₂(X, ._2₂(Y, []_0)), E)) -> if_reach_4_in_3_gggg₆(X, Z, E, L, Y, member_2_in_gg₂(Y, L))
if_reach_4_in_3_gggg₆(X, Z, E, L, Y, member_2_out_gg₂(Y, L)) -> if_reach_4_in_4_gggg₆(X, Z, E, L, Y, delete_3_in_gga₃(Y, L, V1))
delete_3_in_gga₃(X, ._2₂(X, Y), Y) -> delete_3_out_gga₃(X, ._2₂(X, Y), Y)
delete_3_in_gga₃(X, ._2₂(H, T1), ._2₂(H, T2)) -> if_delete_3_in_1_gga₅(X, H, T1, T2, delete_3_in_gga₃(X, T1, T2))
if_delete_3_in_1_gga₅(X, H, T1, T2, delete_3_out_gga₃(X, T1, T2)) -> delete_3_out_gga₃(X, ._2₂(H, T1), ._2₂(H, T2))
if_reach_4_in_4_gggg₆(X, Z, E, L, Y, delete_3_out_gga₃(Y, L, V1)) -> if_reach_4_in_5_gggg₇(X, Z, E, L, Y, V1, reach_4_in_gggg₄(Y, Z, E, V1))
if_reach_4_in_5_gggg₇(X, Z, E, L, Y, V1, reach_4_out_gggg₄(Y, Z, E, V1)) -> reach_4_out_gggg₄(X, Z, E, L)

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

              ↳ PiDP

              ↳ PiDP

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

MEMBER1_2_IN_AG₂(X, ._2₂(H, L)) -> MEMBER1_2_IN_AG₂(X, L)

R is empty.
The argument filtering Pi contains the following mapping:
._2₂(x1, x2) = ._2₂(x1, x2)
MEMBER1_2_IN_AG₂(x1, x2) = MEMBER1_2_IN_AG₁(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

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ PiDP

              ↳ PiDP

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

MEMBER1_2_IN_AG₁(._2₂(H, L)) -> MEMBER1_2_IN_AG₁(L)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {MEMBER1_2_IN_AG₁}.
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:

MEMBER1_2_IN_AG₁(._2₂(H, L)) -> MEMBER1_2_IN_AG₁(L)
The graph contains the following edges 1 > 1

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

              ↳ PiDP

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

MEMBER_2_IN_GG₂(X, ._2₂(H, L)) -> MEMBER_2_IN_GG₂(X, L)

The TRS R consists of the following rules:

reach_4_in_gggg₄(X, Y, E, L) -> if_reach_4_in_1_gggg₅(X, Y, E, L, member_2_in_gg₂(._2₂(X, ._2₂(Y, []_0)), E))
member_2_in_gg₂(H, ._2₂(H, L)) -> member_2_out_gg₂(H, ._2₂(H, L))
member_2_in_gg₂(X, ._2₂(H, L)) -> if_member_2_in_1_gg₄(X, H, L, member_2_in_gg₂(X, L))
if_member_2_in_1_gg₄(X, H, L, member_2_out_gg₂(X, L)) -> member_2_out_gg₂(X, ._2₂(H, L))
if_reach_4_in_1_gggg₅(X, Y, E, L, member_2_out_gg₂(._2₂(X, ._2₂(Y, []_0)), E)) -> reach_4_out_gggg₄(X, Y, E, L)
reach_4_in_gggg₄(X, Z, E, L) -> if_reach_4_in_2_gggg₅(X, Z, E, L, member1_2_in_ag₂(._2₂(X, ._2₂(Y, []_0)), E))
member1_2_in_ag₂(H, ._2₂(H, L)) -> member1_2_out_ag₂(H, ._2₂(H, L))
member1_2_in_ag₂(X, ._2₂(H, L)) -> if_member1_2_in_1_ag₄(X, H, L, member1_2_in_ag₂(X, L))
if_member1_2_in_1_ag₄(X, H, L, member1_2_out_ag₂(X, L)) -> member1_2_out_ag₂(X, ._2₂(H, L))
if_reach_4_in_2_gggg₅(X, Z, E, L, member1_2_out_ag₂(._2₂(X, ._2₂(Y, []_0)), E)) -> if_reach_4_in_3_gggg₆(X, Z, E, L, Y, member_2_in_gg₂(Y, L))
if_reach_4_in_3_gggg₆(X, Z, E, L, Y, member_2_out_gg₂(Y, L)) -> if_reach_4_in_4_gggg₆(X, Z, E, L, Y, delete_3_in_gga₃(Y, L, V1))
delete_3_in_gga₃(X, ._2₂(X, Y), Y) -> delete_3_out_gga₃(X, ._2₂(X, Y), Y)
delete_3_in_gga₃(X, ._2₂(H, T1), ._2₂(H, T2)) -> if_delete_3_in_1_gga₅(X, H, T1, T2, delete_3_in_gga₃(X, T1, T2))
if_delete_3_in_1_gga₅(X, H, T1, T2, delete_3_out_gga₃(X, T1, T2)) -> delete_3_out_gga₃(X, ._2₂(H, T1), ._2₂(H, T2))
if_reach_4_in_4_gggg₆(X, Z, E, L, Y, delete_3_out_gga₃(Y, L, V1)) -> if_reach_4_in_5_gggg₇(X, Z, E, L, Y, V1, reach_4_in_gggg₄(Y, Z, E, V1))
if_reach_4_in_5_gggg₇(X, Z, E, L, Y, V1, reach_4_out_gggg₄(Y, Z, E, V1)) -> reach_4_out_gggg₄(X, Z, E, L)

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

              ↳ PiDP

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

MEMBER_2_IN_GG₂(X, ._2₂(H, L)) -> MEMBER_2_IN_GG₂(X, L)

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

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

MEMBER_2_IN_GG₂(X, ._2₂(H, L)) -> MEMBER_2_IN_GG₂(X, L)

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

MEMBER_2_IN_GG₂(X, ._2₂(H, L)) -> MEMBER_2_IN_GG₂(X, L)
The graph contains the following edges 1 >= 1, 2 > 2

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

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

IF_REACH_4_IN_4_GGGG₆(X, Z, E, L, Y, delete_3_out_gga₃(Y, L, V1)) -> REACH_4_IN_GGGG₄(Y, Z, E, V1)
REACH_4_IN_GGGG₄(X, Z, E, L) -> IF_REACH_4_IN_2_GGGG₅(X, Z, E, L, member1_2_in_ag₂(._2₂(X, ._2₂(Y, []_0)), E))
IF_REACH_4_IN_3_GGGG₆(X, Z, E, L, Y, member_2_out_gg₂(Y, L)) -> IF_REACH_4_IN_4_GGGG₆(X, Z, E, L, Y, delete_3_in_gga₃(Y, L, V1))
IF_REACH_4_IN_2_GGGG₅(X, Z, E, L, member1_2_out_ag₂(._2₂(X, ._2₂(Y, []_0)), E)) -> IF_REACH_4_IN_3_GGGG₆(X, Z, E, L, Y, member_2_in_gg₂(Y, L))

The TRS R consists of the following rules:

reach_4_in_gggg₄(X, Y, E, L) -> if_reach_4_in_1_gggg₅(X, Y, E, L, member_2_in_gg₂(._2₂(X, ._2₂(Y, []_0)), E))
member_2_in_gg₂(H, ._2₂(H, L)) -> member_2_out_gg₂(H, ._2₂(H, L))
member_2_in_gg₂(X, ._2₂(H, L)) -> if_member_2_in_1_gg₄(X, H, L, member_2_in_gg₂(X, L))
if_member_2_in_1_gg₄(X, H, L, member_2_out_gg₂(X, L)) -> member_2_out_gg₂(X, ._2₂(H, L))
if_reach_4_in_1_gggg₅(X, Y, E, L, member_2_out_gg₂(._2₂(X, ._2₂(Y, []_0)), E)) -> reach_4_out_gggg₄(X, Y, E, L)
reach_4_in_gggg₄(X, Z, E, L) -> if_reach_4_in_2_gggg₅(X, Z, E, L, member1_2_in_ag₂(._2₂(X, ._2₂(Y, []_0)), E))
member1_2_in_ag₂(H, ._2₂(H, L)) -> member1_2_out_ag₂(H, ._2₂(H, L))
member1_2_in_ag₂(X, ._2₂(H, L)) -> if_member1_2_in_1_ag₄(X, H, L, member1_2_in_ag₂(X, L))
if_member1_2_in_1_ag₄(X, H, L, member1_2_out_ag₂(X, L)) -> member1_2_out_ag₂(X, ._2₂(H, L))
if_reach_4_in_2_gggg₅(X, Z, E, L, member1_2_out_ag₂(._2₂(X, ._2₂(Y, []_0)), E)) -> if_reach_4_in_3_gggg₆(X, Z, E, L, Y, member_2_in_gg₂(Y, L))
if_reach_4_in_3_gggg₆(X, Z, E, L, Y, member_2_out_gg₂(Y, L)) -> if_reach_4_in_4_gggg₆(X, Z, E, L, Y, delete_3_in_gga₃(Y, L, V1))
delete_3_in_gga₃(X, ._2₂(X, Y), Y) -> delete_3_out_gga₃(X, ._2₂(X, Y), Y)
delete_3_in_gga₃(X, ._2₂(H, T1), ._2₂(H, T2)) -> if_delete_3_in_1_gga₅(X, H, T1, T2, delete_3_in_gga₃(X, T1, T2))
if_delete_3_in_1_gga₅(X, H, T1, T2, delete_3_out_gga₃(X, T1, T2)) -> delete_3_out_gga₃(X, ._2₂(H, T1), ._2₂(H, T2))
if_reach_4_in_4_gggg₆(X, Z, E, L, Y, delete_3_out_gga₃(Y, L, V1)) -> if_reach_4_in_5_gggg₇(X, Z, E, L, Y, V1, reach_4_in_gggg₄(Y, Z, E, V1))
if_reach_4_in_5_gggg₇(X, Z, E, L, Y, V1, reach_4_out_gggg₄(Y, Z, E, V1)) -> reach_4_out_gggg₄(X, Z, E, L)

The argument filtering Pi contains the following mapping:
reach_4_in_gggg₄(x1, x2, x3, x4) = reach_4_in_gggg₄(x1, x2, x3, x4)
._2₂(x1, x2) = ._2₂(x1, x2)
[]_0 = []_0
if_reach_4_in_1_gggg₅(x1, x2, x3, x4, x5) = if_reach_4_in_1_gggg₁(x5)
member_2_in_gg₂(x1, x2) = member_2_in_gg₂(x1, x2)
member_2_out_gg₂(x1, x2) = member_2_out_gg
if_member_2_in_1_gg₄(x1, x2, x3, x4) = if_member_2_in_1_gg₁(x4)
reach_4_out_gggg₄(x1, x2, x3, x4) = reach_4_out_gggg
if_reach_4_in_2_gggg₅(x1, x2, x3, x4, x5) = if_reach_4_in_2_gggg₄(x2, x3, x4, x5)
member1_2_in_ag₂(x1, x2) = member1_2_in_ag₁(x2)
member1_2_out_ag₂(x1, x2) = member1_2_out_ag₁(x1)
if_member1_2_in_1_ag₄(x1, x2, x3, x4) = if_member1_2_in_1_ag₁(x4)
if_reach_4_in_3_gggg₆(x1, x2, x3, x4, x5, x6) = if_reach_4_in_3_gggg₅(x2, x3, x4, x5, x6)
if_reach_4_in_4_gggg₆(x1, x2, x3, x4, x5, x6) = if_reach_4_in_4_gggg₄(x2, x3, x5, x6)
delete_3_in_gga₃(x1, x2, x3) = delete_3_in_gga₂(x1, x2)
delete_3_out_gga₃(x1, x2, x3) = delete_3_out_gga₁(x3)
if_delete_3_in_1_gga₅(x1, x2, x3, x4, x5) = if_delete_3_in_1_gga₂(x2, x5)
if_reach_4_in_5_gggg₇(x1, x2, x3, x4, x5, x6, x7) = if_reach_4_in_5_gggg₁(x7)
IF_REACH_4_IN_4_GGGG₆(x1, x2, x3, x4, x5, x6) = IF_REACH_4_IN_4_GGGG₄(x2, x3, x5, x6)
REACH_4_IN_GGGG₄(x1, x2, x3, x4) = REACH_4_IN_GGGG₄(x1, x2, x3, x4)
IF_REACH_4_IN_2_GGGG₅(x1, x2, x3, x4, x5) = IF_REACH_4_IN_2_GGGG₄(x2, x3, x4, x5)
IF_REACH_4_IN_3_GGGG₆(x1, x2, x3, x4, x5, x6) = IF_REACH_4_IN_3_GGGG₅(x2, x3, x4, x5, x6)

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

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

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

IF_REACH_4_IN_4_GGGG₆(X, Z, E, L, Y, delete_3_out_gga₃(Y, L, V1)) -> REACH_4_IN_GGGG₄(Y, Z, E, V1)
REACH_4_IN_GGGG₄(X, Z, E, L) -> IF_REACH_4_IN_2_GGGG₅(X, Z, E, L, member1_2_in_ag₂(._2₂(X, ._2₂(Y, []_0)), E))
IF_REACH_4_IN_3_GGGG₆(X, Z, E, L, Y, member_2_out_gg₂(Y, L)) -> IF_REACH_4_IN_4_GGGG₆(X, Z, E, L, Y, delete_3_in_gga₃(Y, L, V1))
IF_REACH_4_IN_2_GGGG₅(X, Z, E, L, member1_2_out_ag₂(._2₂(X, ._2₂(Y, []_0)), E)) -> IF_REACH_4_IN_3_GGGG₆(X, Z, E, L, Y, member_2_in_gg₂(Y, L))

The TRS R consists of the following rules:

member1_2_in_ag₂(H, ._2₂(H, L)) -> member1_2_out_ag₂(H, ._2₂(H, L))
member1_2_in_ag₂(X, ._2₂(H, L)) -> if_member1_2_in_1_ag₄(X, H, L, member1_2_in_ag₂(X, L))
delete_3_in_gga₃(X, ._2₂(X, Y), Y) -> delete_3_out_gga₃(X, ._2₂(X, Y), Y)
delete_3_in_gga₃(X, ._2₂(H, T1), ._2₂(H, T2)) -> if_delete_3_in_1_gga₅(X, H, T1, T2, delete_3_in_gga₃(X, T1, T2))
member_2_in_gg₂(H, ._2₂(H, L)) -> member_2_out_gg₂(H, ._2₂(H, L))
member_2_in_gg₂(X, ._2₂(H, L)) -> if_member_2_in_1_gg₄(X, H, L, member_2_in_gg₂(X, L))
if_member1_2_in_1_ag₄(X, H, L, member1_2_out_ag₂(X, L)) -> member1_2_out_ag₂(X, ._2₂(H, L))
if_delete_3_in_1_gga₅(X, H, T1, T2, delete_3_out_gga₃(X, T1, T2)) -> delete_3_out_gga₃(X, ._2₂(H, T1), ._2₂(H, T2))
if_member_2_in_1_gg₄(X, H, L, member_2_out_gg₂(X, L)) -> member_2_out_gg₂(X, ._2₂(H, L))

The argument filtering Pi contains the following mapping:
._2₂(x1, x2) = ._2₂(x1, x2)
[]_0 = []_0
member_2_in_gg₂(x1, x2) = member_2_in_gg₂(x1, x2)
member_2_out_gg₂(x1, x2) = member_2_out_gg
if_member_2_in_1_gg₄(x1, x2, x3, x4) = if_member_2_in_1_gg₁(x4)
member1_2_in_ag₂(x1, x2) = member1_2_in_ag₁(x2)
member1_2_out_ag₂(x1, x2) = member1_2_out_ag₁(x1)
if_member1_2_in_1_ag₄(x1, x2, x3, x4) = if_member1_2_in_1_ag₁(x4)
delete_3_in_gga₃(x1, x2, x3) = delete_3_in_gga₂(x1, x2)
delete_3_out_gga₃(x1, x2, x3) = delete_3_out_gga₁(x3)
if_delete_3_in_1_gga₅(x1, x2, x3, x4, x5) = if_delete_3_in_1_gga₂(x2, x5)
IF_REACH_4_IN_4_GGGG₆(x1, x2, x3, x4, x5, x6) = IF_REACH_4_IN_4_GGGG₄(x2, x3, x5, x6)
REACH_4_IN_GGGG₄(x1, x2, x3, x4) = REACH_4_IN_GGGG₄(x1, x2, x3, x4)
IF_REACH_4_IN_2_GGGG₅(x1, x2, x3, x4, x5) = IF_REACH_4_IN_2_GGGG₄(x2, x3, x4, x5)
IF_REACH_4_IN_3_GGGG₆(x1, x2, x3, x4, x5, x6) = IF_REACH_4_IN_3_GGGG₅(x2, x3, x4, x5, 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

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

IF_REACH_4_IN_4_GGGG₄(Z, E, Y, delete_3_out_gga₁(V1)) -> REACH_4_IN_GGGG₄(Y, Z, E, V1)
REACH_4_IN_GGGG₄(X, Z, E, L) -> IF_REACH_4_IN_2_GGGG₄(Z, E, L, member1_2_in_ag₁(E))
IF_REACH_4_IN_3_GGGG₅(Z, E, L, Y, member_2_out_gg) -> IF_REACH_4_IN_4_GGGG₄(Z, E, Y, delete_3_in_gga₂(Y, L))
IF_REACH_4_IN_2_GGGG₄(Z, E, L, member1_2_out_ag₁(._2₂(X, ._2₂(Y, []_0)))) -> IF_REACH_4_IN_3_GGGG₅(Z, E, L, Y, member_2_in_gg₂(Y, L))

The TRS R consists of the following rules:

member1_2_in_ag₁(._2₂(H, L)) -> member1_2_out_ag₁(H)
member1_2_in_ag₁(._2₂(H, L)) -> if_member1_2_in_1_ag₁(member1_2_in_ag₁(L))
delete_3_in_gga₂(X, ._2₂(X, Y)) -> delete_3_out_gga₁(Y)
delete_3_in_gga₂(X, ._2₂(H, T1)) -> if_delete_3_in_1_gga₂(H, delete_3_in_gga₂(X, T1))
member_2_in_gg₂(H, ._2₂(H, L)) -> member_2_out_gg
member_2_in_gg₂(X, ._2₂(H, L)) -> if_member_2_in_1_gg₁(member_2_in_gg₂(X, L))
if_member1_2_in_1_ag₁(member1_2_out_ag₁(X)) -> member1_2_out_ag₁(X)
if_delete_3_in_1_gga₂(H, delete_3_out_gga₁(T2)) -> delete_3_out_gga₁(._2₂(H, T2))
if_member_2_in_1_gg₁(member_2_out_gg) -> member_2_out_gg

The set Q consists of the following terms:

member1_2_in_ag₁(x0)
delete_3_in_gga₂(x0, x1)
member_2_in_gg₂(x0, x1)
if_member1_2_in_1_ag₁(x0)
if_delete_3_in_1_gga₂(x0, x1)
if_member_2_in_1_gg₁(x0)

We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {REACH_4_IN_GGGG₄, IF_REACH_4_IN_4_GGGG₄, IF_REACH_4_IN_2_GGGG₄, IF_REACH_4_IN_3_GGGG₅}.
We used the following order together with the size-change analysis to show that there are no infinite chains for this DP problem.

Order:Polynomial interpretation:

POL(member1_2_out_ag₁(x₁)) = 1
POL(if_delete_3_in_1_gga₂(x₁, x₂)) = 1 + x₂
POL(._2₂(x₁, x₂)) = 1 + x₂
POL([]_0) = 1
POL(member_2_out_gg) = 1
POL(delete_3_in_gga₂(x₁, x₂)) = x₂
POL(delete_3_out_gga₁(x₁)) = 1 + x₁

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

REACH_4_IN_GGGG₄(X, Z, E, L) -> IF_REACH_4_IN_2_GGGG₄(Z, E, L, member1_2_in_ag₁(E)) (allowed arguments on rhs = {1, 2, 3})
The graph contains the following edges 2 >= 1, 3 >= 2, 4 >= 3
IF_REACH_4_IN_3_GGGG₅(Z, E, L, Y, member_2_out_gg) -> IF_REACH_4_IN_4_GGGG₄(Z, E, Y, delete_3_in_gga₂(Y, L)) (allowed arguments on rhs = {1, 2, 3, 4})
The graph contains the following edges 1 >= 1, 2 >= 2, 4 >= 3, 3 >= 4
IF_REACH_4_IN_2_GGGG₄(Z, E, L, member1_2_out_ag₁(._2₂(X, ._2₂(Y, []_0)))) -> IF_REACH_4_IN_3_GGGG₅(Z, E, L, Y, member_2_in_gg₂(Y, L)) (allowed arguments on rhs = {1, 2, 3, 4})
The graph contains the following edges 1 >= 1, 2 >= 2, 3 >= 3
IF_REACH_4_IN_4_GGGG₄(Z, E, Y, delete_3_out_gga₁(V1)) -> REACH_4_IN_GGGG₄(Y, Z, E, V1) (allowed arguments on rhs = {1, 2, 3, 4})
The graph contains the following edges 3 >= 1, 1 >= 2, 2 >= 3, 4 > 4

We oriented the following set of usable rules.

if_delete_3_in_1_gga₂(H, delete_3_out_gga₁(T2)) -> delete_3_out_gga₁(._2₂(H, T2))
delete_3_in_gga₂(X, ._2₂(X, Y)) -> delete_3_out_gga₁(Y)
delete_3_in_gga₂(X, ._2₂(H, T1)) -> if_delete_3_in_1_gga₂(H, delete_3_in_gga₂(X, T1))