Left Termination proof of /tmp/tmp

Left Termination of the query pattern reach(b,b,b,b) w.r.t. the given Prolog program could not be shown:

↳ PROLOG

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

reach₄(X, Y, Edges, Visited) :- member₂(.₂(X, .₂(Y, {}₀)), Edges).
reach₄(X, Z, Edges, Visited) :- member1₂(.₂(X, .₂(Y, {}₀)), Edges), member₂(Y, Visited), reach₄(Y, Z, Edges, .₂(Y, Visited)).
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).

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)
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, Edges, Visited) -> if_reach_4_in_1_gggg₅(X, Y, Edges, Visited, member_2_in_gg₂(._2₂(X, ._2₂(Y, []_0)), Edges))
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, Edges, Visited, member_2_out_gg₂(._2₂(X, ._2₂(Y, []_0)), Edges)) -> reach_4_out_gggg₄(X, Y, Edges, Visited)
reach_4_in_gggg₄(X, Z, Edges, Visited) -> if_reach_4_in_2_gggg₅(X, Z, Edges, Visited, member1_2_in_ag₂(._2₂(X, ._2₂(Y, []_0)), Edges))
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, Edges, Visited, member1_2_out_ag₂(._2₂(X, ._2₂(Y, []_0)), Edges)) -> if_reach_4_in_3_gggg₆(X, Z, Edges, Visited, Y, member_2_in_gg₂(Y, Visited))
if_reach_4_in_3_gggg₆(X, Z, Edges, Visited, Y, member_2_out_gg₂(Y, Visited)) -> if_reach_4_in_4_gggg₆(X, Z, Edges, Visited, Y, reach_4_in_gggg₄(Y, Z, Edges, ._2₂(Y, Visited)))
if_reach_4_in_4_gggg₆(X, Z, Edges, Visited, Y, reach_4_out_gggg₄(Y, Z, Edges, ._2₂(Y, Visited))) -> reach_4_out_gggg₄(X, Z, Edges, Visited)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of PROLOG

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

  ↳ PrologToPiTRSProof

Pi-finite rewrite system:
The TRS R consists of the following rules:

reach_4_in_gggg₄(X, Y, Edges, Visited) -> if_reach_4_in_1_gggg₅(X, Y, Edges, Visited, member_2_in_gg₂(._2₂(X, ._2₂(Y, []_0)), Edges))
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, Edges, Visited, member_2_out_gg₂(._2₂(X, ._2₂(Y, []_0)), Edges)) -> reach_4_out_gggg₄(X, Y, Edges, Visited)
reach_4_in_gggg₄(X, Z, Edges, Visited) -> if_reach_4_in_2_gggg₅(X, Z, Edges, Visited, member1_2_in_ag₂(._2₂(X, ._2₂(Y, []_0)), Edges))
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, Edges, Visited, member1_2_out_ag₂(._2₂(X, ._2₂(Y, []_0)), Edges)) -> if_reach_4_in_3_gggg₆(X, Z, Edges, Visited, Y, member_2_in_gg₂(Y, Visited))
if_reach_4_in_3_gggg₆(X, Z, Edges, Visited, Y, member_2_out_gg₂(Y, Visited)) -> if_reach_4_in_4_gggg₆(X, Z, Edges, Visited, Y, reach_4_in_gggg₄(Y, Z, Edges, ._2₂(Y, Visited)))
if_reach_4_in_4_gggg₆(X, Z, Edges, Visited, Y, reach_4_out_gggg₄(Y, Z, Edges, ._2₂(Y, Visited))) -> reach_4_out_gggg₄(X, Z, Edges, Visited)

REACH_4_IN_GGGG₄(X, Y, Edges, Visited) -> IF_REACH_4_IN_1_GGGG₅(X, Y, Edges, Visited, member_2_in_gg₂(._2₂(X, ._2₂(Y, []_0)), Edges))
REACH_4_IN_GGGG₄(X, Y, Edges, Visited) -> MEMBER_2_IN_GG₂(._2₂(X, ._2₂(Y, []_0)), Edges)
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, Edges, Visited) -> IF_REACH_4_IN_2_GGGG₅(X, Z, Edges, Visited, member1_2_in_ag₂(._2₂(X, ._2₂(Y, []_0)), Edges))
REACH_4_IN_GGGG₄(X, Z, Edges, Visited) -> MEMBER1_2_IN_AG₂(._2₂(X, ._2₂(Y, []_0)), Edges)
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, Edges, Visited, member1_2_out_ag₂(._2₂(X, ._2₂(Y, []_0)), Edges)) -> IF_REACH_4_IN_3_GGGG₆(X, Z, Edges, Visited, Y, member_2_in_gg₂(Y, Visited))
IF_REACH_4_IN_2_GGGG₅(X, Z, Edges, Visited, member1_2_out_ag₂(._2₂(X, ._2₂(Y, []_0)), Edges)) -> MEMBER_2_IN_GG₂(Y, Visited)
IF_REACH_4_IN_3_GGGG₆(X, Z, Edges, Visited, Y, member_2_out_gg₂(Y, Visited)) -> IF_REACH_4_IN_4_GGGG₆(X, Z, Edges, Visited, Y, reach_4_in_gggg₄(Y, Z, Edges, ._2₂(Y, Visited)))
IF_REACH_4_IN_3_GGGG₆(X, Z, Edges, Visited, Y, member_2_out_gg₂(Y, Visited)) -> REACH_4_IN_GGGG₄(Y, Z, Edges, ._2₂(Y, Visited))

The TRS R consists of the following rules:

reach_4_in_gggg₄(X, Y, Edges, Visited) -> if_reach_4_in_1_gggg₅(X, Y, Edges, Visited, member_2_in_gg₂(._2₂(X, ._2₂(Y, []_0)), Edges))
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, Edges, Visited, member_2_out_gg₂(._2₂(X, ._2₂(Y, []_0)), Edges)) -> reach_4_out_gggg₄(X, Y, Edges, Visited)
reach_4_in_gggg₄(X, Z, Edges, Visited) -> if_reach_4_in_2_gggg₅(X, Z, Edges, Visited, member1_2_in_ag₂(._2₂(X, ._2₂(Y, []_0)), Edges))
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, Edges, Visited, member1_2_out_ag₂(._2₂(X, ._2₂(Y, []_0)), Edges)) -> if_reach_4_in_3_gggg₆(X, Z, Edges, Visited, Y, member_2_in_gg₂(Y, Visited))
if_reach_4_in_3_gggg₆(X, Z, Edges, Visited, Y, member_2_out_gg₂(Y, Visited)) -> if_reach_4_in_4_gggg₆(X, Z, Edges, Visited, Y, reach_4_in_gggg₄(Y, Z, Edges, ._2₂(Y, Visited)))
if_reach_4_in_4_gggg₆(X, Z, Edges, Visited, Y, reach_4_out_gggg₄(Y, Z, Edges, ._2₂(Y, Visited))) -> reach_4_out_gggg₄(X, Z, Edges, Visited)

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_gggg₁(x6)
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_GGGG₁(x6)
REACH_4_IN_GGGG₄(x1, x2, x3, x4) = REACH_4_IN_GGGG₄(x1, x2, x3, x4)
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_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

  ↳ PrologToPiTRSProof

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

REACH_4_IN_GGGG₄(X, Y, Edges, Visited) -> IF_REACH_4_IN_1_GGGG₅(X, Y, Edges, Visited, member_2_in_gg₂(._2₂(X, ._2₂(Y, []_0)), Edges))
REACH_4_IN_GGGG₄(X, Y, Edges, Visited) -> MEMBER_2_IN_GG₂(._2₂(X, ._2₂(Y, []_0)), Edges)
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, Edges, Visited) -> IF_REACH_4_IN_2_GGGG₅(X, Z, Edges, Visited, member1_2_in_ag₂(._2₂(X, ._2₂(Y, []_0)), Edges))
REACH_4_IN_GGGG₄(X, Z, Edges, Visited) -> MEMBER1_2_IN_AG₂(._2₂(X, ._2₂(Y, []_0)), Edges)
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, Edges, Visited, member1_2_out_ag₂(._2₂(X, ._2₂(Y, []_0)), Edges)) -> IF_REACH_4_IN_3_GGGG₆(X, Z, Edges, Visited, Y, member_2_in_gg₂(Y, Visited))
IF_REACH_4_IN_2_GGGG₅(X, Z, Edges, Visited, member1_2_out_ag₂(._2₂(X, ._2₂(Y, []_0)), Edges)) -> MEMBER_2_IN_GG₂(Y, Visited)
IF_REACH_4_IN_3_GGGG₆(X, Z, Edges, Visited, Y, member_2_out_gg₂(Y, Visited)) -> IF_REACH_4_IN_4_GGGG₆(X, Z, Edges, Visited, Y, reach_4_in_gggg₄(Y, Z, Edges, ._2₂(Y, Visited)))
IF_REACH_4_IN_3_GGGG₆(X, Z, Edges, Visited, Y, member_2_out_gg₂(Y, Visited)) -> REACH_4_IN_GGGG₄(Y, Z, Edges, ._2₂(Y, Visited))

The TRS R consists of the following rules:

reach_4_in_gggg₄(X, Y, Edges, Visited) -> if_reach_4_in_1_gggg₅(X, Y, Edges, Visited, member_2_in_gg₂(._2₂(X, ._2₂(Y, []_0)), Edges))
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, Edges, Visited, member_2_out_gg₂(._2₂(X, ._2₂(Y, []_0)), Edges)) -> reach_4_out_gggg₄(X, Y, Edges, Visited)
reach_4_in_gggg₄(X, Z, Edges, Visited) -> if_reach_4_in_2_gggg₅(X, Z, Edges, Visited, member1_2_in_ag₂(._2₂(X, ._2₂(Y, []_0)), Edges))
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, Edges, Visited, member1_2_out_ag₂(._2₂(X, ._2₂(Y, []_0)), Edges)) -> if_reach_4_in_3_gggg₆(X, Z, Edges, Visited, Y, member_2_in_gg₂(Y, Visited))
if_reach_4_in_3_gggg₆(X, Z, Edges, Visited, Y, member_2_out_gg₂(Y, Visited)) -> if_reach_4_in_4_gggg₆(X, Z, Edges, Visited, Y, reach_4_in_gggg₄(Y, Z, Edges, ._2₂(Y, Visited)))
if_reach_4_in_4_gggg₆(X, Z, Edges, Visited, Y, reach_4_out_gggg₄(Y, Z, Edges, ._2₂(Y, Visited))) -> reach_4_out_gggg₄(X, Z, Edges, Visited)

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

              ↳ PiDP

              ↳ PiDP

  ↳ PrologToPiTRSProof

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, Edges, Visited) -> if_reach_4_in_1_gggg₅(X, Y, Edges, Visited, member_2_in_gg₂(._2₂(X, ._2₂(Y, []_0)), Edges))
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, Edges, Visited, member_2_out_gg₂(._2₂(X, ._2₂(Y, []_0)), Edges)) -> reach_4_out_gggg₄(X, Y, Edges, Visited)
reach_4_in_gggg₄(X, Z, Edges, Visited) -> if_reach_4_in_2_gggg₅(X, Z, Edges, Visited, member1_2_in_ag₂(._2₂(X, ._2₂(Y, []_0)), Edges))
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, Edges, Visited, member1_2_out_ag₂(._2₂(X, ._2₂(Y, []_0)), Edges)) -> if_reach_4_in_3_gggg₆(X, Z, Edges, Visited, Y, member_2_in_gg₂(Y, Visited))
if_reach_4_in_3_gggg₆(X, Z, Edges, Visited, Y, member_2_out_gg₂(Y, Visited)) -> if_reach_4_in_4_gggg₆(X, Z, Edges, Visited, Y, reach_4_in_gggg₄(Y, Z, Edges, ._2₂(Y, Visited)))
if_reach_4_in_4_gggg₆(X, Z, Edges, Visited, Y, reach_4_out_gggg₄(Y, Z, Edges, ._2₂(Y, Visited))) -> reach_4_out_gggg₄(X, Z, Edges, Visited)

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

              ↳ PiDP

              ↳ PiDP

  ↳ PrologToPiTRSProof

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

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ PiDP

              ↳ PiDP

  ↳ PrologToPiTRSProof

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

                ↳ UsableRulesProof

              ↳ PiDP

  ↳ PrologToPiTRSProof

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, Edges, Visited) -> if_reach_4_in_1_gggg₅(X, Y, Edges, Visited, member_2_in_gg₂(._2₂(X, ._2₂(Y, []_0)), Edges))
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, Edges, Visited, member_2_out_gg₂(._2₂(X, ._2₂(Y, []_0)), Edges)) -> reach_4_out_gggg₄(X, Y, Edges, Visited)
reach_4_in_gggg₄(X, Z, Edges, Visited) -> if_reach_4_in_2_gggg₅(X, Z, Edges, Visited, member1_2_in_ag₂(._2₂(X, ._2₂(Y, []_0)), Edges))
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, Edges, Visited, member1_2_out_ag₂(._2₂(X, ._2₂(Y, []_0)), Edges)) -> if_reach_4_in_3_gggg₆(X, Z, Edges, Visited, Y, member_2_in_gg₂(Y, Visited))
if_reach_4_in_3_gggg₆(X, Z, Edges, Visited, Y, member_2_out_gg₂(Y, Visited)) -> if_reach_4_in_4_gggg₆(X, Z, Edges, Visited, Y, reach_4_in_gggg₄(Y, Z, Edges, ._2₂(Y, Visited)))
if_reach_4_in_4_gggg₆(X, Z, Edges, Visited, Y, reach_4_out_gggg₄(Y, Z, Edges, ._2₂(Y, Visited))) -> reach_4_out_gggg₄(X, Z, Edges, Visited)

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

              ↳ PiDP

  ↳ PrologToPiTRSProof

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

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ PiDP

  ↳ PrologToPiTRSProof

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

                ↳ UsableRulesProof

  ↳ PrologToPiTRSProof

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

REACH_4_IN_GGGG₄(X, Z, Edges, Visited) -> IF_REACH_4_IN_2_GGGG₅(X, Z, Edges, Visited, member1_2_in_ag₂(._2₂(X, ._2₂(Y, []_0)), Edges))
IF_REACH_4_IN_3_GGGG₆(X, Z, Edges, Visited, Y, member_2_out_gg₂(Y, Visited)) -> REACH_4_IN_GGGG₄(Y, Z, Edges, ._2₂(Y, Visited))
IF_REACH_4_IN_2_GGGG₅(X, Z, Edges, Visited, member1_2_out_ag₂(._2₂(X, ._2₂(Y, []_0)), Edges)) -> IF_REACH_4_IN_3_GGGG₆(X, Z, Edges, Visited, Y, member_2_in_gg₂(Y, Visited))

The TRS R consists of the following rules:

reach_4_in_gggg₄(X, Y, Edges, Visited) -> if_reach_4_in_1_gggg₅(X, Y, Edges, Visited, member_2_in_gg₂(._2₂(X, ._2₂(Y, []_0)), Edges))
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, Edges, Visited, member_2_out_gg₂(._2₂(X, ._2₂(Y, []_0)), Edges)) -> reach_4_out_gggg₄(X, Y, Edges, Visited)
reach_4_in_gggg₄(X, Z, Edges, Visited) -> if_reach_4_in_2_gggg₅(X, Z, Edges, Visited, member1_2_in_ag₂(._2₂(X, ._2₂(Y, []_0)), Edges))
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, Edges, Visited, member1_2_out_ag₂(._2₂(X, ._2₂(Y, []_0)), Edges)) -> if_reach_4_in_3_gggg₆(X, Z, Edges, Visited, Y, member_2_in_gg₂(Y, Visited))
if_reach_4_in_3_gggg₆(X, Z, Edges, Visited, Y, member_2_out_gg₂(Y, Visited)) -> if_reach_4_in_4_gggg₆(X, Z, Edges, Visited, Y, reach_4_in_gggg₄(Y, Z, Edges, ._2₂(Y, Visited)))
if_reach_4_in_4_gggg₆(X, Z, Edges, Visited, Y, reach_4_out_gggg₄(Y, Z, Edges, ._2₂(Y, Visited))) -> reach_4_out_gggg₄(X, Z, Edges, Visited)

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_gggg₁(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

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

  ↳ PrologToPiTRSProof

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

REACH_4_IN_GGGG₄(X, Z, Edges, Visited) -> IF_REACH_4_IN_2_GGGG₅(X, Z, Edges, Visited, member1_2_in_ag₂(._2₂(X, ._2₂(Y, []_0)), Edges))
IF_REACH_4_IN_3_GGGG₆(X, Z, Edges, Visited, Y, member_2_out_gg₂(Y, Visited)) -> REACH_4_IN_GGGG₄(Y, Z, Edges, ._2₂(Y, Visited))
IF_REACH_4_IN_2_GGGG₅(X, Z, Edges, Visited, member1_2_out_ag₂(._2₂(X, ._2₂(Y, []_0)), Edges)) -> IF_REACH_4_IN_3_GGGG₆(X, Z, Edges, Visited, Y, member_2_in_gg₂(Y, Visited))

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))
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_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)
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

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

  ↳ PrologToPiTRSProof

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

REACH_4_IN_GGGG₄(X, Z, Edges, Visited) -> IF_REACH_4_IN_2_GGGG₄(Z, Edges, Visited, member1_2_in_ag₁(Edges))
IF_REACH_4_IN_3_GGGG₅(Z, Edges, Visited, Y, member_2_out_gg) -> REACH_4_IN_GGGG₄(Y, Z, Edges, ._2₂(Y, Visited))
IF_REACH_4_IN_2_GGGG₄(Z, Edges, Visited, member1_2_out_ag₁(._2₂(X, ._2₂(Y, []_0)))) -> IF_REACH_4_IN_3_GGGG₅(Z, Edges, Visited, Y, member_2_in_gg₂(Y, Visited))

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))
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_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)
member_2_in_gg₂(x0, x1)
if_member1_2_in_1_ag₁(x0)
if_member_2_in_1_gg₁(x0)

We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {IF_REACH_4_IN_2_GGGG₄, REACH_4_IN_GGGG₄, IF_REACH_4_IN_3_GGGG₅}.
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)
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, Edges, Visited) -> if_reach_4_in_1_gggg₅(X, Y, Edges, Visited, member_2_in_gg₂(._2₂(X, ._2₂(Y, []_0)), Edges))
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, Edges, Visited, member_2_out_gg₂(._2₂(X, ._2₂(Y, []_0)), Edges)) -> reach_4_out_gggg₄(X, Y, Edges, Visited)
reach_4_in_gggg₄(X, Z, Edges, Visited) -> if_reach_4_in_2_gggg₅(X, Z, Edges, Visited, member1_2_in_ag₂(._2₂(X, ._2₂(Y, []_0)), Edges))
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, Edges, Visited, member1_2_out_ag₂(._2₂(X, ._2₂(Y, []_0)), Edges)) -> if_reach_4_in_3_gggg₆(X, Z, Edges, Visited, Y, member_2_in_gg₂(Y, Visited))
if_reach_4_in_3_gggg₆(X, Z, Edges, Visited, Y, member_2_out_gg₂(Y, Visited)) -> if_reach_4_in_4_gggg₆(X, Z, Edges, Visited, Y, reach_4_in_gggg₄(Y, Z, Edges, ._2₂(Y, Visited)))
if_reach_4_in_4_gggg₆(X, Z, Edges, Visited, Y, reach_4_out_gggg₄(Y, Z, Edges, ._2₂(Y, Visited))) -> reach_4_out_gggg₄(X, Z, Edges, Visited)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of PROLOG

↳ PROLOG

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

Pi-finite rewrite system:
The TRS R consists of the following rules:

reach_4_in_gggg₄(X, Y, Edges, Visited) -> if_reach_4_in_1_gggg₅(X, Y, Edges, Visited, member_2_in_gg₂(._2₂(X, ._2₂(Y, []_0)), Edges))
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, Edges, Visited, member_2_out_gg₂(._2₂(X, ._2₂(Y, []_0)), Edges)) -> reach_4_out_gggg₄(X, Y, Edges, Visited)
reach_4_in_gggg₄(X, Z, Edges, Visited) -> if_reach_4_in_2_gggg₅(X, Z, Edges, Visited, member1_2_in_ag₂(._2₂(X, ._2₂(Y, []_0)), Edges))
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, Edges, Visited, member1_2_out_ag₂(._2₂(X, ._2₂(Y, []_0)), Edges)) -> if_reach_4_in_3_gggg₆(X, Z, Edges, Visited, Y, member_2_in_gg₂(Y, Visited))
if_reach_4_in_3_gggg₆(X, Z, Edges, Visited, Y, member_2_out_gg₂(Y, Visited)) -> if_reach_4_in_4_gggg₆(X, Z, Edges, Visited, Y, reach_4_in_gggg₄(Y, Z, Edges, ._2₂(Y, Visited)))
if_reach_4_in_4_gggg₆(X, Z, Edges, Visited, Y, reach_4_out_gggg₄(Y, Z, Edges, ._2₂(Y, Visited))) -> reach_4_out_gggg₄(X, Z, Edges, Visited)

REACH_4_IN_GGGG₄(X, Y, Edges, Visited) -> IF_REACH_4_IN_1_GGGG₅(X, Y, Edges, Visited, member_2_in_gg₂(._2₂(X, ._2₂(Y, []_0)), Edges))
REACH_4_IN_GGGG₄(X, Y, Edges, Visited) -> MEMBER_2_IN_GG₂(._2₂(X, ._2₂(Y, []_0)), Edges)
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, Edges, Visited) -> IF_REACH_4_IN_2_GGGG₅(X, Z, Edges, Visited, member1_2_in_ag₂(._2₂(X, ._2₂(Y, []_0)), Edges))
REACH_4_IN_GGGG₄(X, Z, Edges, Visited) -> MEMBER1_2_IN_AG₂(._2₂(X, ._2₂(Y, []_0)), Edges)
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, Edges, Visited, member1_2_out_ag₂(._2₂(X, ._2₂(Y, []_0)), Edges)) -> IF_REACH_4_IN_3_GGGG₆(X, Z, Edges, Visited, Y, member_2_in_gg₂(Y, Visited))
IF_REACH_4_IN_2_GGGG₅(X, Z, Edges, Visited, member1_2_out_ag₂(._2₂(X, ._2₂(Y, []_0)), Edges)) -> MEMBER_2_IN_GG₂(Y, Visited)
IF_REACH_4_IN_3_GGGG₆(X, Z, Edges, Visited, Y, member_2_out_gg₂(Y, Visited)) -> IF_REACH_4_IN_4_GGGG₆(X, Z, Edges, Visited, Y, reach_4_in_gggg₄(Y, Z, Edges, ._2₂(Y, Visited)))
IF_REACH_4_IN_3_GGGG₆(X, Z, Edges, Visited, Y, member_2_out_gg₂(Y, Visited)) -> REACH_4_IN_GGGG₄(Y, Z, Edges, ._2₂(Y, Visited))

The TRS R consists of the following rules:

reach_4_in_gggg₄(X, Y, Edges, Visited) -> if_reach_4_in_1_gggg₅(X, Y, Edges, Visited, member_2_in_gg₂(._2₂(X, ._2₂(Y, []_0)), Edges))
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, Edges, Visited, member_2_out_gg₂(._2₂(X, ._2₂(Y, []_0)), Edges)) -> reach_4_out_gggg₄(X, Y, Edges, Visited)
reach_4_in_gggg₄(X, Z, Edges, Visited) -> if_reach_4_in_2_gggg₅(X, Z, Edges, Visited, member1_2_in_ag₂(._2₂(X, ._2₂(Y, []_0)), Edges))
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, Edges, Visited, member1_2_out_ag₂(._2₂(X, ._2₂(Y, []_0)), Edges)) -> if_reach_4_in_3_gggg₆(X, Z, Edges, Visited, Y, member_2_in_gg₂(Y, Visited))
if_reach_4_in_3_gggg₆(X, Z, Edges, Visited, Y, member_2_out_gg₂(Y, Visited)) -> if_reach_4_in_4_gggg₆(X, Z, Edges, Visited, Y, reach_4_in_gggg₄(Y, Z, Edges, ._2₂(Y, Visited)))
if_reach_4_in_4_gggg₆(X, Z, Edges, Visited, Y, reach_4_out_gggg₄(Y, Z, Edges, ._2₂(Y, Visited))) -> reach_4_out_gggg₄(X, Z, Edges, Visited)

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_gggg₅(x1, x2, x3, x4, x5)
member_2_in_gg₂(x1, x2) = member_2_in_gg₂(x1, x2)
member_2_out_gg₂(x1, x2) = member_2_out_gg₂(x1, x2)
if_member_2_in_1_gg₄(x1, x2, x3, x4) = if_member_2_in_1_gg₄(x1, x2, x3, x4)
reach_4_out_gggg₄(x1, x2, x3, x4) = reach_4_out_gggg₄(x1, x2, x3, x4)
if_reach_4_in_2_gggg₅(x1, x2, x3, x4, x5) = if_reach_4_in_2_gggg₅(x1, x2, x3, x4, x5)
member1_2_in_ag₂(x1, x2) = member1_2_in_ag₁(x2)
member1_2_out_ag₂(x1, x2) = member1_2_out_ag₂(x1, x2)
if_member1_2_in_1_ag₄(x1, x2, x3, x4) = if_member1_2_in_1_ag₃(x2, x3, x4)
if_reach_4_in_3_gggg₆(x1, x2, x3, x4, x5, x6) = if_reach_4_in_3_gggg₆(x1, x2, x3, x4, x5, x6)
if_reach_4_in_4_gggg₆(x1, x2, x3, x4, x5, x6) = if_reach_4_in_4_gggg₅(x1, x2, x3, x4, x6)
IF_MEMBER_2_IN_1_GG₄(x1, x2, x3, x4) = IF_MEMBER_2_IN_1_GG₄(x1, x2, x3, x4)
IF_REACH_4_IN_4_GGGG₆(x1, x2, x3, x4, x5, x6) = IF_REACH_4_IN_4_GGGG₅(x1, x2, x3, x4, x6)
REACH_4_IN_GGGG₄(x1, x2, x3, x4) = REACH_4_IN_GGGG₄(x1, x2, x3, x4)
IF_REACH_4_IN_1_GGGG₅(x1, x2, x3, x4, x5) = IF_REACH_4_IN_1_GGGG₅(x1, x2, x3, x4, 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_MEMBER1_2_IN_1_AG₄(x1, x2, x3, x4) = IF_MEMBER1_2_IN_1_AG₃(x2, x3, x4)
IF_REACH_4_IN_2_GGGG₅(x1, x2, x3, x4, x5) = IF_REACH_4_IN_2_GGGG₅(x1, x2, x3, x4, x5)
IF_REACH_4_IN_3_GGGG₆(x1, x2, x3, x4, x5, x6) = IF_REACH_4_IN_3_GGGG₆(x1, x2, x3, x4, x5, x6)

We have to consider all (P,R,Pi)-chains

↳ PROLOG

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

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

REACH_4_IN_GGGG₄(X, Y, Edges, Visited) -> IF_REACH_4_IN_1_GGGG₅(X, Y, Edges, Visited, member_2_in_gg₂(._2₂(X, ._2₂(Y, []_0)), Edges))
REACH_4_IN_GGGG₄(X, Y, Edges, Visited) -> MEMBER_2_IN_GG₂(._2₂(X, ._2₂(Y, []_0)), Edges)
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, Edges, Visited) -> IF_REACH_4_IN_2_GGGG₅(X, Z, Edges, Visited, member1_2_in_ag₂(._2₂(X, ._2₂(Y, []_0)), Edges))
REACH_4_IN_GGGG₄(X, Z, Edges, Visited) -> MEMBER1_2_IN_AG₂(._2₂(X, ._2₂(Y, []_0)), Edges)
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, Edges, Visited, member1_2_out_ag₂(._2₂(X, ._2₂(Y, []_0)), Edges)) -> IF_REACH_4_IN_3_GGGG₆(X, Z, Edges, Visited, Y, member_2_in_gg₂(Y, Visited))
IF_REACH_4_IN_2_GGGG₅(X, Z, Edges, Visited, member1_2_out_ag₂(._2₂(X, ._2₂(Y, []_0)), Edges)) -> MEMBER_2_IN_GG₂(Y, Visited)
IF_REACH_4_IN_3_GGGG₆(X, Z, Edges, Visited, Y, member_2_out_gg₂(Y, Visited)) -> IF_REACH_4_IN_4_GGGG₆(X, Z, Edges, Visited, Y, reach_4_in_gggg₄(Y, Z, Edges, ._2₂(Y, Visited)))
IF_REACH_4_IN_3_GGGG₆(X, Z, Edges, Visited, Y, member_2_out_gg₂(Y, Visited)) -> REACH_4_IN_GGGG₄(Y, Z, Edges, ._2₂(Y, Visited))

The TRS R consists of the following rules:

reach_4_in_gggg₄(X, Y, Edges, Visited) -> if_reach_4_in_1_gggg₅(X, Y, Edges, Visited, member_2_in_gg₂(._2₂(X, ._2₂(Y, []_0)), Edges))
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, Edges, Visited, member_2_out_gg₂(._2₂(X, ._2₂(Y, []_0)), Edges)) -> reach_4_out_gggg₄(X, Y, Edges, Visited)
reach_4_in_gggg₄(X, Z, Edges, Visited) -> if_reach_4_in_2_gggg₅(X, Z, Edges, Visited, member1_2_in_ag₂(._2₂(X, ._2₂(Y, []_0)), Edges))
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, Edges, Visited, member1_2_out_ag₂(._2₂(X, ._2₂(Y, []_0)), Edges)) -> if_reach_4_in_3_gggg₆(X, Z, Edges, Visited, Y, member_2_in_gg₂(Y, Visited))
if_reach_4_in_3_gggg₆(X, Z, Edges, Visited, Y, member_2_out_gg₂(Y, Visited)) -> if_reach_4_in_4_gggg₆(X, Z, Edges, Visited, Y, reach_4_in_gggg₄(Y, Z, Edges, ._2₂(Y, Visited)))
if_reach_4_in_4_gggg₆(X, Z, Edges, Visited, Y, reach_4_out_gggg₄(Y, Z, Edges, ._2₂(Y, Visited))) -> reach_4_out_gggg₄(X, Z, Edges, Visited)

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_gggg₅(x1, x2, x3, x4, x5)
member_2_in_gg₂(x1, x2) = member_2_in_gg₂(x1, x2)
member_2_out_gg₂(x1, x2) = member_2_out_gg₂(x1, x2)
if_member_2_in_1_gg₄(x1, x2, x3, x4) = if_member_2_in_1_gg₄(x1, x2, x3, x4)
reach_4_out_gggg₄(x1, x2, x3, x4) = reach_4_out_gggg₄(x1, x2, x3, x4)
if_reach_4_in_2_gggg₅(x1, x2, x3, x4, x5) = if_reach_4_in_2_gggg₅(x1, x2, x3, x4, x5)
member1_2_in_ag₂(x1, x2) = member1_2_in_ag₁(x2)
member1_2_out_ag₂(x1, x2) = member1_2_out_ag₂(x1, x2)
if_member1_2_in_1_ag₄(x1, x2, x3, x4) = if_member1_2_in_1_ag₃(x2, x3, x4)
if_reach_4_in_3_gggg₆(x1, x2, x3, x4, x5, x6) = if_reach_4_in_3_gggg₆(x1, x2, x3, x4, x5, x6)
if_reach_4_in_4_gggg₆(x1, x2, x3, x4, x5, x6) = if_reach_4_in_4_gggg₅(x1, x2, x3, x4, x6)
IF_MEMBER_2_IN_1_GG₄(x1, x2, x3, x4) = IF_MEMBER_2_IN_1_GG₄(x1, x2, x3, x4)
IF_REACH_4_IN_4_GGGG₆(x1, x2, x3, x4, x5, x6) = IF_REACH_4_IN_4_GGGG₅(x1, x2, x3, x4, x6)
REACH_4_IN_GGGG₄(x1, x2, x3, x4) = REACH_4_IN_GGGG₄(x1, x2, x3, x4)
IF_REACH_4_IN_1_GGGG₅(x1, x2, x3, x4, x5) = IF_REACH_4_IN_1_GGGG₅(x1, x2, x3, x4, 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_MEMBER1_2_IN_1_AG₄(x1, x2, x3, x4) = IF_MEMBER1_2_IN_1_AG₃(x2, x3, x4)
IF_REACH_4_IN_2_GGGG₅(x1, x2, x3, x4, x5) = IF_REACH_4_IN_2_GGGG₅(x1, x2, x3, x4, x5)
IF_REACH_4_IN_3_GGGG₆(x1, x2, x3, x4, x5, x6) = IF_REACH_4_IN_3_GGGG₆(x1, x2, x3, x4, x5, x6)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph contains 3 SCCs with 7 less nodes.

↳ PROLOG

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ 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, Edges, Visited) -> if_reach_4_in_1_gggg₅(X, Y, Edges, Visited, member_2_in_gg₂(._2₂(X, ._2₂(Y, []_0)), Edges))
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, Edges, Visited, member_2_out_gg₂(._2₂(X, ._2₂(Y, []_0)), Edges)) -> reach_4_out_gggg₄(X, Y, Edges, Visited)
reach_4_in_gggg₄(X, Z, Edges, Visited) -> if_reach_4_in_2_gggg₅(X, Z, Edges, Visited, member1_2_in_ag₂(._2₂(X, ._2₂(Y, []_0)), Edges))
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, Edges, Visited, member1_2_out_ag₂(._2₂(X, ._2₂(Y, []_0)), Edges)) -> if_reach_4_in_3_gggg₆(X, Z, Edges, Visited, Y, member_2_in_gg₂(Y, Visited))
if_reach_4_in_3_gggg₆(X, Z, Edges, Visited, Y, member_2_out_gg₂(Y, Visited)) -> if_reach_4_in_4_gggg₆(X, Z, Edges, Visited, Y, reach_4_in_gggg₄(Y, Z, Edges, ._2₂(Y, Visited)))
if_reach_4_in_4_gggg₆(X, Z, Edges, Visited, Y, reach_4_out_gggg₄(Y, Z, Edges, ._2₂(Y, Visited))) -> reach_4_out_gggg₄(X, Z, Edges, Visited)

↳ PROLOG

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ 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, Edges, Visited) -> if_reach_4_in_1_gggg₅(X, Y, Edges, Visited, member_2_in_gg₂(._2₂(X, ._2₂(Y, []_0)), Edges))
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, Edges, Visited, member_2_out_gg₂(._2₂(X, ._2₂(Y, []_0)), Edges)) -> reach_4_out_gggg₄(X, Y, Edges, Visited)
reach_4_in_gggg₄(X, Z, Edges, Visited) -> if_reach_4_in_2_gggg₅(X, Z, Edges, Visited, member1_2_in_ag₂(._2₂(X, ._2₂(Y, []_0)), Edges))
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, Edges, Visited, member1_2_out_ag₂(._2₂(X, ._2₂(Y, []_0)), Edges)) -> if_reach_4_in_3_gggg₆(X, Z, Edges, Visited, Y, member_2_in_gg₂(Y, Visited))
if_reach_4_in_3_gggg₆(X, Z, Edges, Visited, Y, member_2_out_gg₂(Y, Visited)) -> if_reach_4_in_4_gggg₆(X, Z, Edges, Visited, Y, reach_4_in_gggg₄(Y, Z, Edges, ._2₂(Y, Visited)))
if_reach_4_in_4_gggg₆(X, Z, Edges, Visited, Y, reach_4_out_gggg₄(Y, Z, Edges, ._2₂(Y, Visited))) -> reach_4_out_gggg₄(X, Z, Edges, Visited)

↳ PROLOG

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

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

REACH_4_IN_GGGG₄(X, Z, Edges, Visited) -> IF_REACH_4_IN_2_GGGG₅(X, Z, Edges, Visited, member1_2_in_ag₂(._2₂(X, ._2₂(Y, []_0)), Edges))
IF_REACH_4_IN_3_GGGG₆(X, Z, Edges, Visited, Y, member_2_out_gg₂(Y, Visited)) -> REACH_4_IN_GGGG₄(Y, Z, Edges, ._2₂(Y, Visited))
IF_REACH_4_IN_2_GGGG₅(X, Z, Edges, Visited, member1_2_out_ag₂(._2₂(X, ._2₂(Y, []_0)), Edges)) -> IF_REACH_4_IN_3_GGGG₆(X, Z, Edges, Visited, Y, member_2_in_gg₂(Y, Visited))

The TRS R consists of the following rules:

reach_4_in_gggg₄(X, Y, Edges, Visited) -> if_reach_4_in_1_gggg₅(X, Y, Edges, Visited, member_2_in_gg₂(._2₂(X, ._2₂(Y, []_0)), Edges))
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, Edges, Visited, member_2_out_gg₂(._2₂(X, ._2₂(Y, []_0)), Edges)) -> reach_4_out_gggg₄(X, Y, Edges, Visited)
reach_4_in_gggg₄(X, Z, Edges, Visited) -> if_reach_4_in_2_gggg₅(X, Z, Edges, Visited, member1_2_in_ag₂(._2₂(X, ._2₂(Y, []_0)), Edges))
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, Edges, Visited, member1_2_out_ag₂(._2₂(X, ._2₂(Y, []_0)), Edges)) -> if_reach_4_in_3_gggg₆(X, Z, Edges, Visited, Y, member_2_in_gg₂(Y, Visited))
if_reach_4_in_3_gggg₆(X, Z, Edges, Visited, Y, member_2_out_gg₂(Y, Visited)) -> if_reach_4_in_4_gggg₆(X, Z, Edges, Visited, Y, reach_4_in_gggg₄(Y, Z, Edges, ._2₂(Y, Visited)))
if_reach_4_in_4_gggg₆(X, Z, Edges, Visited, Y, reach_4_out_gggg₄(Y, Z, Edges, ._2₂(Y, Visited))) -> reach_4_out_gggg₄(X, Z, Edges, Visited)

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_gggg₅(x1, x2, x3, x4, x5)
member_2_in_gg₂(x1, x2) = member_2_in_gg₂(x1, x2)
member_2_out_gg₂(x1, x2) = member_2_out_gg₂(x1, x2)
if_member_2_in_1_gg₄(x1, x2, x3, x4) = if_member_2_in_1_gg₄(x1, x2, x3, x4)
reach_4_out_gggg₄(x1, x2, x3, x4) = reach_4_out_gggg₄(x1, x2, x3, x4)
if_reach_4_in_2_gggg₅(x1, x2, x3, x4, x5) = if_reach_4_in_2_gggg₅(x1, x2, x3, x4, x5)
member1_2_in_ag₂(x1, x2) = member1_2_in_ag₁(x2)
member1_2_out_ag₂(x1, x2) = member1_2_out_ag₂(x1, x2)
if_member1_2_in_1_ag₄(x1, x2, x3, x4) = if_member1_2_in_1_ag₃(x2, x3, x4)
if_reach_4_in_3_gggg₆(x1, x2, x3, x4, x5, x6) = if_reach_4_in_3_gggg₆(x1, x2, x3, x4, x5, x6)
if_reach_4_in_4_gggg₆(x1, x2, x3, x4, x5, x6) = if_reach_4_in_4_gggg₅(x1, x2, x3, x4, 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_GGGG₅(x1, x2, x3, x4, x5)
IF_REACH_4_IN_3_GGGG₆(x1, x2, x3, x4, x5, x6) = IF_REACH_4_IN_3_GGGG₆(x1, x2, x3, x4, x5, x6)

We have to consider all (P,R,Pi)-chains