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



↳ PROLOG
  ↳ UnrequestedClauseRemoverProof

overlap2(Xs, Ys) :- member22(X, Xs), member12(X, Ys).
hasaorb1(Xs) :- overlap2(Xs, .2(a0, .2(b0, {}0))).
member12(X, .2(Y, Xs)) :- member12(X, Xs).
member12(X, .2(X, Xs)).
member22(X, .2(Y, Xs)) :- member22(X, Xs).
member22(X, .2(X, Xs)).


The clause

hasaorb1(Xs) :- overlap2(Xs, .2(a0, .2(b0, {}0))).

can be ignored, as it is not needed by any of the given querys.

Deleting this clauses results in the following prolog program:

overlap2(Xs, Ys) :- member22(X, Xs), member12(X, Ys).
member12(X, .2(Y, Xs)) :- member12(X, Xs).
member12(X, .2(X, Xs)).
member22(X, .2(Y, Xs)) :- member22(X, Xs).
member22(X, .2(X, Xs)).



↳ PROLOG
  ↳ UnrequestedClauseRemoverProof
    ↳ PROLOG
      ↳ PrologToPiTRSProof

overlap2(Xs, Ys) :- member22(X, Xs), member12(X, Ys).
member12(X, .2(Y, Xs)) :- member12(X, Xs).
member12(X, .2(X, Xs)).
member22(X, .2(Y, Xs)) :- member22(X, Xs).
member22(X, .2(X, Xs)).


With regard to the inferred argument filtering the predicates were used in the following modes:
overlap2: (b,b)
member22: (f,b)
member12: (b,b)
Transforming PROLOG into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:


overlap_2_in_gg2(Xs, Ys) -> if_overlap_2_in_1_gg3(Xs, Ys, member2_2_in_ag2(X, Xs))
member2_2_in_ag2(X, ._22(Y, Xs)) -> if_member2_2_in_1_ag4(X, Y, Xs, member2_2_in_ag2(X, Xs))
member2_2_in_ag2(X, ._22(X, Xs)) -> member2_2_out_ag2(X, ._22(X, Xs))
if_member2_2_in_1_ag4(X, Y, Xs, member2_2_out_ag2(X, Xs)) -> member2_2_out_ag2(X, ._22(Y, Xs))
if_overlap_2_in_1_gg3(Xs, Ys, member2_2_out_ag2(X, Xs)) -> if_overlap_2_in_2_gg4(Xs, Ys, X, member1_2_in_gg2(X, Ys))
member1_2_in_gg2(X, ._22(Y, Xs)) -> if_member1_2_in_1_gg4(X, Y, Xs, member1_2_in_gg2(X, Xs))
member1_2_in_gg2(X, ._22(X, Xs)) -> member1_2_out_gg2(X, ._22(X, Xs))
if_member1_2_in_1_gg4(X, Y, Xs, member1_2_out_gg2(X, Xs)) -> member1_2_out_gg2(X, ._22(Y, Xs))
if_overlap_2_in_2_gg4(Xs, Ys, X, member1_2_out_gg2(X, Ys)) -> overlap_2_out_gg2(Xs, Ys)

The argument filtering Pi contains the following mapping:
overlap_2_in_gg2(x1, x2)  =  overlap_2_in_gg2(x1, x2)
._22(x1, x2)  =  ._22(x1, x2)
if_overlap_2_in_1_gg3(x1, x2, x3)  =  if_overlap_2_in_1_gg2(x2, x3)
member2_2_in_ag2(x1, x2)  =  member2_2_in_ag1(x2)
if_member2_2_in_1_ag4(x1, x2, x3, x4)  =  if_member2_2_in_1_ag1(x4)
member2_2_out_ag2(x1, x2)  =  member2_2_out_ag1(x1)
if_overlap_2_in_2_gg4(x1, x2, x3, x4)  =  if_overlap_2_in_2_gg1(x4)
member1_2_in_gg2(x1, x2)  =  member1_2_in_gg2(x1, x2)
if_member1_2_in_1_gg4(x1, x2, x3, x4)  =  if_member1_2_in_1_gg1(x4)
member1_2_out_gg2(x1, x2)  =  member1_2_out_gg
overlap_2_out_gg2(x1, x2)  =  overlap_2_out_gg

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of PROLOG



↳ PROLOG
  ↳ UnrequestedClauseRemoverProof
    ↳ PROLOG
      ↳ PrologToPiTRSProof
        ↳ PiTRS
          ↳ DependencyPairsProof

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

overlap_2_in_gg2(Xs, Ys) -> if_overlap_2_in_1_gg3(Xs, Ys, member2_2_in_ag2(X, Xs))
member2_2_in_ag2(X, ._22(Y, Xs)) -> if_member2_2_in_1_ag4(X, Y, Xs, member2_2_in_ag2(X, Xs))
member2_2_in_ag2(X, ._22(X, Xs)) -> member2_2_out_ag2(X, ._22(X, Xs))
if_member2_2_in_1_ag4(X, Y, Xs, member2_2_out_ag2(X, Xs)) -> member2_2_out_ag2(X, ._22(Y, Xs))
if_overlap_2_in_1_gg3(Xs, Ys, member2_2_out_ag2(X, Xs)) -> if_overlap_2_in_2_gg4(Xs, Ys, X, member1_2_in_gg2(X, Ys))
member1_2_in_gg2(X, ._22(Y, Xs)) -> if_member1_2_in_1_gg4(X, Y, Xs, member1_2_in_gg2(X, Xs))
member1_2_in_gg2(X, ._22(X, Xs)) -> member1_2_out_gg2(X, ._22(X, Xs))
if_member1_2_in_1_gg4(X, Y, Xs, member1_2_out_gg2(X, Xs)) -> member1_2_out_gg2(X, ._22(Y, Xs))
if_overlap_2_in_2_gg4(Xs, Ys, X, member1_2_out_gg2(X, Ys)) -> overlap_2_out_gg2(Xs, Ys)

The argument filtering Pi contains the following mapping:
overlap_2_in_gg2(x1, x2)  =  overlap_2_in_gg2(x1, x2)
._22(x1, x2)  =  ._22(x1, x2)
if_overlap_2_in_1_gg3(x1, x2, x3)  =  if_overlap_2_in_1_gg2(x2, x3)
member2_2_in_ag2(x1, x2)  =  member2_2_in_ag1(x2)
if_member2_2_in_1_ag4(x1, x2, x3, x4)  =  if_member2_2_in_1_ag1(x4)
member2_2_out_ag2(x1, x2)  =  member2_2_out_ag1(x1)
if_overlap_2_in_2_gg4(x1, x2, x3, x4)  =  if_overlap_2_in_2_gg1(x4)
member1_2_in_gg2(x1, x2)  =  member1_2_in_gg2(x1, x2)
if_member1_2_in_1_gg4(x1, x2, x3, x4)  =  if_member1_2_in_1_gg1(x4)
member1_2_out_gg2(x1, x2)  =  member1_2_out_gg
overlap_2_out_gg2(x1, x2)  =  overlap_2_out_gg


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

OVERLAP_2_IN_GG2(Xs, Ys) -> IF_OVERLAP_2_IN_1_GG3(Xs, Ys, member2_2_in_ag2(X, Xs))
OVERLAP_2_IN_GG2(Xs, Ys) -> MEMBER2_2_IN_AG2(X, Xs)
MEMBER2_2_IN_AG2(X, ._22(Y, Xs)) -> IF_MEMBER2_2_IN_1_AG4(X, Y, Xs, member2_2_in_ag2(X, Xs))
MEMBER2_2_IN_AG2(X, ._22(Y, Xs)) -> MEMBER2_2_IN_AG2(X, Xs)
IF_OVERLAP_2_IN_1_GG3(Xs, Ys, member2_2_out_ag2(X, Xs)) -> IF_OVERLAP_2_IN_2_GG4(Xs, Ys, X, member1_2_in_gg2(X, Ys))
IF_OVERLAP_2_IN_1_GG3(Xs, Ys, member2_2_out_ag2(X, Xs)) -> MEMBER1_2_IN_GG2(X, Ys)
MEMBER1_2_IN_GG2(X, ._22(Y, Xs)) -> IF_MEMBER1_2_IN_1_GG4(X, Y, Xs, member1_2_in_gg2(X, Xs))
MEMBER1_2_IN_GG2(X, ._22(Y, Xs)) -> MEMBER1_2_IN_GG2(X, Xs)

The TRS R consists of the following rules:

overlap_2_in_gg2(Xs, Ys) -> if_overlap_2_in_1_gg3(Xs, Ys, member2_2_in_ag2(X, Xs))
member2_2_in_ag2(X, ._22(Y, Xs)) -> if_member2_2_in_1_ag4(X, Y, Xs, member2_2_in_ag2(X, Xs))
member2_2_in_ag2(X, ._22(X, Xs)) -> member2_2_out_ag2(X, ._22(X, Xs))
if_member2_2_in_1_ag4(X, Y, Xs, member2_2_out_ag2(X, Xs)) -> member2_2_out_ag2(X, ._22(Y, Xs))
if_overlap_2_in_1_gg3(Xs, Ys, member2_2_out_ag2(X, Xs)) -> if_overlap_2_in_2_gg4(Xs, Ys, X, member1_2_in_gg2(X, Ys))
member1_2_in_gg2(X, ._22(Y, Xs)) -> if_member1_2_in_1_gg4(X, Y, Xs, member1_2_in_gg2(X, Xs))
member1_2_in_gg2(X, ._22(X, Xs)) -> member1_2_out_gg2(X, ._22(X, Xs))
if_member1_2_in_1_gg4(X, Y, Xs, member1_2_out_gg2(X, Xs)) -> member1_2_out_gg2(X, ._22(Y, Xs))
if_overlap_2_in_2_gg4(Xs, Ys, X, member1_2_out_gg2(X, Ys)) -> overlap_2_out_gg2(Xs, Ys)

The argument filtering Pi contains the following mapping:
overlap_2_in_gg2(x1, x2)  =  overlap_2_in_gg2(x1, x2)
._22(x1, x2)  =  ._22(x1, x2)
if_overlap_2_in_1_gg3(x1, x2, x3)  =  if_overlap_2_in_1_gg2(x2, x3)
member2_2_in_ag2(x1, x2)  =  member2_2_in_ag1(x2)
if_member2_2_in_1_ag4(x1, x2, x3, x4)  =  if_member2_2_in_1_ag1(x4)
member2_2_out_ag2(x1, x2)  =  member2_2_out_ag1(x1)
if_overlap_2_in_2_gg4(x1, x2, x3, x4)  =  if_overlap_2_in_2_gg1(x4)
member1_2_in_gg2(x1, x2)  =  member1_2_in_gg2(x1, x2)
if_member1_2_in_1_gg4(x1, x2, x3, x4)  =  if_member1_2_in_1_gg1(x4)
member1_2_out_gg2(x1, x2)  =  member1_2_out_gg
overlap_2_out_gg2(x1, x2)  =  overlap_2_out_gg
IF_MEMBER2_2_IN_1_AG4(x1, x2, x3, x4)  =  IF_MEMBER2_2_IN_1_AG1(x4)
IF_OVERLAP_2_IN_1_GG3(x1, x2, x3)  =  IF_OVERLAP_2_IN_1_GG2(x2, x3)
OVERLAP_2_IN_GG2(x1, x2)  =  OVERLAP_2_IN_GG2(x1, x2)
MEMBER1_2_IN_GG2(x1, x2)  =  MEMBER1_2_IN_GG2(x1, x2)
IF_OVERLAP_2_IN_2_GG4(x1, x2, x3, x4)  =  IF_OVERLAP_2_IN_2_GG1(x4)
IF_MEMBER1_2_IN_1_GG4(x1, x2, x3, x4)  =  IF_MEMBER1_2_IN_1_GG1(x4)
MEMBER2_2_IN_AG2(x1, x2)  =  MEMBER2_2_IN_AG1(x2)

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

↳ PROLOG
  ↳ UnrequestedClauseRemoverProof
    ↳ PROLOG
      ↳ PrologToPiTRSProof
        ↳ PiTRS
          ↳ DependencyPairsProof
            ↳ PiDP
              ↳ DependencyGraphProof

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

OVERLAP_2_IN_GG2(Xs, Ys) -> IF_OVERLAP_2_IN_1_GG3(Xs, Ys, member2_2_in_ag2(X, Xs))
OVERLAP_2_IN_GG2(Xs, Ys) -> MEMBER2_2_IN_AG2(X, Xs)
MEMBER2_2_IN_AG2(X, ._22(Y, Xs)) -> IF_MEMBER2_2_IN_1_AG4(X, Y, Xs, member2_2_in_ag2(X, Xs))
MEMBER2_2_IN_AG2(X, ._22(Y, Xs)) -> MEMBER2_2_IN_AG2(X, Xs)
IF_OVERLAP_2_IN_1_GG3(Xs, Ys, member2_2_out_ag2(X, Xs)) -> IF_OVERLAP_2_IN_2_GG4(Xs, Ys, X, member1_2_in_gg2(X, Ys))
IF_OVERLAP_2_IN_1_GG3(Xs, Ys, member2_2_out_ag2(X, Xs)) -> MEMBER1_2_IN_GG2(X, Ys)
MEMBER1_2_IN_GG2(X, ._22(Y, Xs)) -> IF_MEMBER1_2_IN_1_GG4(X, Y, Xs, member1_2_in_gg2(X, Xs))
MEMBER1_2_IN_GG2(X, ._22(Y, Xs)) -> MEMBER1_2_IN_GG2(X, Xs)

The TRS R consists of the following rules:

overlap_2_in_gg2(Xs, Ys) -> if_overlap_2_in_1_gg3(Xs, Ys, member2_2_in_ag2(X, Xs))
member2_2_in_ag2(X, ._22(Y, Xs)) -> if_member2_2_in_1_ag4(X, Y, Xs, member2_2_in_ag2(X, Xs))
member2_2_in_ag2(X, ._22(X, Xs)) -> member2_2_out_ag2(X, ._22(X, Xs))
if_member2_2_in_1_ag4(X, Y, Xs, member2_2_out_ag2(X, Xs)) -> member2_2_out_ag2(X, ._22(Y, Xs))
if_overlap_2_in_1_gg3(Xs, Ys, member2_2_out_ag2(X, Xs)) -> if_overlap_2_in_2_gg4(Xs, Ys, X, member1_2_in_gg2(X, Ys))
member1_2_in_gg2(X, ._22(Y, Xs)) -> if_member1_2_in_1_gg4(X, Y, Xs, member1_2_in_gg2(X, Xs))
member1_2_in_gg2(X, ._22(X, Xs)) -> member1_2_out_gg2(X, ._22(X, Xs))
if_member1_2_in_1_gg4(X, Y, Xs, member1_2_out_gg2(X, Xs)) -> member1_2_out_gg2(X, ._22(Y, Xs))
if_overlap_2_in_2_gg4(Xs, Ys, X, member1_2_out_gg2(X, Ys)) -> overlap_2_out_gg2(Xs, Ys)

The argument filtering Pi contains the following mapping:
overlap_2_in_gg2(x1, x2)  =  overlap_2_in_gg2(x1, x2)
._22(x1, x2)  =  ._22(x1, x2)
if_overlap_2_in_1_gg3(x1, x2, x3)  =  if_overlap_2_in_1_gg2(x2, x3)
member2_2_in_ag2(x1, x2)  =  member2_2_in_ag1(x2)
if_member2_2_in_1_ag4(x1, x2, x3, x4)  =  if_member2_2_in_1_ag1(x4)
member2_2_out_ag2(x1, x2)  =  member2_2_out_ag1(x1)
if_overlap_2_in_2_gg4(x1, x2, x3, x4)  =  if_overlap_2_in_2_gg1(x4)
member1_2_in_gg2(x1, x2)  =  member1_2_in_gg2(x1, x2)
if_member1_2_in_1_gg4(x1, x2, x3, x4)  =  if_member1_2_in_1_gg1(x4)
member1_2_out_gg2(x1, x2)  =  member1_2_out_gg
overlap_2_out_gg2(x1, x2)  =  overlap_2_out_gg
IF_MEMBER2_2_IN_1_AG4(x1, x2, x3, x4)  =  IF_MEMBER2_2_IN_1_AG1(x4)
IF_OVERLAP_2_IN_1_GG3(x1, x2, x3)  =  IF_OVERLAP_2_IN_1_GG2(x2, x3)
OVERLAP_2_IN_GG2(x1, x2)  =  OVERLAP_2_IN_GG2(x1, x2)
MEMBER1_2_IN_GG2(x1, x2)  =  MEMBER1_2_IN_GG2(x1, x2)
IF_OVERLAP_2_IN_2_GG4(x1, x2, x3, x4)  =  IF_OVERLAP_2_IN_2_GG1(x4)
IF_MEMBER1_2_IN_1_GG4(x1, x2, x3, x4)  =  IF_MEMBER1_2_IN_1_GG1(x4)
MEMBER2_2_IN_AG2(x1, x2)  =  MEMBER2_2_IN_AG1(x2)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph contains 2 SCCs with 6 less nodes.

↳ PROLOG
  ↳ UnrequestedClauseRemoverProof
    ↳ PROLOG
      ↳ PrologToPiTRSProof
        ↳ PiTRS
          ↳ DependencyPairsProof
            ↳ PiDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ PiDP
                    ↳ UsableRulesProof
                  ↳ PiDP

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

MEMBER1_2_IN_GG2(X, ._22(Y, Xs)) -> MEMBER1_2_IN_GG2(X, Xs)

The TRS R consists of the following rules:

overlap_2_in_gg2(Xs, Ys) -> if_overlap_2_in_1_gg3(Xs, Ys, member2_2_in_ag2(X, Xs))
member2_2_in_ag2(X, ._22(Y, Xs)) -> if_member2_2_in_1_ag4(X, Y, Xs, member2_2_in_ag2(X, Xs))
member2_2_in_ag2(X, ._22(X, Xs)) -> member2_2_out_ag2(X, ._22(X, Xs))
if_member2_2_in_1_ag4(X, Y, Xs, member2_2_out_ag2(X, Xs)) -> member2_2_out_ag2(X, ._22(Y, Xs))
if_overlap_2_in_1_gg3(Xs, Ys, member2_2_out_ag2(X, Xs)) -> if_overlap_2_in_2_gg4(Xs, Ys, X, member1_2_in_gg2(X, Ys))
member1_2_in_gg2(X, ._22(Y, Xs)) -> if_member1_2_in_1_gg4(X, Y, Xs, member1_2_in_gg2(X, Xs))
member1_2_in_gg2(X, ._22(X, Xs)) -> member1_2_out_gg2(X, ._22(X, Xs))
if_member1_2_in_1_gg4(X, Y, Xs, member1_2_out_gg2(X, Xs)) -> member1_2_out_gg2(X, ._22(Y, Xs))
if_overlap_2_in_2_gg4(Xs, Ys, X, member1_2_out_gg2(X, Ys)) -> overlap_2_out_gg2(Xs, Ys)

The argument filtering Pi contains the following mapping:
overlap_2_in_gg2(x1, x2)  =  overlap_2_in_gg2(x1, x2)
._22(x1, x2)  =  ._22(x1, x2)
if_overlap_2_in_1_gg3(x1, x2, x3)  =  if_overlap_2_in_1_gg2(x2, x3)
member2_2_in_ag2(x1, x2)  =  member2_2_in_ag1(x2)
if_member2_2_in_1_ag4(x1, x2, x3, x4)  =  if_member2_2_in_1_ag1(x4)
member2_2_out_ag2(x1, x2)  =  member2_2_out_ag1(x1)
if_overlap_2_in_2_gg4(x1, x2, x3, x4)  =  if_overlap_2_in_2_gg1(x4)
member1_2_in_gg2(x1, x2)  =  member1_2_in_gg2(x1, x2)
if_member1_2_in_1_gg4(x1, x2, x3, x4)  =  if_member1_2_in_1_gg1(x4)
member1_2_out_gg2(x1, x2)  =  member1_2_out_gg
overlap_2_out_gg2(x1, x2)  =  overlap_2_out_gg
MEMBER1_2_IN_GG2(x1, x2)  =  MEMBER1_2_IN_GG2(x1, x2)

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

↳ PROLOG
  ↳ UnrequestedClauseRemoverProof
    ↳ PROLOG
      ↳ PrologToPiTRSProof
        ↳ PiTRS
          ↳ DependencyPairsProof
            ↳ PiDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ PiDP
                    ↳ UsableRulesProof
                      ↳ PiDP
                        ↳ PiDPToQDPProof
                  ↳ PiDP

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

MEMBER1_2_IN_GG2(X, ._22(Y, Xs)) -> MEMBER1_2_IN_GG2(X, Xs)

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
  ↳ UnrequestedClauseRemoverProof
    ↳ PROLOG
      ↳ PrologToPiTRSProof
        ↳ PiTRS
          ↳ DependencyPairsProof
            ↳ PiDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ PiDP
                    ↳ UsableRulesProof
                      ↳ PiDP
                        ↳ PiDPToQDPProof
                          ↳ QDP
                            ↳ QDPSizeChangeProof
                  ↳ PiDP

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

MEMBER1_2_IN_GG2(X, ._22(Y, Xs)) -> MEMBER1_2_IN_GG2(X, Xs)

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_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: 



↳ PROLOG
  ↳ UnrequestedClauseRemoverProof
    ↳ PROLOG
      ↳ PrologToPiTRSProof
        ↳ PiTRS
          ↳ DependencyPairsProof
            ↳ PiDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ PiDP
                  ↳ PiDP
                    ↳ UsableRulesProof

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

MEMBER2_2_IN_AG2(X, ._22(Y, Xs)) -> MEMBER2_2_IN_AG2(X, Xs)

The TRS R consists of the following rules:

overlap_2_in_gg2(Xs, Ys) -> if_overlap_2_in_1_gg3(Xs, Ys, member2_2_in_ag2(X, Xs))
member2_2_in_ag2(X, ._22(Y, Xs)) -> if_member2_2_in_1_ag4(X, Y, Xs, member2_2_in_ag2(X, Xs))
member2_2_in_ag2(X, ._22(X, Xs)) -> member2_2_out_ag2(X, ._22(X, Xs))
if_member2_2_in_1_ag4(X, Y, Xs, member2_2_out_ag2(X, Xs)) -> member2_2_out_ag2(X, ._22(Y, Xs))
if_overlap_2_in_1_gg3(Xs, Ys, member2_2_out_ag2(X, Xs)) -> if_overlap_2_in_2_gg4(Xs, Ys, X, member1_2_in_gg2(X, Ys))
member1_2_in_gg2(X, ._22(Y, Xs)) -> if_member1_2_in_1_gg4(X, Y, Xs, member1_2_in_gg2(X, Xs))
member1_2_in_gg2(X, ._22(X, Xs)) -> member1_2_out_gg2(X, ._22(X, Xs))
if_member1_2_in_1_gg4(X, Y, Xs, member1_2_out_gg2(X, Xs)) -> member1_2_out_gg2(X, ._22(Y, Xs))
if_overlap_2_in_2_gg4(Xs, Ys, X, member1_2_out_gg2(X, Ys)) -> overlap_2_out_gg2(Xs, Ys)

The argument filtering Pi contains the following mapping:
overlap_2_in_gg2(x1, x2)  =  overlap_2_in_gg2(x1, x2)
._22(x1, x2)  =  ._22(x1, x2)
if_overlap_2_in_1_gg3(x1, x2, x3)  =  if_overlap_2_in_1_gg2(x2, x3)
member2_2_in_ag2(x1, x2)  =  member2_2_in_ag1(x2)
if_member2_2_in_1_ag4(x1, x2, x3, x4)  =  if_member2_2_in_1_ag1(x4)
member2_2_out_ag2(x1, x2)  =  member2_2_out_ag1(x1)
if_overlap_2_in_2_gg4(x1, x2, x3, x4)  =  if_overlap_2_in_2_gg1(x4)
member1_2_in_gg2(x1, x2)  =  member1_2_in_gg2(x1, x2)
if_member1_2_in_1_gg4(x1, x2, x3, x4)  =  if_member1_2_in_1_gg1(x4)
member1_2_out_gg2(x1, x2)  =  member1_2_out_gg
overlap_2_out_gg2(x1, x2)  =  overlap_2_out_gg
MEMBER2_2_IN_AG2(x1, x2)  =  MEMBER2_2_IN_AG1(x2)

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

↳ PROLOG
  ↳ UnrequestedClauseRemoverProof
    ↳ PROLOG
      ↳ PrologToPiTRSProof
        ↳ PiTRS
          ↳ DependencyPairsProof
            ↳ PiDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ PiDP
                  ↳ PiDP
                    ↳ UsableRulesProof
                      ↳ PiDP
                        ↳ PiDPToQDPProof

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

MEMBER2_2_IN_AG2(X, ._22(Y, Xs)) -> MEMBER2_2_IN_AG2(X, Xs)

R is empty.
The argument filtering Pi contains the following mapping:
._22(x1, x2)  =  ._22(x1, x2)
MEMBER2_2_IN_AG2(x1, x2)  =  MEMBER2_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
  ↳ UnrequestedClauseRemoverProof
    ↳ PROLOG
      ↳ PrologToPiTRSProof
        ↳ PiTRS
          ↳ DependencyPairsProof
            ↳ PiDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ PiDP
                  ↳ PiDP
                    ↳ UsableRulesProof
                      ↳ PiDP
                        ↳ PiDPToQDPProof
                          ↳ QDP
                            ↳ QDPSizeChangeProof

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

MEMBER2_2_IN_AG1(._22(Y, Xs)) -> MEMBER2_2_IN_AG1(Xs)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {MEMBER2_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: