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



PROLOG
  ↳ UnrequestedClauseRemoverProof

member2(X, .2(Y, Xs)) :- member2(X, Xs).
member2(X, .2(X, Xs)).
subset2(.2(X, Xs), Ys) :- member2(X, Ys), subset2(Xs, Ys).
subset2({}0, Ys).
member12(X, .2(Y, Xs)) :- member12(X, Xs).
member12(X, .2(X, Xs)).
subset12(.2(X, Xs), Ys) :- member12(X, Ys), subset12(Xs, Ys).
subset12({}0, Ys).


The clauses

member12(X, .2(Y, Xs)) :- member12(X, Xs).
member12(X, .2(X, Xs)).
subset12(.2(X, Xs), Ys) :- member12(X, Ys), subset12(Xs, Ys).
subset12({}0, Ys).

can be ignored, as they are not needed by any of the given querys.

Deleting these clauses results in the following prolog program:

member2(X, .2(Y, Xs)) :- member2(X, Xs).
member2(X, .2(X, Xs)).
subset2(.2(X, Xs), Ys) :- member2(X, Ys), subset2(Xs, Ys).
subset2({}0, Ys).



↳ PROLOG
  ↳ UnrequestedClauseRemoverProof
PROLOG
      ↳ PrologToPiTRSProof

member2(X, .2(Y, Xs)) :- member2(X, Xs).
member2(X, .2(X, Xs)).
subset2(.2(X, Xs), Ys) :- member2(X, Ys), subset2(Xs, Ys).
subset2({}0, Ys).


With regard to the inferred argument filtering the predicates were used in the following modes:
subset2: (b,b)
member2: (b,b)
Transforming PROLOG into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:


subset_2_in_gg2(._22(X, Xs), Ys) -> if_subset_2_in_1_gg4(X, Xs, Ys, member_2_in_gg2(X, Ys))
member_2_in_gg2(X, ._22(Y, Xs)) -> if_member_2_in_1_gg4(X, Y, Xs, member_2_in_gg2(X, Xs))
member_2_in_gg2(X, ._22(X, Xs)) -> member_2_out_gg2(X, ._22(X, Xs))
if_member_2_in_1_gg4(X, Y, Xs, member_2_out_gg2(X, Xs)) -> member_2_out_gg2(X, ._22(Y, Xs))
if_subset_2_in_1_gg4(X, Xs, Ys, member_2_out_gg2(X, Ys)) -> if_subset_2_in_2_gg4(X, Xs, Ys, subset_2_in_gg2(Xs, Ys))
subset_2_in_gg2([]_0, Ys) -> subset_2_out_gg2([]_0, Ys)
if_subset_2_in_2_gg4(X, Xs, Ys, subset_2_out_gg2(Xs, Ys)) -> subset_2_out_gg2(._22(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset_2_in_gg2(x1, x2)  =  subset_2_in_gg2(x1, x2)
._22(x1, x2)  =  ._22(x1, x2)
[]_0  =  []_0
if_subset_2_in_1_gg4(x1, x2, x3, x4)  =  if_subset_2_in_1_gg3(x2, x3, x4)
member_2_in_gg2(x1, x2)  =  member_2_in_gg2(x1, x2)
if_member_2_in_1_gg4(x1, x2, x3, x4)  =  if_member_2_in_1_gg1(x4)
member_2_out_gg2(x1, x2)  =  member_2_out_gg
if_subset_2_in_2_gg4(x1, x2, x3, x4)  =  if_subset_2_in_2_gg1(x4)
subset_2_out_gg2(x1, x2)  =  subset_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:

subset_2_in_gg2(._22(X, Xs), Ys) -> if_subset_2_in_1_gg4(X, Xs, Ys, member_2_in_gg2(X, Ys))
member_2_in_gg2(X, ._22(Y, Xs)) -> if_member_2_in_1_gg4(X, Y, Xs, member_2_in_gg2(X, Xs))
member_2_in_gg2(X, ._22(X, Xs)) -> member_2_out_gg2(X, ._22(X, Xs))
if_member_2_in_1_gg4(X, Y, Xs, member_2_out_gg2(X, Xs)) -> member_2_out_gg2(X, ._22(Y, Xs))
if_subset_2_in_1_gg4(X, Xs, Ys, member_2_out_gg2(X, Ys)) -> if_subset_2_in_2_gg4(X, Xs, Ys, subset_2_in_gg2(Xs, Ys))
subset_2_in_gg2([]_0, Ys) -> subset_2_out_gg2([]_0, Ys)
if_subset_2_in_2_gg4(X, Xs, Ys, subset_2_out_gg2(Xs, Ys)) -> subset_2_out_gg2(._22(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset_2_in_gg2(x1, x2)  =  subset_2_in_gg2(x1, x2)
._22(x1, x2)  =  ._22(x1, x2)
[]_0  =  []_0
if_subset_2_in_1_gg4(x1, x2, x3, x4)  =  if_subset_2_in_1_gg3(x2, x3, x4)
member_2_in_gg2(x1, x2)  =  member_2_in_gg2(x1, x2)
if_member_2_in_1_gg4(x1, x2, x3, x4)  =  if_member_2_in_1_gg1(x4)
member_2_out_gg2(x1, x2)  =  member_2_out_gg
if_subset_2_in_2_gg4(x1, x2, x3, x4)  =  if_subset_2_in_2_gg1(x4)
subset_2_out_gg2(x1, x2)  =  subset_2_out_gg


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

SUBSET_2_IN_GG2(._22(X, Xs), Ys) -> IF_SUBSET_2_IN_1_GG4(X, Xs, Ys, member_2_in_gg2(X, Ys))
SUBSET_2_IN_GG2(._22(X, Xs), Ys) -> MEMBER_2_IN_GG2(X, Ys)
MEMBER_2_IN_GG2(X, ._22(Y, Xs)) -> IF_MEMBER_2_IN_1_GG4(X, Y, Xs, member_2_in_gg2(X, Xs))
MEMBER_2_IN_GG2(X, ._22(Y, Xs)) -> MEMBER_2_IN_GG2(X, Xs)
IF_SUBSET_2_IN_1_GG4(X, Xs, Ys, member_2_out_gg2(X, Ys)) -> IF_SUBSET_2_IN_2_GG4(X, Xs, Ys, subset_2_in_gg2(Xs, Ys))
IF_SUBSET_2_IN_1_GG4(X, Xs, Ys, member_2_out_gg2(X, Ys)) -> SUBSET_2_IN_GG2(Xs, Ys)

The TRS R consists of the following rules:

subset_2_in_gg2(._22(X, Xs), Ys) -> if_subset_2_in_1_gg4(X, Xs, Ys, member_2_in_gg2(X, Ys))
member_2_in_gg2(X, ._22(Y, Xs)) -> if_member_2_in_1_gg4(X, Y, Xs, member_2_in_gg2(X, Xs))
member_2_in_gg2(X, ._22(X, Xs)) -> member_2_out_gg2(X, ._22(X, Xs))
if_member_2_in_1_gg4(X, Y, Xs, member_2_out_gg2(X, Xs)) -> member_2_out_gg2(X, ._22(Y, Xs))
if_subset_2_in_1_gg4(X, Xs, Ys, member_2_out_gg2(X, Ys)) -> if_subset_2_in_2_gg4(X, Xs, Ys, subset_2_in_gg2(Xs, Ys))
subset_2_in_gg2([]_0, Ys) -> subset_2_out_gg2([]_0, Ys)
if_subset_2_in_2_gg4(X, Xs, Ys, subset_2_out_gg2(Xs, Ys)) -> subset_2_out_gg2(._22(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset_2_in_gg2(x1, x2)  =  subset_2_in_gg2(x1, x2)
._22(x1, x2)  =  ._22(x1, x2)
[]_0  =  []_0
if_subset_2_in_1_gg4(x1, x2, x3, x4)  =  if_subset_2_in_1_gg3(x2, x3, x4)
member_2_in_gg2(x1, x2)  =  member_2_in_gg2(x1, x2)
if_member_2_in_1_gg4(x1, x2, x3, x4)  =  if_member_2_in_1_gg1(x4)
member_2_out_gg2(x1, x2)  =  member_2_out_gg
if_subset_2_in_2_gg4(x1, x2, x3, x4)  =  if_subset_2_in_2_gg1(x4)
subset_2_out_gg2(x1, x2)  =  subset_2_out_gg
IF_MEMBER_2_IN_1_GG4(x1, x2, x3, x4)  =  IF_MEMBER_2_IN_1_GG1(x4)
IF_SUBSET_2_IN_1_GG4(x1, x2, x3, x4)  =  IF_SUBSET_2_IN_1_GG3(x2, x3, x4)
MEMBER_2_IN_GG2(x1, x2)  =  MEMBER_2_IN_GG2(x1, x2)
SUBSET_2_IN_GG2(x1, x2)  =  SUBSET_2_IN_GG2(x1, x2)
IF_SUBSET_2_IN_2_GG4(x1, x2, x3, x4)  =  IF_SUBSET_2_IN_2_GG1(x4)

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:

SUBSET_2_IN_GG2(._22(X, Xs), Ys) -> IF_SUBSET_2_IN_1_GG4(X, Xs, Ys, member_2_in_gg2(X, Ys))
SUBSET_2_IN_GG2(._22(X, Xs), Ys) -> MEMBER_2_IN_GG2(X, Ys)
MEMBER_2_IN_GG2(X, ._22(Y, Xs)) -> IF_MEMBER_2_IN_1_GG4(X, Y, Xs, member_2_in_gg2(X, Xs))
MEMBER_2_IN_GG2(X, ._22(Y, Xs)) -> MEMBER_2_IN_GG2(X, Xs)
IF_SUBSET_2_IN_1_GG4(X, Xs, Ys, member_2_out_gg2(X, Ys)) -> IF_SUBSET_2_IN_2_GG4(X, Xs, Ys, subset_2_in_gg2(Xs, Ys))
IF_SUBSET_2_IN_1_GG4(X, Xs, Ys, member_2_out_gg2(X, Ys)) -> SUBSET_2_IN_GG2(Xs, Ys)

The TRS R consists of the following rules:

subset_2_in_gg2(._22(X, Xs), Ys) -> if_subset_2_in_1_gg4(X, Xs, Ys, member_2_in_gg2(X, Ys))
member_2_in_gg2(X, ._22(Y, Xs)) -> if_member_2_in_1_gg4(X, Y, Xs, member_2_in_gg2(X, Xs))
member_2_in_gg2(X, ._22(X, Xs)) -> member_2_out_gg2(X, ._22(X, Xs))
if_member_2_in_1_gg4(X, Y, Xs, member_2_out_gg2(X, Xs)) -> member_2_out_gg2(X, ._22(Y, Xs))
if_subset_2_in_1_gg4(X, Xs, Ys, member_2_out_gg2(X, Ys)) -> if_subset_2_in_2_gg4(X, Xs, Ys, subset_2_in_gg2(Xs, Ys))
subset_2_in_gg2([]_0, Ys) -> subset_2_out_gg2([]_0, Ys)
if_subset_2_in_2_gg4(X, Xs, Ys, subset_2_out_gg2(Xs, Ys)) -> subset_2_out_gg2(._22(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset_2_in_gg2(x1, x2)  =  subset_2_in_gg2(x1, x2)
._22(x1, x2)  =  ._22(x1, x2)
[]_0  =  []_0
if_subset_2_in_1_gg4(x1, x2, x3, x4)  =  if_subset_2_in_1_gg3(x2, x3, x4)
member_2_in_gg2(x1, x2)  =  member_2_in_gg2(x1, x2)
if_member_2_in_1_gg4(x1, x2, x3, x4)  =  if_member_2_in_1_gg1(x4)
member_2_out_gg2(x1, x2)  =  member_2_out_gg
if_subset_2_in_2_gg4(x1, x2, x3, x4)  =  if_subset_2_in_2_gg1(x4)
subset_2_out_gg2(x1, x2)  =  subset_2_out_gg
IF_MEMBER_2_IN_1_GG4(x1, x2, x3, x4)  =  IF_MEMBER_2_IN_1_GG1(x4)
IF_SUBSET_2_IN_1_GG4(x1, x2, x3, x4)  =  IF_SUBSET_2_IN_1_GG3(x2, x3, x4)
MEMBER_2_IN_GG2(x1, x2)  =  MEMBER_2_IN_GG2(x1, x2)
SUBSET_2_IN_GG2(x1, x2)  =  SUBSET_2_IN_GG2(x1, x2)
IF_SUBSET_2_IN_2_GG4(x1, x2, x3, x4)  =  IF_SUBSET_2_IN_2_GG1(x4)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph contains 2 SCCs with 3 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:

MEMBER_2_IN_GG2(X, ._22(Y, Xs)) -> MEMBER_2_IN_GG2(X, Xs)

The TRS R consists of the following rules:

subset_2_in_gg2(._22(X, Xs), Ys) -> if_subset_2_in_1_gg4(X, Xs, Ys, member_2_in_gg2(X, Ys))
member_2_in_gg2(X, ._22(Y, Xs)) -> if_member_2_in_1_gg4(X, Y, Xs, member_2_in_gg2(X, Xs))
member_2_in_gg2(X, ._22(X, Xs)) -> member_2_out_gg2(X, ._22(X, Xs))
if_member_2_in_1_gg4(X, Y, Xs, member_2_out_gg2(X, Xs)) -> member_2_out_gg2(X, ._22(Y, Xs))
if_subset_2_in_1_gg4(X, Xs, Ys, member_2_out_gg2(X, Ys)) -> if_subset_2_in_2_gg4(X, Xs, Ys, subset_2_in_gg2(Xs, Ys))
subset_2_in_gg2([]_0, Ys) -> subset_2_out_gg2([]_0, Ys)
if_subset_2_in_2_gg4(X, Xs, Ys, subset_2_out_gg2(Xs, Ys)) -> subset_2_out_gg2(._22(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset_2_in_gg2(x1, x2)  =  subset_2_in_gg2(x1, x2)
._22(x1, x2)  =  ._22(x1, x2)
[]_0  =  []_0
if_subset_2_in_1_gg4(x1, x2, x3, x4)  =  if_subset_2_in_1_gg3(x2, x3, x4)
member_2_in_gg2(x1, x2)  =  member_2_in_gg2(x1, x2)
if_member_2_in_1_gg4(x1, x2, x3, x4)  =  if_member_2_in_1_gg1(x4)
member_2_out_gg2(x1, x2)  =  member_2_out_gg
if_subset_2_in_2_gg4(x1, x2, x3, x4)  =  if_subset_2_in_2_gg1(x4)
subset_2_out_gg2(x1, x2)  =  subset_2_out_gg
MEMBER_2_IN_GG2(x1, x2)  =  MEMBER_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:

MEMBER_2_IN_GG2(X, ._22(Y, Xs)) -> MEMBER_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:

MEMBER_2_IN_GG2(X, ._22(Y, Xs)) -> MEMBER_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 {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:



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

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

IF_SUBSET_2_IN_1_GG4(X, Xs, Ys, member_2_out_gg2(X, Ys)) -> SUBSET_2_IN_GG2(Xs, Ys)
SUBSET_2_IN_GG2(._22(X, Xs), Ys) -> IF_SUBSET_2_IN_1_GG4(X, Xs, Ys, member_2_in_gg2(X, Ys))

The TRS R consists of the following rules:

subset_2_in_gg2(._22(X, Xs), Ys) -> if_subset_2_in_1_gg4(X, Xs, Ys, member_2_in_gg2(X, Ys))
member_2_in_gg2(X, ._22(Y, Xs)) -> if_member_2_in_1_gg4(X, Y, Xs, member_2_in_gg2(X, Xs))
member_2_in_gg2(X, ._22(X, Xs)) -> member_2_out_gg2(X, ._22(X, Xs))
if_member_2_in_1_gg4(X, Y, Xs, member_2_out_gg2(X, Xs)) -> member_2_out_gg2(X, ._22(Y, Xs))
if_subset_2_in_1_gg4(X, Xs, Ys, member_2_out_gg2(X, Ys)) -> if_subset_2_in_2_gg4(X, Xs, Ys, subset_2_in_gg2(Xs, Ys))
subset_2_in_gg2([]_0, Ys) -> subset_2_out_gg2([]_0, Ys)
if_subset_2_in_2_gg4(X, Xs, Ys, subset_2_out_gg2(Xs, Ys)) -> subset_2_out_gg2(._22(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset_2_in_gg2(x1, x2)  =  subset_2_in_gg2(x1, x2)
._22(x1, x2)  =  ._22(x1, x2)
[]_0  =  []_0
if_subset_2_in_1_gg4(x1, x2, x3, x4)  =  if_subset_2_in_1_gg3(x2, x3, x4)
member_2_in_gg2(x1, x2)  =  member_2_in_gg2(x1, x2)
if_member_2_in_1_gg4(x1, x2, x3, x4)  =  if_member_2_in_1_gg1(x4)
member_2_out_gg2(x1, x2)  =  member_2_out_gg
if_subset_2_in_2_gg4(x1, x2, x3, x4)  =  if_subset_2_in_2_gg1(x4)
subset_2_out_gg2(x1, x2)  =  subset_2_out_gg
IF_SUBSET_2_IN_1_GG4(x1, x2, x3, x4)  =  IF_SUBSET_2_IN_1_GG3(x2, x3, x4)
SUBSET_2_IN_GG2(x1, x2)  =  SUBSET_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
                  ↳ PiDP
                    ↳ UsableRulesProof
PiDP
                        ↳ PiDPToQDPProof

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

IF_SUBSET_2_IN_1_GG4(X, Xs, Ys, member_2_out_gg2(X, Ys)) -> SUBSET_2_IN_GG2(Xs, Ys)
SUBSET_2_IN_GG2(._22(X, Xs), Ys) -> IF_SUBSET_2_IN_1_GG4(X, Xs, Ys, member_2_in_gg2(X, Ys))

The TRS R consists of the following rules:

member_2_in_gg2(X, ._22(Y, Xs)) -> if_member_2_in_1_gg4(X, Y, Xs, member_2_in_gg2(X, Xs))
member_2_in_gg2(X, ._22(X, Xs)) -> member_2_out_gg2(X, ._22(X, Xs))
if_member_2_in_1_gg4(X, Y, Xs, member_2_out_gg2(X, Xs)) -> member_2_out_gg2(X, ._22(Y, Xs))

The argument filtering Pi contains the following mapping:
._22(x1, x2)  =  ._22(x1, x2)
member_2_in_gg2(x1, x2)  =  member_2_in_gg2(x1, x2)
if_member_2_in_1_gg4(x1, x2, x3, x4)  =  if_member_2_in_1_gg1(x4)
member_2_out_gg2(x1, x2)  =  member_2_out_gg
IF_SUBSET_2_IN_1_GG4(x1, x2, x3, x4)  =  IF_SUBSET_2_IN_1_GG3(x2, x3, x4)
SUBSET_2_IN_GG2(x1, x2)  =  SUBSET_2_IN_GG2(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
  ↳ 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:

IF_SUBSET_2_IN_1_GG3(Xs, Ys, member_2_out_gg) -> SUBSET_2_IN_GG2(Xs, Ys)
SUBSET_2_IN_GG2(._22(X, Xs), Ys) -> IF_SUBSET_2_IN_1_GG3(Xs, Ys, member_2_in_gg2(X, Ys))

The TRS R consists of the following rules:

member_2_in_gg2(X, ._22(Y, Xs)) -> if_member_2_in_1_gg1(member_2_in_gg2(X, Xs))
member_2_in_gg2(X, ._22(X, Xs)) -> member_2_out_gg
if_member_2_in_1_gg1(member_2_out_gg) -> member_2_out_gg

The set Q consists of the following terms:

member_2_in_gg2(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 {SUBSET_2_IN_GG2, IF_SUBSET_2_IN_1_GG3}.
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: