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



↳ PROLOG
  ↳ PrologToPiTRSProof
  ↳ 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: (f,b)
member2: (f,b)
Transforming PROLOG into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:


subset_2_in_ag2(._22(X, Xs), Ys) -> if_subset_2_in_1_ag4(X, Xs, Ys, member_2_in_ag2(X, Ys))
member_2_in_ag2(X, ._22(Y, Xs)) -> if_member_2_in_1_ag4(X, Y, Xs, member_2_in_ag2(X, Xs))
member_2_in_ag2(X, ._22(X, Xs)) -> member_2_out_ag2(X, ._22(X, Xs))
if_member_2_in_1_ag4(X, Y, Xs, member_2_out_ag2(X, Xs)) -> member_2_out_ag2(X, ._22(Y, Xs))
if_subset_2_in_1_ag4(X, Xs, Ys, member_2_out_ag2(X, Ys)) -> if_subset_2_in_2_ag4(X, Xs, Ys, subset_2_in_ag2(Xs, Ys))
subset_2_in_ag2([]_0, Ys) -> subset_2_out_ag2([]_0, Ys)
if_subset_2_in_2_ag4(X, Xs, Ys, subset_2_out_ag2(Xs, Ys)) -> subset_2_out_ag2(._22(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset_2_in_ag2(x1, x2)  =  subset_2_in_ag1(x2)
._22(x1, x2)  =  ._22(x1, x2)
[]_0  =  []_0
if_subset_2_in_1_ag4(x1, x2, x3, x4)  =  if_subset_2_in_1_ag2(x3, x4)
member_2_in_ag2(x1, x2)  =  member_2_in_ag1(x2)
if_member_2_in_1_ag4(x1, x2, x3, x4)  =  if_member_2_in_1_ag1(x4)
member_2_out_ag2(x1, x2)  =  member_2_out_ag1(x1)
if_subset_2_in_2_ag4(x1, x2, x3, x4)  =  if_subset_2_in_2_ag2(x1, x4)
subset_2_out_ag2(x1, x2)  =  subset_2_out_ag1(x1)

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:

subset_2_in_ag2(._22(X, Xs), Ys) -> if_subset_2_in_1_ag4(X, Xs, Ys, member_2_in_ag2(X, Ys))
member_2_in_ag2(X, ._22(Y, Xs)) -> if_member_2_in_1_ag4(X, Y, Xs, member_2_in_ag2(X, Xs))
member_2_in_ag2(X, ._22(X, Xs)) -> member_2_out_ag2(X, ._22(X, Xs))
if_member_2_in_1_ag4(X, Y, Xs, member_2_out_ag2(X, Xs)) -> member_2_out_ag2(X, ._22(Y, Xs))
if_subset_2_in_1_ag4(X, Xs, Ys, member_2_out_ag2(X, Ys)) -> if_subset_2_in_2_ag4(X, Xs, Ys, subset_2_in_ag2(Xs, Ys))
subset_2_in_ag2([]_0, Ys) -> subset_2_out_ag2([]_0, Ys)
if_subset_2_in_2_ag4(X, Xs, Ys, subset_2_out_ag2(Xs, Ys)) -> subset_2_out_ag2(._22(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset_2_in_ag2(x1, x2)  =  subset_2_in_ag1(x2)
._22(x1, x2)  =  ._22(x1, x2)
[]_0  =  []_0
if_subset_2_in_1_ag4(x1, x2, x3, x4)  =  if_subset_2_in_1_ag2(x3, x4)
member_2_in_ag2(x1, x2)  =  member_2_in_ag1(x2)
if_member_2_in_1_ag4(x1, x2, x3, x4)  =  if_member_2_in_1_ag1(x4)
member_2_out_ag2(x1, x2)  =  member_2_out_ag1(x1)
if_subset_2_in_2_ag4(x1, x2, x3, x4)  =  if_subset_2_in_2_ag2(x1, x4)
subset_2_out_ag2(x1, x2)  =  subset_2_out_ag1(x1)


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

SUBSET_2_IN_AG2(._22(X, Xs), Ys) -> IF_SUBSET_2_IN_1_AG4(X, Xs, Ys, member_2_in_ag2(X, Ys))
SUBSET_2_IN_AG2(._22(X, Xs), Ys) -> MEMBER_2_IN_AG2(X, Ys)
MEMBER_2_IN_AG2(X, ._22(Y, Xs)) -> IF_MEMBER_2_IN_1_AG4(X, Y, Xs, member_2_in_ag2(X, Xs))
MEMBER_2_IN_AG2(X, ._22(Y, Xs)) -> MEMBER_2_IN_AG2(X, Xs)
IF_SUBSET_2_IN_1_AG4(X, Xs, Ys, member_2_out_ag2(X, Ys)) -> IF_SUBSET_2_IN_2_AG4(X, Xs, Ys, subset_2_in_ag2(Xs, Ys))
IF_SUBSET_2_IN_1_AG4(X, Xs, Ys, member_2_out_ag2(X, Ys)) -> SUBSET_2_IN_AG2(Xs, Ys)

The TRS R consists of the following rules:

subset_2_in_ag2(._22(X, Xs), Ys) -> if_subset_2_in_1_ag4(X, Xs, Ys, member_2_in_ag2(X, Ys))
member_2_in_ag2(X, ._22(Y, Xs)) -> if_member_2_in_1_ag4(X, Y, Xs, member_2_in_ag2(X, Xs))
member_2_in_ag2(X, ._22(X, Xs)) -> member_2_out_ag2(X, ._22(X, Xs))
if_member_2_in_1_ag4(X, Y, Xs, member_2_out_ag2(X, Xs)) -> member_2_out_ag2(X, ._22(Y, Xs))
if_subset_2_in_1_ag4(X, Xs, Ys, member_2_out_ag2(X, Ys)) -> if_subset_2_in_2_ag4(X, Xs, Ys, subset_2_in_ag2(Xs, Ys))
subset_2_in_ag2([]_0, Ys) -> subset_2_out_ag2([]_0, Ys)
if_subset_2_in_2_ag4(X, Xs, Ys, subset_2_out_ag2(Xs, Ys)) -> subset_2_out_ag2(._22(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset_2_in_ag2(x1, x2)  =  subset_2_in_ag1(x2)
._22(x1, x2)  =  ._22(x1, x2)
[]_0  =  []_0
if_subset_2_in_1_ag4(x1, x2, x3, x4)  =  if_subset_2_in_1_ag2(x3, x4)
member_2_in_ag2(x1, x2)  =  member_2_in_ag1(x2)
if_member_2_in_1_ag4(x1, x2, x3, x4)  =  if_member_2_in_1_ag1(x4)
member_2_out_ag2(x1, x2)  =  member_2_out_ag1(x1)
if_subset_2_in_2_ag4(x1, x2, x3, x4)  =  if_subset_2_in_2_ag2(x1, x4)
subset_2_out_ag2(x1, x2)  =  subset_2_out_ag1(x1)
IF_SUBSET_2_IN_2_AG4(x1, x2, x3, x4)  =  IF_SUBSET_2_IN_2_AG2(x1, x4)
IF_MEMBER_2_IN_1_AG4(x1, x2, x3, x4)  =  IF_MEMBER_2_IN_1_AG1(x4)
MEMBER_2_IN_AG2(x1, x2)  =  MEMBER_2_IN_AG1(x2)
IF_SUBSET_2_IN_1_AG4(x1, x2, x3, x4)  =  IF_SUBSET_2_IN_1_AG2(x3, x4)
SUBSET_2_IN_AG2(x1, x2)  =  SUBSET_2_IN_AG1(x2)

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:

SUBSET_2_IN_AG2(._22(X, Xs), Ys) -> IF_SUBSET_2_IN_1_AG4(X, Xs, Ys, member_2_in_ag2(X, Ys))
SUBSET_2_IN_AG2(._22(X, Xs), Ys) -> MEMBER_2_IN_AG2(X, Ys)
MEMBER_2_IN_AG2(X, ._22(Y, Xs)) -> IF_MEMBER_2_IN_1_AG4(X, Y, Xs, member_2_in_ag2(X, Xs))
MEMBER_2_IN_AG2(X, ._22(Y, Xs)) -> MEMBER_2_IN_AG2(X, Xs)
IF_SUBSET_2_IN_1_AG4(X, Xs, Ys, member_2_out_ag2(X, Ys)) -> IF_SUBSET_2_IN_2_AG4(X, Xs, Ys, subset_2_in_ag2(Xs, Ys))
IF_SUBSET_2_IN_1_AG4(X, Xs, Ys, member_2_out_ag2(X, Ys)) -> SUBSET_2_IN_AG2(Xs, Ys)

The TRS R consists of the following rules:

subset_2_in_ag2(._22(X, Xs), Ys) -> if_subset_2_in_1_ag4(X, Xs, Ys, member_2_in_ag2(X, Ys))
member_2_in_ag2(X, ._22(Y, Xs)) -> if_member_2_in_1_ag4(X, Y, Xs, member_2_in_ag2(X, Xs))
member_2_in_ag2(X, ._22(X, Xs)) -> member_2_out_ag2(X, ._22(X, Xs))
if_member_2_in_1_ag4(X, Y, Xs, member_2_out_ag2(X, Xs)) -> member_2_out_ag2(X, ._22(Y, Xs))
if_subset_2_in_1_ag4(X, Xs, Ys, member_2_out_ag2(X, Ys)) -> if_subset_2_in_2_ag4(X, Xs, Ys, subset_2_in_ag2(Xs, Ys))
subset_2_in_ag2([]_0, Ys) -> subset_2_out_ag2([]_0, Ys)
if_subset_2_in_2_ag4(X, Xs, Ys, subset_2_out_ag2(Xs, Ys)) -> subset_2_out_ag2(._22(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset_2_in_ag2(x1, x2)  =  subset_2_in_ag1(x2)
._22(x1, x2)  =  ._22(x1, x2)
[]_0  =  []_0
if_subset_2_in_1_ag4(x1, x2, x3, x4)  =  if_subset_2_in_1_ag2(x3, x4)
member_2_in_ag2(x1, x2)  =  member_2_in_ag1(x2)
if_member_2_in_1_ag4(x1, x2, x3, x4)  =  if_member_2_in_1_ag1(x4)
member_2_out_ag2(x1, x2)  =  member_2_out_ag1(x1)
if_subset_2_in_2_ag4(x1, x2, x3, x4)  =  if_subset_2_in_2_ag2(x1, x4)
subset_2_out_ag2(x1, x2)  =  subset_2_out_ag1(x1)
IF_SUBSET_2_IN_2_AG4(x1, x2, x3, x4)  =  IF_SUBSET_2_IN_2_AG2(x1, x4)
IF_MEMBER_2_IN_1_AG4(x1, x2, x3, x4)  =  IF_MEMBER_2_IN_1_AG1(x4)
MEMBER_2_IN_AG2(x1, x2)  =  MEMBER_2_IN_AG1(x2)
IF_SUBSET_2_IN_1_AG4(x1, x2, x3, x4)  =  IF_SUBSET_2_IN_1_AG2(x3, x4)
SUBSET_2_IN_AG2(x1, x2)  =  SUBSET_2_IN_AG1(x2)

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

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

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

MEMBER_2_IN_AG2(X, ._22(Y, Xs)) -> MEMBER_2_IN_AG2(X, Xs)

The TRS R consists of the following rules:

subset_2_in_ag2(._22(X, Xs), Ys) -> if_subset_2_in_1_ag4(X, Xs, Ys, member_2_in_ag2(X, Ys))
member_2_in_ag2(X, ._22(Y, Xs)) -> if_member_2_in_1_ag4(X, Y, Xs, member_2_in_ag2(X, Xs))
member_2_in_ag2(X, ._22(X, Xs)) -> member_2_out_ag2(X, ._22(X, Xs))
if_member_2_in_1_ag4(X, Y, Xs, member_2_out_ag2(X, Xs)) -> member_2_out_ag2(X, ._22(Y, Xs))
if_subset_2_in_1_ag4(X, Xs, Ys, member_2_out_ag2(X, Ys)) -> if_subset_2_in_2_ag4(X, Xs, Ys, subset_2_in_ag2(Xs, Ys))
subset_2_in_ag2([]_0, Ys) -> subset_2_out_ag2([]_0, Ys)
if_subset_2_in_2_ag4(X, Xs, Ys, subset_2_out_ag2(Xs, Ys)) -> subset_2_out_ag2(._22(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset_2_in_ag2(x1, x2)  =  subset_2_in_ag1(x2)
._22(x1, x2)  =  ._22(x1, x2)
[]_0  =  []_0
if_subset_2_in_1_ag4(x1, x2, x3, x4)  =  if_subset_2_in_1_ag2(x3, x4)
member_2_in_ag2(x1, x2)  =  member_2_in_ag1(x2)
if_member_2_in_1_ag4(x1, x2, x3, x4)  =  if_member_2_in_1_ag1(x4)
member_2_out_ag2(x1, x2)  =  member_2_out_ag1(x1)
if_subset_2_in_2_ag4(x1, x2, x3, x4)  =  if_subset_2_in_2_ag2(x1, x4)
subset_2_out_ag2(x1, x2)  =  subset_2_out_ag1(x1)
MEMBER_2_IN_AG2(x1, x2)  =  MEMBER_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
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
              ↳ PiDP
  ↳ PrologToPiTRSProof

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

MEMBER_2_IN_AG2(X, ._22(Y, Xs)) -> MEMBER_2_IN_AG2(X, Xs)

R is empty.
The argument filtering Pi contains the following mapping:
._22(x1, x2)  =  ._22(x1, x2)
MEMBER_2_IN_AG2(x1, x2)  =  MEMBER_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
  ↳ PrologToPiTRSProof

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

MEMBER_2_IN_AG1(._22(Y, Xs)) -> MEMBER_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 {MEMBER_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: 



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

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

IF_SUBSET_2_IN_1_AG4(X, Xs, Ys, member_2_out_ag2(X, Ys)) -> SUBSET_2_IN_AG2(Xs, Ys)
SUBSET_2_IN_AG2(._22(X, Xs), Ys) -> IF_SUBSET_2_IN_1_AG4(X, Xs, Ys, member_2_in_ag2(X, Ys))

The TRS R consists of the following rules:

subset_2_in_ag2(._22(X, Xs), Ys) -> if_subset_2_in_1_ag4(X, Xs, Ys, member_2_in_ag2(X, Ys))
member_2_in_ag2(X, ._22(Y, Xs)) -> if_member_2_in_1_ag4(X, Y, Xs, member_2_in_ag2(X, Xs))
member_2_in_ag2(X, ._22(X, Xs)) -> member_2_out_ag2(X, ._22(X, Xs))
if_member_2_in_1_ag4(X, Y, Xs, member_2_out_ag2(X, Xs)) -> member_2_out_ag2(X, ._22(Y, Xs))
if_subset_2_in_1_ag4(X, Xs, Ys, member_2_out_ag2(X, Ys)) -> if_subset_2_in_2_ag4(X, Xs, Ys, subset_2_in_ag2(Xs, Ys))
subset_2_in_ag2([]_0, Ys) -> subset_2_out_ag2([]_0, Ys)
if_subset_2_in_2_ag4(X, Xs, Ys, subset_2_out_ag2(Xs, Ys)) -> subset_2_out_ag2(._22(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset_2_in_ag2(x1, x2)  =  subset_2_in_ag1(x2)
._22(x1, x2)  =  ._22(x1, x2)
[]_0  =  []_0
if_subset_2_in_1_ag4(x1, x2, x3, x4)  =  if_subset_2_in_1_ag2(x3, x4)
member_2_in_ag2(x1, x2)  =  member_2_in_ag1(x2)
if_member_2_in_1_ag4(x1, x2, x3, x4)  =  if_member_2_in_1_ag1(x4)
member_2_out_ag2(x1, x2)  =  member_2_out_ag1(x1)
if_subset_2_in_2_ag4(x1, x2, x3, x4)  =  if_subset_2_in_2_ag2(x1, x4)
subset_2_out_ag2(x1, x2)  =  subset_2_out_ag1(x1)
IF_SUBSET_2_IN_1_AG4(x1, x2, x3, x4)  =  IF_SUBSET_2_IN_1_AG2(x3, x4)
SUBSET_2_IN_AG2(x1, x2)  =  SUBSET_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
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
  ↳ PrologToPiTRSProof

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

IF_SUBSET_2_IN_1_AG4(X, Xs, Ys, member_2_out_ag2(X, Ys)) -> SUBSET_2_IN_AG2(Xs, Ys)
SUBSET_2_IN_AG2(._22(X, Xs), Ys) -> IF_SUBSET_2_IN_1_AG4(X, Xs, Ys, member_2_in_ag2(X, Ys))

The TRS R consists of the following rules:

member_2_in_ag2(X, ._22(Y, Xs)) -> if_member_2_in_1_ag4(X, Y, Xs, member_2_in_ag2(X, Xs))
member_2_in_ag2(X, ._22(X, Xs)) -> member_2_out_ag2(X, ._22(X, Xs))
if_member_2_in_1_ag4(X, Y, Xs, member_2_out_ag2(X, Xs)) -> member_2_out_ag2(X, ._22(Y, Xs))

The argument filtering Pi contains the following mapping:
._22(x1, x2)  =  ._22(x1, x2)
member_2_in_ag2(x1, x2)  =  member_2_in_ag1(x2)
if_member_2_in_1_ag4(x1, x2, x3, x4)  =  if_member_2_in_1_ag1(x4)
member_2_out_ag2(x1, x2)  =  member_2_out_ag1(x1)
IF_SUBSET_2_IN_1_AG4(x1, x2, x3, x4)  =  IF_SUBSET_2_IN_1_AG2(x3, x4)
SUBSET_2_IN_AG2(x1, x2)  =  SUBSET_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
  ↳ PrologToPiTRSProof

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

IF_SUBSET_2_IN_1_AG2(Ys, member_2_out_ag1(X)) -> SUBSET_2_IN_AG1(Ys)
SUBSET_2_IN_AG1(Ys) -> IF_SUBSET_2_IN_1_AG2(Ys, member_2_in_ag1(Ys))

The TRS R consists of the following rules:

member_2_in_ag1(._22(Y, Xs)) -> if_member_2_in_1_ag1(member_2_in_ag1(Xs))
member_2_in_ag1(._22(X, Xs)) -> member_2_out_ag1(X)
if_member_2_in_1_ag1(member_2_out_ag1(X)) -> member_2_out_ag1(X)

The set Q consists of the following terms:

member_2_in_ag1(x0)
if_member_2_in_1_ag1(x0)

We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {SUBSET_2_IN_AG1, IF_SUBSET_2_IN_1_AG2}.
With regard to the inferred argument filtering the predicates were used in the following modes:
subset2: (f,b)
member2: (f,b)
Transforming PROLOG into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

subset_2_in_ag2(._22(X, Xs), Ys) -> if_subset_2_in_1_ag4(X, Xs, Ys, member_2_in_ag2(X, Ys))
member_2_in_ag2(X, ._22(Y, Xs)) -> if_member_2_in_1_ag4(X, Y, Xs, member_2_in_ag2(X, Xs))
member_2_in_ag2(X, ._22(X, Xs)) -> member_2_out_ag2(X, ._22(X, Xs))
if_member_2_in_1_ag4(X, Y, Xs, member_2_out_ag2(X, Xs)) -> member_2_out_ag2(X, ._22(Y, Xs))
if_subset_2_in_1_ag4(X, Xs, Ys, member_2_out_ag2(X, Ys)) -> if_subset_2_in_2_ag4(X, Xs, Ys, subset_2_in_ag2(Xs, Ys))
subset_2_in_ag2([]_0, Ys) -> subset_2_out_ag2([]_0, Ys)
if_subset_2_in_2_ag4(X, Xs, Ys, subset_2_out_ag2(Xs, Ys)) -> subset_2_out_ag2(._22(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset_2_in_ag2(x1, x2)  =  subset_2_in_ag1(x2)
._22(x1, x2)  =  ._22(x1, x2)
[]_0  =  []_0
if_subset_2_in_1_ag4(x1, x2, x3, x4)  =  if_subset_2_in_1_ag2(x3, x4)
member_2_in_ag2(x1, x2)  =  member_2_in_ag1(x2)
if_member_2_in_1_ag4(x1, x2, x3, x4)  =  if_member_2_in_1_ag3(x2, x3, x4)
member_2_out_ag2(x1, x2)  =  member_2_out_ag2(x1, x2)
if_subset_2_in_2_ag4(x1, x2, x3, x4)  =  if_subset_2_in_2_ag3(x1, x3, x4)
subset_2_out_ag2(x1, x2)  =  subset_2_out_ag2(x1, x2)

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:

subset_2_in_ag2(._22(X, Xs), Ys) -> if_subset_2_in_1_ag4(X, Xs, Ys, member_2_in_ag2(X, Ys))
member_2_in_ag2(X, ._22(Y, Xs)) -> if_member_2_in_1_ag4(X, Y, Xs, member_2_in_ag2(X, Xs))
member_2_in_ag2(X, ._22(X, Xs)) -> member_2_out_ag2(X, ._22(X, Xs))
if_member_2_in_1_ag4(X, Y, Xs, member_2_out_ag2(X, Xs)) -> member_2_out_ag2(X, ._22(Y, Xs))
if_subset_2_in_1_ag4(X, Xs, Ys, member_2_out_ag2(X, Ys)) -> if_subset_2_in_2_ag4(X, Xs, Ys, subset_2_in_ag2(Xs, Ys))
subset_2_in_ag2([]_0, Ys) -> subset_2_out_ag2([]_0, Ys)
if_subset_2_in_2_ag4(X, Xs, Ys, subset_2_out_ag2(Xs, Ys)) -> subset_2_out_ag2(._22(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset_2_in_ag2(x1, x2)  =  subset_2_in_ag1(x2)
._22(x1, x2)  =  ._22(x1, x2)
[]_0  =  []_0
if_subset_2_in_1_ag4(x1, x2, x3, x4)  =  if_subset_2_in_1_ag2(x3, x4)
member_2_in_ag2(x1, x2)  =  member_2_in_ag1(x2)
if_member_2_in_1_ag4(x1, x2, x3, x4)  =  if_member_2_in_1_ag3(x2, x3, x4)
member_2_out_ag2(x1, x2)  =  member_2_out_ag2(x1, x2)
if_subset_2_in_2_ag4(x1, x2, x3, x4)  =  if_subset_2_in_2_ag3(x1, x3, x4)
subset_2_out_ag2(x1, x2)  =  subset_2_out_ag2(x1, x2)


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

SUBSET_2_IN_AG2(._22(X, Xs), Ys) -> IF_SUBSET_2_IN_1_AG4(X, Xs, Ys, member_2_in_ag2(X, Ys))
SUBSET_2_IN_AG2(._22(X, Xs), Ys) -> MEMBER_2_IN_AG2(X, Ys)
MEMBER_2_IN_AG2(X, ._22(Y, Xs)) -> IF_MEMBER_2_IN_1_AG4(X, Y, Xs, member_2_in_ag2(X, Xs))
MEMBER_2_IN_AG2(X, ._22(Y, Xs)) -> MEMBER_2_IN_AG2(X, Xs)
IF_SUBSET_2_IN_1_AG4(X, Xs, Ys, member_2_out_ag2(X, Ys)) -> IF_SUBSET_2_IN_2_AG4(X, Xs, Ys, subset_2_in_ag2(Xs, Ys))
IF_SUBSET_2_IN_1_AG4(X, Xs, Ys, member_2_out_ag2(X, Ys)) -> SUBSET_2_IN_AG2(Xs, Ys)

The TRS R consists of the following rules:

subset_2_in_ag2(._22(X, Xs), Ys) -> if_subset_2_in_1_ag4(X, Xs, Ys, member_2_in_ag2(X, Ys))
member_2_in_ag2(X, ._22(Y, Xs)) -> if_member_2_in_1_ag4(X, Y, Xs, member_2_in_ag2(X, Xs))
member_2_in_ag2(X, ._22(X, Xs)) -> member_2_out_ag2(X, ._22(X, Xs))
if_member_2_in_1_ag4(X, Y, Xs, member_2_out_ag2(X, Xs)) -> member_2_out_ag2(X, ._22(Y, Xs))
if_subset_2_in_1_ag4(X, Xs, Ys, member_2_out_ag2(X, Ys)) -> if_subset_2_in_2_ag4(X, Xs, Ys, subset_2_in_ag2(Xs, Ys))
subset_2_in_ag2([]_0, Ys) -> subset_2_out_ag2([]_0, Ys)
if_subset_2_in_2_ag4(X, Xs, Ys, subset_2_out_ag2(Xs, Ys)) -> subset_2_out_ag2(._22(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset_2_in_ag2(x1, x2)  =  subset_2_in_ag1(x2)
._22(x1, x2)  =  ._22(x1, x2)
[]_0  =  []_0
if_subset_2_in_1_ag4(x1, x2, x3, x4)  =  if_subset_2_in_1_ag2(x3, x4)
member_2_in_ag2(x1, x2)  =  member_2_in_ag1(x2)
if_member_2_in_1_ag4(x1, x2, x3, x4)  =  if_member_2_in_1_ag3(x2, x3, x4)
member_2_out_ag2(x1, x2)  =  member_2_out_ag2(x1, x2)
if_subset_2_in_2_ag4(x1, x2, x3, x4)  =  if_subset_2_in_2_ag3(x1, x3, x4)
subset_2_out_ag2(x1, x2)  =  subset_2_out_ag2(x1, x2)
IF_SUBSET_2_IN_2_AG4(x1, x2, x3, x4)  =  IF_SUBSET_2_IN_2_AG3(x1, x3, x4)
IF_MEMBER_2_IN_1_AG4(x1, x2, x3, x4)  =  IF_MEMBER_2_IN_1_AG3(x2, x3, x4)
MEMBER_2_IN_AG2(x1, x2)  =  MEMBER_2_IN_AG1(x2)
IF_SUBSET_2_IN_1_AG4(x1, x2, x3, x4)  =  IF_SUBSET_2_IN_1_AG2(x3, x4)
SUBSET_2_IN_AG2(x1, x2)  =  SUBSET_2_IN_AG1(x2)

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:

SUBSET_2_IN_AG2(._22(X, Xs), Ys) -> IF_SUBSET_2_IN_1_AG4(X, Xs, Ys, member_2_in_ag2(X, Ys))
SUBSET_2_IN_AG2(._22(X, Xs), Ys) -> MEMBER_2_IN_AG2(X, Ys)
MEMBER_2_IN_AG2(X, ._22(Y, Xs)) -> IF_MEMBER_2_IN_1_AG4(X, Y, Xs, member_2_in_ag2(X, Xs))
MEMBER_2_IN_AG2(X, ._22(Y, Xs)) -> MEMBER_2_IN_AG2(X, Xs)
IF_SUBSET_2_IN_1_AG4(X, Xs, Ys, member_2_out_ag2(X, Ys)) -> IF_SUBSET_2_IN_2_AG4(X, Xs, Ys, subset_2_in_ag2(Xs, Ys))
IF_SUBSET_2_IN_1_AG4(X, Xs, Ys, member_2_out_ag2(X, Ys)) -> SUBSET_2_IN_AG2(Xs, Ys)

The TRS R consists of the following rules:

subset_2_in_ag2(._22(X, Xs), Ys) -> if_subset_2_in_1_ag4(X, Xs, Ys, member_2_in_ag2(X, Ys))
member_2_in_ag2(X, ._22(Y, Xs)) -> if_member_2_in_1_ag4(X, Y, Xs, member_2_in_ag2(X, Xs))
member_2_in_ag2(X, ._22(X, Xs)) -> member_2_out_ag2(X, ._22(X, Xs))
if_member_2_in_1_ag4(X, Y, Xs, member_2_out_ag2(X, Xs)) -> member_2_out_ag2(X, ._22(Y, Xs))
if_subset_2_in_1_ag4(X, Xs, Ys, member_2_out_ag2(X, Ys)) -> if_subset_2_in_2_ag4(X, Xs, Ys, subset_2_in_ag2(Xs, Ys))
subset_2_in_ag2([]_0, Ys) -> subset_2_out_ag2([]_0, Ys)
if_subset_2_in_2_ag4(X, Xs, Ys, subset_2_out_ag2(Xs, Ys)) -> subset_2_out_ag2(._22(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset_2_in_ag2(x1, x2)  =  subset_2_in_ag1(x2)
._22(x1, x2)  =  ._22(x1, x2)
[]_0  =  []_0
if_subset_2_in_1_ag4(x1, x2, x3, x4)  =  if_subset_2_in_1_ag2(x3, x4)
member_2_in_ag2(x1, x2)  =  member_2_in_ag1(x2)
if_member_2_in_1_ag4(x1, x2, x3, x4)  =  if_member_2_in_1_ag3(x2, x3, x4)
member_2_out_ag2(x1, x2)  =  member_2_out_ag2(x1, x2)
if_subset_2_in_2_ag4(x1, x2, x3, x4)  =  if_subset_2_in_2_ag3(x1, x3, x4)
subset_2_out_ag2(x1, x2)  =  subset_2_out_ag2(x1, x2)
IF_SUBSET_2_IN_2_AG4(x1, x2, x3, x4)  =  IF_SUBSET_2_IN_2_AG3(x1, x3, x4)
IF_MEMBER_2_IN_1_AG4(x1, x2, x3, x4)  =  IF_MEMBER_2_IN_1_AG3(x2, x3, x4)
MEMBER_2_IN_AG2(x1, x2)  =  MEMBER_2_IN_AG1(x2)
IF_SUBSET_2_IN_1_AG4(x1, x2, x3, x4)  =  IF_SUBSET_2_IN_1_AG2(x3, x4)
SUBSET_2_IN_AG2(x1, x2)  =  SUBSET_2_IN_AG1(x2)

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

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

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

MEMBER_2_IN_AG2(X, ._22(Y, Xs)) -> MEMBER_2_IN_AG2(X, Xs)

The TRS R consists of the following rules:

subset_2_in_ag2(._22(X, Xs), Ys) -> if_subset_2_in_1_ag4(X, Xs, Ys, member_2_in_ag2(X, Ys))
member_2_in_ag2(X, ._22(Y, Xs)) -> if_member_2_in_1_ag4(X, Y, Xs, member_2_in_ag2(X, Xs))
member_2_in_ag2(X, ._22(X, Xs)) -> member_2_out_ag2(X, ._22(X, Xs))
if_member_2_in_1_ag4(X, Y, Xs, member_2_out_ag2(X, Xs)) -> member_2_out_ag2(X, ._22(Y, Xs))
if_subset_2_in_1_ag4(X, Xs, Ys, member_2_out_ag2(X, Ys)) -> if_subset_2_in_2_ag4(X, Xs, Ys, subset_2_in_ag2(Xs, Ys))
subset_2_in_ag2([]_0, Ys) -> subset_2_out_ag2([]_0, Ys)
if_subset_2_in_2_ag4(X, Xs, Ys, subset_2_out_ag2(Xs, Ys)) -> subset_2_out_ag2(._22(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset_2_in_ag2(x1, x2)  =  subset_2_in_ag1(x2)
._22(x1, x2)  =  ._22(x1, x2)
[]_0  =  []_0
if_subset_2_in_1_ag4(x1, x2, x3, x4)  =  if_subset_2_in_1_ag2(x3, x4)
member_2_in_ag2(x1, x2)  =  member_2_in_ag1(x2)
if_member_2_in_1_ag4(x1, x2, x3, x4)  =  if_member_2_in_1_ag3(x2, x3, x4)
member_2_out_ag2(x1, x2)  =  member_2_out_ag2(x1, x2)
if_subset_2_in_2_ag4(x1, x2, x3, x4)  =  if_subset_2_in_2_ag3(x1, x3, x4)
subset_2_out_ag2(x1, x2)  =  subset_2_out_ag2(x1, x2)
MEMBER_2_IN_AG2(x1, x2)  =  MEMBER_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
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
              ↳ PiDP

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

MEMBER_2_IN_AG2(X, ._22(Y, Xs)) -> MEMBER_2_IN_AG2(X, Xs)

R is empty.
The argument filtering Pi contains the following mapping:
._22(x1, x2)  =  ._22(x1, x2)
MEMBER_2_IN_AG2(x1, x2)  =  MEMBER_2_IN_AG1(x2)

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

↳ PROLOG
  ↳ PrologToPiTRSProof
  ↳ 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_AG4(X, Xs, Ys, member_2_out_ag2(X, Ys)) -> SUBSET_2_IN_AG2(Xs, Ys)
SUBSET_2_IN_AG2(._22(X, Xs), Ys) -> IF_SUBSET_2_IN_1_AG4(X, Xs, Ys, member_2_in_ag2(X, Ys))

The TRS R consists of the following rules:

subset_2_in_ag2(._22(X, Xs), Ys) -> if_subset_2_in_1_ag4(X, Xs, Ys, member_2_in_ag2(X, Ys))
member_2_in_ag2(X, ._22(Y, Xs)) -> if_member_2_in_1_ag4(X, Y, Xs, member_2_in_ag2(X, Xs))
member_2_in_ag2(X, ._22(X, Xs)) -> member_2_out_ag2(X, ._22(X, Xs))
if_member_2_in_1_ag4(X, Y, Xs, member_2_out_ag2(X, Xs)) -> member_2_out_ag2(X, ._22(Y, Xs))
if_subset_2_in_1_ag4(X, Xs, Ys, member_2_out_ag2(X, Ys)) -> if_subset_2_in_2_ag4(X, Xs, Ys, subset_2_in_ag2(Xs, Ys))
subset_2_in_ag2([]_0, Ys) -> subset_2_out_ag2([]_0, Ys)
if_subset_2_in_2_ag4(X, Xs, Ys, subset_2_out_ag2(Xs, Ys)) -> subset_2_out_ag2(._22(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
subset_2_in_ag2(x1, x2)  =  subset_2_in_ag1(x2)
._22(x1, x2)  =  ._22(x1, x2)
[]_0  =  []_0
if_subset_2_in_1_ag4(x1, x2, x3, x4)  =  if_subset_2_in_1_ag2(x3, x4)
member_2_in_ag2(x1, x2)  =  member_2_in_ag1(x2)
if_member_2_in_1_ag4(x1, x2, x3, x4)  =  if_member_2_in_1_ag3(x2, x3, x4)
member_2_out_ag2(x1, x2)  =  member_2_out_ag2(x1, x2)
if_subset_2_in_2_ag4(x1, x2, x3, x4)  =  if_subset_2_in_2_ag3(x1, x3, x4)
subset_2_out_ag2(x1, x2)  =  subset_2_out_ag2(x1, x2)
IF_SUBSET_2_IN_1_AG4(x1, x2, x3, x4)  =  IF_SUBSET_2_IN_1_AG2(x3, x4)
SUBSET_2_IN_AG2(x1, x2)  =  SUBSET_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
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP

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

IF_SUBSET_2_IN_1_AG4(X, Xs, Ys, member_2_out_ag2(X, Ys)) -> SUBSET_2_IN_AG2(Xs, Ys)
SUBSET_2_IN_AG2(._22(X, Xs), Ys) -> IF_SUBSET_2_IN_1_AG4(X, Xs, Ys, member_2_in_ag2(X, Ys))

The TRS R consists of the following rules:

member_2_in_ag2(X, ._22(Y, Xs)) -> if_member_2_in_1_ag4(X, Y, Xs, member_2_in_ag2(X, Xs))
member_2_in_ag2(X, ._22(X, Xs)) -> member_2_out_ag2(X, ._22(X, Xs))
if_member_2_in_1_ag4(X, Y, Xs, member_2_out_ag2(X, Xs)) -> member_2_out_ag2(X, ._22(Y, Xs))

The argument filtering Pi contains the following mapping:
._22(x1, x2)  =  ._22(x1, x2)
member_2_in_ag2(x1, x2)  =  member_2_in_ag1(x2)
if_member_2_in_1_ag4(x1, x2, x3, x4)  =  if_member_2_in_1_ag3(x2, x3, x4)
member_2_out_ag2(x1, x2)  =  member_2_out_ag2(x1, x2)
IF_SUBSET_2_IN_1_AG4(x1, x2, x3, x4)  =  IF_SUBSET_2_IN_1_AG2(x3, x4)
SUBSET_2_IN_AG2(x1, x2)  =  SUBSET_2_IN_AG1(x2)

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