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



↳ PROLOG
  ↳ PrologToPiTRSProof

select3(X, .2(X, Xs), Xs).
select3(X, .2(Y, Xs), .2(Y, Zs)) :- select3(X, Xs, Zs).


With regard to the inferred argument filtering the predicates were used in the following modes:
select3: (f,f,b)
Transforming PROLOG into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:


select_3_in_aag3(X, ._22(X, Xs), Xs) -> select_3_out_aag3(X, ._22(X, Xs), Xs)
select_3_in_aag3(X, ._22(Y, Xs), ._22(Y, Zs)) -> if_select_3_in_1_aag5(X, Y, Xs, Zs, select_3_in_aag3(X, Xs, Zs))
if_select_3_in_1_aag5(X, Y, Xs, Zs, select_3_out_aag3(X, Xs, Zs)) -> select_3_out_aag3(X, ._22(Y, Xs), ._22(Y, Zs))

The argument filtering Pi contains the following mapping:
select_3_in_aag3(x1, x2, x3)  =  select_3_in_aag1(x3)
._22(x1, x2)  =  ._21(x2)
select_3_out_aag3(x1, x2, x3)  =  select_3_out_aag1(x2)
if_select_3_in_1_aag5(x1, x2, x3, x4, x5)  =  if_select_3_in_1_aag1(x5)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of PROLOG



↳ PROLOG
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof

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

select_3_in_aag3(X, ._22(X, Xs), Xs) -> select_3_out_aag3(X, ._22(X, Xs), Xs)
select_3_in_aag3(X, ._22(Y, Xs), ._22(Y, Zs)) -> if_select_3_in_1_aag5(X, Y, Xs, Zs, select_3_in_aag3(X, Xs, Zs))
if_select_3_in_1_aag5(X, Y, Xs, Zs, select_3_out_aag3(X, Xs, Zs)) -> select_3_out_aag3(X, ._22(Y, Xs), ._22(Y, Zs))

The argument filtering Pi contains the following mapping:
select_3_in_aag3(x1, x2, x3)  =  select_3_in_aag1(x3)
._22(x1, x2)  =  ._21(x2)
select_3_out_aag3(x1, x2, x3)  =  select_3_out_aag1(x2)
if_select_3_in_1_aag5(x1, x2, x3, x4, x5)  =  if_select_3_in_1_aag1(x5)


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

SELECT_3_IN_AAG3(X, ._22(Y, Xs), ._22(Y, Zs)) -> IF_SELECT_3_IN_1_AAG5(X, Y, Xs, Zs, select_3_in_aag3(X, Xs, Zs))
SELECT_3_IN_AAG3(X, ._22(Y, Xs), ._22(Y, Zs)) -> SELECT_3_IN_AAG3(X, Xs, Zs)

The TRS R consists of the following rules:

select_3_in_aag3(X, ._22(X, Xs), Xs) -> select_3_out_aag3(X, ._22(X, Xs), Xs)
select_3_in_aag3(X, ._22(Y, Xs), ._22(Y, Zs)) -> if_select_3_in_1_aag5(X, Y, Xs, Zs, select_3_in_aag3(X, Xs, Zs))
if_select_3_in_1_aag5(X, Y, Xs, Zs, select_3_out_aag3(X, Xs, Zs)) -> select_3_out_aag3(X, ._22(Y, Xs), ._22(Y, Zs))

The argument filtering Pi contains the following mapping:
select_3_in_aag3(x1, x2, x3)  =  select_3_in_aag1(x3)
._22(x1, x2)  =  ._21(x2)
select_3_out_aag3(x1, x2, x3)  =  select_3_out_aag1(x2)
if_select_3_in_1_aag5(x1, x2, x3, x4, x5)  =  if_select_3_in_1_aag1(x5)
IF_SELECT_3_IN_1_AAG5(x1, x2, x3, x4, x5)  =  IF_SELECT_3_IN_1_AAG1(x5)
SELECT_3_IN_AAG3(x1, x2, x3)  =  SELECT_3_IN_AAG1(x3)

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

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

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

SELECT_3_IN_AAG3(X, ._22(Y, Xs), ._22(Y, Zs)) -> IF_SELECT_3_IN_1_AAG5(X, Y, Xs, Zs, select_3_in_aag3(X, Xs, Zs))
SELECT_3_IN_AAG3(X, ._22(Y, Xs), ._22(Y, Zs)) -> SELECT_3_IN_AAG3(X, Xs, Zs)

The TRS R consists of the following rules:

select_3_in_aag3(X, ._22(X, Xs), Xs) -> select_3_out_aag3(X, ._22(X, Xs), Xs)
select_3_in_aag3(X, ._22(Y, Xs), ._22(Y, Zs)) -> if_select_3_in_1_aag5(X, Y, Xs, Zs, select_3_in_aag3(X, Xs, Zs))
if_select_3_in_1_aag5(X, Y, Xs, Zs, select_3_out_aag3(X, Xs, Zs)) -> select_3_out_aag3(X, ._22(Y, Xs), ._22(Y, Zs))

The argument filtering Pi contains the following mapping:
select_3_in_aag3(x1, x2, x3)  =  select_3_in_aag1(x3)
._22(x1, x2)  =  ._21(x2)
select_3_out_aag3(x1, x2, x3)  =  select_3_out_aag1(x2)
if_select_3_in_1_aag5(x1, x2, x3, x4, x5)  =  if_select_3_in_1_aag1(x5)
IF_SELECT_3_IN_1_AAG5(x1, x2, x3, x4, x5)  =  IF_SELECT_3_IN_1_AAG1(x5)
SELECT_3_IN_AAG3(x1, x2, x3)  =  SELECT_3_IN_AAG1(x3)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph contains 1 SCC with 1 less node.

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

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

SELECT_3_IN_AAG3(X, ._22(Y, Xs), ._22(Y, Zs)) -> SELECT_3_IN_AAG3(X, Xs, Zs)

The TRS R consists of the following rules:

select_3_in_aag3(X, ._22(X, Xs), Xs) -> select_3_out_aag3(X, ._22(X, Xs), Xs)
select_3_in_aag3(X, ._22(Y, Xs), ._22(Y, Zs)) -> if_select_3_in_1_aag5(X, Y, Xs, Zs, select_3_in_aag3(X, Xs, Zs))
if_select_3_in_1_aag5(X, Y, Xs, Zs, select_3_out_aag3(X, Xs, Zs)) -> select_3_out_aag3(X, ._22(Y, Xs), ._22(Y, Zs))

The argument filtering Pi contains the following mapping:
select_3_in_aag3(x1, x2, x3)  =  select_3_in_aag1(x3)
._22(x1, x2)  =  ._21(x2)
select_3_out_aag3(x1, x2, x3)  =  select_3_out_aag1(x2)
if_select_3_in_1_aag5(x1, x2, x3, x4, x5)  =  if_select_3_in_1_aag1(x5)
SELECT_3_IN_AAG3(x1, x2, x3)  =  SELECT_3_IN_AAG1(x3)

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
            ↳ PiDP
              ↳ UsableRulesProof
                ↳ PiDP
                  ↳ PiDPToQDPProof

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

SELECT_3_IN_AAG3(X, ._22(Y, Xs), ._22(Y, Zs)) -> SELECT_3_IN_AAG3(X, Xs, Zs)

R is empty.
The argument filtering Pi contains the following mapping:
._22(x1, x2)  =  ._21(x2)
SELECT_3_IN_AAG3(x1, x2, x3)  =  SELECT_3_IN_AAG1(x3)

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
            ↳ PiDP
              ↳ UsableRulesProof
                ↳ PiDP
                  ↳ PiDPToQDPProof
                    ↳ QDP
                      ↳ QDPSizeChangeProof

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

SELECT_3_IN_AAG1(._21(Zs)) -> SELECT_3_IN_AAG1(Zs)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {SELECT_3_IN_AAG1}.
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: