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



↳ PROLOG
  ↳ PrologToPiTRSProof

reverse3({}0, X, X).
reverse3(.2(X, Y), Z, U) :- reverse3(Y, Z, .2(X, U)).


With regard to the inferred argument filtering the predicates were used in the following modes:
reverse3: (b,f,b)
Transforming PROLOG into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:


reverse_3_in_gag3([]_0, X, X) -> reverse_3_out_gag3([]_0, X, X)
reverse_3_in_gag3(._22(X, Y), Z, U) -> if_reverse_3_in_1_gag5(X, Y, Z, U, reverse_3_in_gag3(Y, Z, ._22(X, U)))
if_reverse_3_in_1_gag5(X, Y, Z, U, reverse_3_out_gag3(Y, Z, ._22(X, U))) -> reverse_3_out_gag3(._22(X, Y), Z, U)

The argument filtering Pi contains the following mapping:
reverse_3_in_gag3(x1, x2, x3)  =  reverse_3_in_gag2(x1, x3)
[]_0  =  []_0
._22(x1, x2)  =  ._22(x1, x2)
reverse_3_out_gag3(x1, x2, x3)  =  reverse_3_out_gag1(x2)
if_reverse_3_in_1_gag5(x1, x2, x3, x4, x5)  =  if_reverse_3_in_1_gag1(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:

reverse_3_in_gag3([]_0, X, X) -> reverse_3_out_gag3([]_0, X, X)
reverse_3_in_gag3(._22(X, Y), Z, U) -> if_reverse_3_in_1_gag5(X, Y, Z, U, reverse_3_in_gag3(Y, Z, ._22(X, U)))
if_reverse_3_in_1_gag5(X, Y, Z, U, reverse_3_out_gag3(Y, Z, ._22(X, U))) -> reverse_3_out_gag3(._22(X, Y), Z, U)

The argument filtering Pi contains the following mapping:
reverse_3_in_gag3(x1, x2, x3)  =  reverse_3_in_gag2(x1, x3)
[]_0  =  []_0
._22(x1, x2)  =  ._22(x1, x2)
reverse_3_out_gag3(x1, x2, x3)  =  reverse_3_out_gag1(x2)
if_reverse_3_in_1_gag5(x1, x2, x3, x4, x5)  =  if_reverse_3_in_1_gag1(x5)


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

REVERSE_3_IN_GAG3(._22(X, Y), Z, U) -> IF_REVERSE_3_IN_1_GAG5(X, Y, Z, U, reverse_3_in_gag3(Y, Z, ._22(X, U)))
REVERSE_3_IN_GAG3(._22(X, Y), Z, U) -> REVERSE_3_IN_GAG3(Y, Z, ._22(X, U))

The TRS R consists of the following rules:

reverse_3_in_gag3([]_0, X, X) -> reverse_3_out_gag3([]_0, X, X)
reverse_3_in_gag3(._22(X, Y), Z, U) -> if_reverse_3_in_1_gag5(X, Y, Z, U, reverse_3_in_gag3(Y, Z, ._22(X, U)))
if_reverse_3_in_1_gag5(X, Y, Z, U, reverse_3_out_gag3(Y, Z, ._22(X, U))) -> reverse_3_out_gag3(._22(X, Y), Z, U)

The argument filtering Pi contains the following mapping:
reverse_3_in_gag3(x1, x2, x3)  =  reverse_3_in_gag2(x1, x3)
[]_0  =  []_0
._22(x1, x2)  =  ._22(x1, x2)
reverse_3_out_gag3(x1, x2, x3)  =  reverse_3_out_gag1(x2)
if_reverse_3_in_1_gag5(x1, x2, x3, x4, x5)  =  if_reverse_3_in_1_gag1(x5)
IF_REVERSE_3_IN_1_GAG5(x1, x2, x3, x4, x5)  =  IF_REVERSE_3_IN_1_GAG1(x5)
REVERSE_3_IN_GAG3(x1, x2, x3)  =  REVERSE_3_IN_GAG2(x1, 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:

REVERSE_3_IN_GAG3(._22(X, Y), Z, U) -> IF_REVERSE_3_IN_1_GAG5(X, Y, Z, U, reverse_3_in_gag3(Y, Z, ._22(X, U)))
REVERSE_3_IN_GAG3(._22(X, Y), Z, U) -> REVERSE_3_IN_GAG3(Y, Z, ._22(X, U))

The TRS R consists of the following rules:

reverse_3_in_gag3([]_0, X, X) -> reverse_3_out_gag3([]_0, X, X)
reverse_3_in_gag3(._22(X, Y), Z, U) -> if_reverse_3_in_1_gag5(X, Y, Z, U, reverse_3_in_gag3(Y, Z, ._22(X, U)))
if_reverse_3_in_1_gag5(X, Y, Z, U, reverse_3_out_gag3(Y, Z, ._22(X, U))) -> reverse_3_out_gag3(._22(X, Y), Z, U)

The argument filtering Pi contains the following mapping:
reverse_3_in_gag3(x1, x2, x3)  =  reverse_3_in_gag2(x1, x3)
[]_0  =  []_0
._22(x1, x2)  =  ._22(x1, x2)
reverse_3_out_gag3(x1, x2, x3)  =  reverse_3_out_gag1(x2)
if_reverse_3_in_1_gag5(x1, x2, x3, x4, x5)  =  if_reverse_3_in_1_gag1(x5)
IF_REVERSE_3_IN_1_GAG5(x1, x2, x3, x4, x5)  =  IF_REVERSE_3_IN_1_GAG1(x5)
REVERSE_3_IN_GAG3(x1, x2, x3)  =  REVERSE_3_IN_GAG2(x1, 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:

REVERSE_3_IN_GAG3(._22(X, Y), Z, U) -> REVERSE_3_IN_GAG3(Y, Z, ._22(X, U))

The TRS R consists of the following rules:

reverse_3_in_gag3([]_0, X, X) -> reverse_3_out_gag3([]_0, X, X)
reverse_3_in_gag3(._22(X, Y), Z, U) -> if_reverse_3_in_1_gag5(X, Y, Z, U, reverse_3_in_gag3(Y, Z, ._22(X, U)))
if_reverse_3_in_1_gag5(X, Y, Z, U, reverse_3_out_gag3(Y, Z, ._22(X, U))) -> reverse_3_out_gag3(._22(X, Y), Z, U)

The argument filtering Pi contains the following mapping:
reverse_3_in_gag3(x1, x2, x3)  =  reverse_3_in_gag2(x1, x3)
[]_0  =  []_0
._22(x1, x2)  =  ._22(x1, x2)
reverse_3_out_gag3(x1, x2, x3)  =  reverse_3_out_gag1(x2)
if_reverse_3_in_1_gag5(x1, x2, x3, x4, x5)  =  if_reverse_3_in_1_gag1(x5)
REVERSE_3_IN_GAG3(x1, x2, x3)  =  REVERSE_3_IN_GAG2(x1, 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:

REVERSE_3_IN_GAG3(._22(X, Y), Z, U) -> REVERSE_3_IN_GAG3(Y, Z, ._22(X, U))

R is empty.
The argument filtering Pi contains the following mapping:
._22(x1, x2)  =  ._22(x1, x2)
REVERSE_3_IN_GAG3(x1, x2, x3)  =  REVERSE_3_IN_GAG2(x1, 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:

REVERSE_3_IN_GAG2(._22(X, Y), U) -> REVERSE_3_IN_GAG2(Y, ._22(X, U))

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {REVERSE_3_IN_GAG2}.
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: