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



↳ PROLOG
  ↳ PrologToPiTRSProof

reverse2(L, LR) :- revacc3(L, LR, {}0).
revacc3({}0, L, L).
revacc3(.2(EL, T), R, A) :- revacc3(T, R, .2(EL, A)).


With regard to the inferred argument filtering the predicates were used in the following modes:
reverse2: (b,f)
revacc3: (b,f,b)
Transforming PROLOG into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:


reverse_2_in_ga2(L, LR) -> if_reverse_2_in_1_ga3(L, LR, revacc_3_in_gag3(L, LR, []_0))
revacc_3_in_gag3([]_0, L, L) -> revacc_3_out_gag3([]_0, L, L)
revacc_3_in_gag3(._22(EL, T), R, A) -> if_revacc_3_in_1_gag5(EL, T, R, A, revacc_3_in_gag3(T, R, ._22(EL, A)))
if_revacc_3_in_1_gag5(EL, T, R, A, revacc_3_out_gag3(T, R, ._22(EL, A))) -> revacc_3_out_gag3(._22(EL, T), R, A)
if_reverse_2_in_1_ga3(L, LR, revacc_3_out_gag3(L, LR, []_0)) -> reverse_2_out_ga2(L, LR)

The argument filtering Pi contains the following mapping:
reverse_2_in_ga2(x1, x2)  =  reverse_2_in_ga1(x1)
[]_0  =  []_0
._22(x1, x2)  =  ._22(x1, x2)
if_reverse_2_in_1_ga3(x1, x2, x3)  =  if_reverse_2_in_1_ga1(x3)
revacc_3_in_gag3(x1, x2, x3)  =  revacc_3_in_gag2(x1, x3)
revacc_3_out_gag3(x1, x2, x3)  =  revacc_3_out_gag1(x2)
if_revacc_3_in_1_gag5(x1, x2, x3, x4, x5)  =  if_revacc_3_in_1_gag1(x5)
reverse_2_out_ga2(x1, x2)  =  reverse_2_out_ga1(x2)

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_2_in_ga2(L, LR) -> if_reverse_2_in_1_ga3(L, LR, revacc_3_in_gag3(L, LR, []_0))
revacc_3_in_gag3([]_0, L, L) -> revacc_3_out_gag3([]_0, L, L)
revacc_3_in_gag3(._22(EL, T), R, A) -> if_revacc_3_in_1_gag5(EL, T, R, A, revacc_3_in_gag3(T, R, ._22(EL, A)))
if_revacc_3_in_1_gag5(EL, T, R, A, revacc_3_out_gag3(T, R, ._22(EL, A))) -> revacc_3_out_gag3(._22(EL, T), R, A)
if_reverse_2_in_1_ga3(L, LR, revacc_3_out_gag3(L, LR, []_0)) -> reverse_2_out_ga2(L, LR)

The argument filtering Pi contains the following mapping:
reverse_2_in_ga2(x1, x2)  =  reverse_2_in_ga1(x1)
[]_0  =  []_0
._22(x1, x2)  =  ._22(x1, x2)
if_reverse_2_in_1_ga3(x1, x2, x3)  =  if_reverse_2_in_1_ga1(x3)
revacc_3_in_gag3(x1, x2, x3)  =  revacc_3_in_gag2(x1, x3)
revacc_3_out_gag3(x1, x2, x3)  =  revacc_3_out_gag1(x2)
if_revacc_3_in_1_gag5(x1, x2, x3, x4, x5)  =  if_revacc_3_in_1_gag1(x5)
reverse_2_out_ga2(x1, x2)  =  reverse_2_out_ga1(x2)


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

REVERSE_2_IN_GA2(L, LR) -> IF_REVERSE_2_IN_1_GA3(L, LR, revacc_3_in_gag3(L, LR, []_0))
REVERSE_2_IN_GA2(L, LR) -> REVACC_3_IN_GAG3(L, LR, []_0)
REVACC_3_IN_GAG3(._22(EL, T), R, A) -> IF_REVACC_3_IN_1_GAG5(EL, T, R, A, revacc_3_in_gag3(T, R, ._22(EL, A)))
REVACC_3_IN_GAG3(._22(EL, T), R, A) -> REVACC_3_IN_GAG3(T, R, ._22(EL, A))

The TRS R consists of the following rules:

reverse_2_in_ga2(L, LR) -> if_reverse_2_in_1_ga3(L, LR, revacc_3_in_gag3(L, LR, []_0))
revacc_3_in_gag3([]_0, L, L) -> revacc_3_out_gag3([]_0, L, L)
revacc_3_in_gag3(._22(EL, T), R, A) -> if_revacc_3_in_1_gag5(EL, T, R, A, revacc_3_in_gag3(T, R, ._22(EL, A)))
if_revacc_3_in_1_gag5(EL, T, R, A, revacc_3_out_gag3(T, R, ._22(EL, A))) -> revacc_3_out_gag3(._22(EL, T), R, A)
if_reverse_2_in_1_ga3(L, LR, revacc_3_out_gag3(L, LR, []_0)) -> reverse_2_out_ga2(L, LR)

The argument filtering Pi contains the following mapping:
reverse_2_in_ga2(x1, x2)  =  reverse_2_in_ga1(x1)
[]_0  =  []_0
._22(x1, x2)  =  ._22(x1, x2)
if_reverse_2_in_1_ga3(x1, x2, x3)  =  if_reverse_2_in_1_ga1(x3)
revacc_3_in_gag3(x1, x2, x3)  =  revacc_3_in_gag2(x1, x3)
revacc_3_out_gag3(x1, x2, x3)  =  revacc_3_out_gag1(x2)
if_revacc_3_in_1_gag5(x1, x2, x3, x4, x5)  =  if_revacc_3_in_1_gag1(x5)
reverse_2_out_ga2(x1, x2)  =  reverse_2_out_ga1(x2)
REVACC_3_IN_GAG3(x1, x2, x3)  =  REVACC_3_IN_GAG2(x1, x3)
REVERSE_2_IN_GA2(x1, x2)  =  REVERSE_2_IN_GA1(x1)
IF_REVERSE_2_IN_1_GA3(x1, x2, x3)  =  IF_REVERSE_2_IN_1_GA1(x3)
IF_REVACC_3_IN_1_GAG5(x1, x2, x3, x4, x5)  =  IF_REVACC_3_IN_1_GAG1(x5)

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_2_IN_GA2(L, LR) -> IF_REVERSE_2_IN_1_GA3(L, LR, revacc_3_in_gag3(L, LR, []_0))
REVERSE_2_IN_GA2(L, LR) -> REVACC_3_IN_GAG3(L, LR, []_0)
REVACC_3_IN_GAG3(._22(EL, T), R, A) -> IF_REVACC_3_IN_1_GAG5(EL, T, R, A, revacc_3_in_gag3(T, R, ._22(EL, A)))
REVACC_3_IN_GAG3(._22(EL, T), R, A) -> REVACC_3_IN_GAG3(T, R, ._22(EL, A))

The TRS R consists of the following rules:

reverse_2_in_ga2(L, LR) -> if_reverse_2_in_1_ga3(L, LR, revacc_3_in_gag3(L, LR, []_0))
revacc_3_in_gag3([]_0, L, L) -> revacc_3_out_gag3([]_0, L, L)
revacc_3_in_gag3(._22(EL, T), R, A) -> if_revacc_3_in_1_gag5(EL, T, R, A, revacc_3_in_gag3(T, R, ._22(EL, A)))
if_revacc_3_in_1_gag5(EL, T, R, A, revacc_3_out_gag3(T, R, ._22(EL, A))) -> revacc_3_out_gag3(._22(EL, T), R, A)
if_reverse_2_in_1_ga3(L, LR, revacc_3_out_gag3(L, LR, []_0)) -> reverse_2_out_ga2(L, LR)

The argument filtering Pi contains the following mapping:
reverse_2_in_ga2(x1, x2)  =  reverse_2_in_ga1(x1)
[]_0  =  []_0
._22(x1, x2)  =  ._22(x1, x2)
if_reverse_2_in_1_ga3(x1, x2, x3)  =  if_reverse_2_in_1_ga1(x3)
revacc_3_in_gag3(x1, x2, x3)  =  revacc_3_in_gag2(x1, x3)
revacc_3_out_gag3(x1, x2, x3)  =  revacc_3_out_gag1(x2)
if_revacc_3_in_1_gag5(x1, x2, x3, x4, x5)  =  if_revacc_3_in_1_gag1(x5)
reverse_2_out_ga2(x1, x2)  =  reverse_2_out_ga1(x2)
REVACC_3_IN_GAG3(x1, x2, x3)  =  REVACC_3_IN_GAG2(x1, x3)
REVERSE_2_IN_GA2(x1, x2)  =  REVERSE_2_IN_GA1(x1)
IF_REVERSE_2_IN_1_GA3(x1, x2, x3)  =  IF_REVERSE_2_IN_1_GA1(x3)
IF_REVACC_3_IN_1_GAG5(x1, x2, x3, x4, x5)  =  IF_REVACC_3_IN_1_GAG1(x5)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph contains 1 SCC with 3 less nodes.

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

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

REVACC_3_IN_GAG3(._22(EL, T), R, A) -> REVACC_3_IN_GAG3(T, R, ._22(EL, A))

The TRS R consists of the following rules:

reverse_2_in_ga2(L, LR) -> if_reverse_2_in_1_ga3(L, LR, revacc_3_in_gag3(L, LR, []_0))
revacc_3_in_gag3([]_0, L, L) -> revacc_3_out_gag3([]_0, L, L)
revacc_3_in_gag3(._22(EL, T), R, A) -> if_revacc_3_in_1_gag5(EL, T, R, A, revacc_3_in_gag3(T, R, ._22(EL, A)))
if_revacc_3_in_1_gag5(EL, T, R, A, revacc_3_out_gag3(T, R, ._22(EL, A))) -> revacc_3_out_gag3(._22(EL, T), R, A)
if_reverse_2_in_1_ga3(L, LR, revacc_3_out_gag3(L, LR, []_0)) -> reverse_2_out_ga2(L, LR)

The argument filtering Pi contains the following mapping:
reverse_2_in_ga2(x1, x2)  =  reverse_2_in_ga1(x1)
[]_0  =  []_0
._22(x1, x2)  =  ._22(x1, x2)
if_reverse_2_in_1_ga3(x1, x2, x3)  =  if_reverse_2_in_1_ga1(x3)
revacc_3_in_gag3(x1, x2, x3)  =  revacc_3_in_gag2(x1, x3)
revacc_3_out_gag3(x1, x2, x3)  =  revacc_3_out_gag1(x2)
if_revacc_3_in_1_gag5(x1, x2, x3, x4, x5)  =  if_revacc_3_in_1_gag1(x5)
reverse_2_out_ga2(x1, x2)  =  reverse_2_out_ga1(x2)
REVACC_3_IN_GAG3(x1, x2, x3)  =  REVACC_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:

REVACC_3_IN_GAG3(._22(EL, T), R, A) -> REVACC_3_IN_GAG3(T, R, ._22(EL, A))

R is empty.
The argument filtering Pi contains the following mapping:
._22(x1, x2)  =  ._22(x1, x2)
REVACC_3_IN_GAG3(x1, x2, x3)  =  REVACC_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:

REVACC_3_IN_GAG2(._22(EL, T), A) -> REVACC_3_IN_GAG2(T, ._22(EL, A))

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