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



PROLOG
  ↳ PrologToPiTRSProof

palindrome1(Xs) :- reverse2(Xs, Xs).
reverse2(X1s, X2s) :- reverse33(X1s, {}0, X2s).
reverse33(.2(X, X1s), X2s, Ys) :- reverse33(X1s, .2(X, X2s), Ys).
reverse33({}0, Xs, Xs).


With regard to the inferred argument filtering the predicates were used in the following modes:
palindrome1: (b)
reverse2: (b,b)
reverse33: (b,b,b)
Transforming PROLOG into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:


palindrome_1_in_g1(Xs) -> if_palindrome_1_in_1_g2(Xs, reverse_2_in_gg2(Xs, Xs))
reverse_2_in_gg2(X1s, X2s) -> if_reverse_2_in_1_gg3(X1s, X2s, reverse3_3_in_ggg3(X1s, []_0, X2s))
reverse3_3_in_ggg3(._22(X, X1s), X2s, Ys) -> if_reverse3_3_in_1_ggg5(X, X1s, X2s, Ys, reverse3_3_in_ggg3(X1s, ._22(X, X2s), Ys))
reverse3_3_in_ggg3([]_0, Xs, Xs) -> reverse3_3_out_ggg3([]_0, Xs, Xs)
if_reverse3_3_in_1_ggg5(X, X1s, X2s, Ys, reverse3_3_out_ggg3(X1s, ._22(X, X2s), Ys)) -> reverse3_3_out_ggg3(._22(X, X1s), X2s, Ys)
if_reverse_2_in_1_gg3(X1s, X2s, reverse3_3_out_ggg3(X1s, []_0, X2s)) -> reverse_2_out_gg2(X1s, X2s)
if_palindrome_1_in_1_g2(Xs, reverse_2_out_gg2(Xs, Xs)) -> palindrome_1_out_g1(Xs)

The argument filtering Pi contains the following mapping:
palindrome_1_in_g1(x1)  =  palindrome_1_in_g1(x1)
[]_0  =  []_0
._22(x1, x2)  =  ._22(x1, x2)
if_palindrome_1_in_1_g2(x1, x2)  =  if_palindrome_1_in_1_g1(x2)
reverse_2_in_gg2(x1, x2)  =  reverse_2_in_gg2(x1, x2)
if_reverse_2_in_1_gg3(x1, x2, x3)  =  if_reverse_2_in_1_gg1(x3)
reverse3_3_in_ggg3(x1, x2, x3)  =  reverse3_3_in_ggg3(x1, x2, x3)
if_reverse3_3_in_1_ggg5(x1, x2, x3, x4, x5)  =  if_reverse3_3_in_1_ggg1(x5)
reverse3_3_out_ggg3(x1, x2, x3)  =  reverse3_3_out_ggg
reverse_2_out_gg2(x1, x2)  =  reverse_2_out_gg
palindrome_1_out_g1(x1)  =  palindrome_1_out_g

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:

palindrome_1_in_g1(Xs) -> if_palindrome_1_in_1_g2(Xs, reverse_2_in_gg2(Xs, Xs))
reverse_2_in_gg2(X1s, X2s) -> if_reverse_2_in_1_gg3(X1s, X2s, reverse3_3_in_ggg3(X1s, []_0, X2s))
reverse3_3_in_ggg3(._22(X, X1s), X2s, Ys) -> if_reverse3_3_in_1_ggg5(X, X1s, X2s, Ys, reverse3_3_in_ggg3(X1s, ._22(X, X2s), Ys))
reverse3_3_in_ggg3([]_0, Xs, Xs) -> reverse3_3_out_ggg3([]_0, Xs, Xs)
if_reverse3_3_in_1_ggg5(X, X1s, X2s, Ys, reverse3_3_out_ggg3(X1s, ._22(X, X2s), Ys)) -> reverse3_3_out_ggg3(._22(X, X1s), X2s, Ys)
if_reverse_2_in_1_gg3(X1s, X2s, reverse3_3_out_ggg3(X1s, []_0, X2s)) -> reverse_2_out_gg2(X1s, X2s)
if_palindrome_1_in_1_g2(Xs, reverse_2_out_gg2(Xs, Xs)) -> palindrome_1_out_g1(Xs)

The argument filtering Pi contains the following mapping:
palindrome_1_in_g1(x1)  =  palindrome_1_in_g1(x1)
[]_0  =  []_0
._22(x1, x2)  =  ._22(x1, x2)
if_palindrome_1_in_1_g2(x1, x2)  =  if_palindrome_1_in_1_g1(x2)
reverse_2_in_gg2(x1, x2)  =  reverse_2_in_gg2(x1, x2)
if_reverse_2_in_1_gg3(x1, x2, x3)  =  if_reverse_2_in_1_gg1(x3)
reverse3_3_in_ggg3(x1, x2, x3)  =  reverse3_3_in_ggg3(x1, x2, x3)
if_reverse3_3_in_1_ggg5(x1, x2, x3, x4, x5)  =  if_reverse3_3_in_1_ggg1(x5)
reverse3_3_out_ggg3(x1, x2, x3)  =  reverse3_3_out_ggg
reverse_2_out_gg2(x1, x2)  =  reverse_2_out_gg
palindrome_1_out_g1(x1)  =  palindrome_1_out_g


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

PALINDROME_1_IN_G1(Xs) -> IF_PALINDROME_1_IN_1_G2(Xs, reverse_2_in_gg2(Xs, Xs))
PALINDROME_1_IN_G1(Xs) -> REVERSE_2_IN_GG2(Xs, Xs)
REVERSE_2_IN_GG2(X1s, X2s) -> IF_REVERSE_2_IN_1_GG3(X1s, X2s, reverse3_3_in_ggg3(X1s, []_0, X2s))
REVERSE_2_IN_GG2(X1s, X2s) -> REVERSE3_3_IN_GGG3(X1s, []_0, X2s)
REVERSE3_3_IN_GGG3(._22(X, X1s), X2s, Ys) -> IF_REVERSE3_3_IN_1_GGG5(X, X1s, X2s, Ys, reverse3_3_in_ggg3(X1s, ._22(X, X2s), Ys))
REVERSE3_3_IN_GGG3(._22(X, X1s), X2s, Ys) -> REVERSE3_3_IN_GGG3(X1s, ._22(X, X2s), Ys)

The TRS R consists of the following rules:

palindrome_1_in_g1(Xs) -> if_palindrome_1_in_1_g2(Xs, reverse_2_in_gg2(Xs, Xs))
reverse_2_in_gg2(X1s, X2s) -> if_reverse_2_in_1_gg3(X1s, X2s, reverse3_3_in_ggg3(X1s, []_0, X2s))
reverse3_3_in_ggg3(._22(X, X1s), X2s, Ys) -> if_reverse3_3_in_1_ggg5(X, X1s, X2s, Ys, reverse3_3_in_ggg3(X1s, ._22(X, X2s), Ys))
reverse3_3_in_ggg3([]_0, Xs, Xs) -> reverse3_3_out_ggg3([]_0, Xs, Xs)
if_reverse3_3_in_1_ggg5(X, X1s, X2s, Ys, reverse3_3_out_ggg3(X1s, ._22(X, X2s), Ys)) -> reverse3_3_out_ggg3(._22(X, X1s), X2s, Ys)
if_reverse_2_in_1_gg3(X1s, X2s, reverse3_3_out_ggg3(X1s, []_0, X2s)) -> reverse_2_out_gg2(X1s, X2s)
if_palindrome_1_in_1_g2(Xs, reverse_2_out_gg2(Xs, Xs)) -> palindrome_1_out_g1(Xs)

The argument filtering Pi contains the following mapping:
palindrome_1_in_g1(x1)  =  palindrome_1_in_g1(x1)
[]_0  =  []_0
._22(x1, x2)  =  ._22(x1, x2)
if_palindrome_1_in_1_g2(x1, x2)  =  if_palindrome_1_in_1_g1(x2)
reverse_2_in_gg2(x1, x2)  =  reverse_2_in_gg2(x1, x2)
if_reverse_2_in_1_gg3(x1, x2, x3)  =  if_reverse_2_in_1_gg1(x3)
reverse3_3_in_ggg3(x1, x2, x3)  =  reverse3_3_in_ggg3(x1, x2, x3)
if_reverse3_3_in_1_ggg5(x1, x2, x3, x4, x5)  =  if_reverse3_3_in_1_ggg1(x5)
reverse3_3_out_ggg3(x1, x2, x3)  =  reverse3_3_out_ggg
reverse_2_out_gg2(x1, x2)  =  reverse_2_out_gg
palindrome_1_out_g1(x1)  =  palindrome_1_out_g
IF_REVERSE3_3_IN_1_GGG5(x1, x2, x3, x4, x5)  =  IF_REVERSE3_3_IN_1_GGG1(x5)
IF_PALINDROME_1_IN_1_G2(x1, x2)  =  IF_PALINDROME_1_IN_1_G1(x2)
REVERSE_2_IN_GG2(x1, x2)  =  REVERSE_2_IN_GG2(x1, x2)
REVERSE3_3_IN_GGG3(x1, x2, x3)  =  REVERSE3_3_IN_GGG3(x1, x2, x3)
IF_REVERSE_2_IN_1_GG3(x1, x2, x3)  =  IF_REVERSE_2_IN_1_GG1(x3)
PALINDROME_1_IN_G1(x1)  =  PALINDROME_1_IN_G1(x1)

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:

PALINDROME_1_IN_G1(Xs) -> IF_PALINDROME_1_IN_1_G2(Xs, reverse_2_in_gg2(Xs, Xs))
PALINDROME_1_IN_G1(Xs) -> REVERSE_2_IN_GG2(Xs, Xs)
REVERSE_2_IN_GG2(X1s, X2s) -> IF_REVERSE_2_IN_1_GG3(X1s, X2s, reverse3_3_in_ggg3(X1s, []_0, X2s))
REVERSE_2_IN_GG2(X1s, X2s) -> REVERSE3_3_IN_GGG3(X1s, []_0, X2s)
REVERSE3_3_IN_GGG3(._22(X, X1s), X2s, Ys) -> IF_REVERSE3_3_IN_1_GGG5(X, X1s, X2s, Ys, reverse3_3_in_ggg3(X1s, ._22(X, X2s), Ys))
REVERSE3_3_IN_GGG3(._22(X, X1s), X2s, Ys) -> REVERSE3_3_IN_GGG3(X1s, ._22(X, X2s), Ys)

The TRS R consists of the following rules:

palindrome_1_in_g1(Xs) -> if_palindrome_1_in_1_g2(Xs, reverse_2_in_gg2(Xs, Xs))
reverse_2_in_gg2(X1s, X2s) -> if_reverse_2_in_1_gg3(X1s, X2s, reverse3_3_in_ggg3(X1s, []_0, X2s))
reverse3_3_in_ggg3(._22(X, X1s), X2s, Ys) -> if_reverse3_3_in_1_ggg5(X, X1s, X2s, Ys, reverse3_3_in_ggg3(X1s, ._22(X, X2s), Ys))
reverse3_3_in_ggg3([]_0, Xs, Xs) -> reverse3_3_out_ggg3([]_0, Xs, Xs)
if_reverse3_3_in_1_ggg5(X, X1s, X2s, Ys, reverse3_3_out_ggg3(X1s, ._22(X, X2s), Ys)) -> reverse3_3_out_ggg3(._22(X, X1s), X2s, Ys)
if_reverse_2_in_1_gg3(X1s, X2s, reverse3_3_out_ggg3(X1s, []_0, X2s)) -> reverse_2_out_gg2(X1s, X2s)
if_palindrome_1_in_1_g2(Xs, reverse_2_out_gg2(Xs, Xs)) -> palindrome_1_out_g1(Xs)

The argument filtering Pi contains the following mapping:
palindrome_1_in_g1(x1)  =  palindrome_1_in_g1(x1)
[]_0  =  []_0
._22(x1, x2)  =  ._22(x1, x2)
if_palindrome_1_in_1_g2(x1, x2)  =  if_palindrome_1_in_1_g1(x2)
reverse_2_in_gg2(x1, x2)  =  reverse_2_in_gg2(x1, x2)
if_reverse_2_in_1_gg3(x1, x2, x3)  =  if_reverse_2_in_1_gg1(x3)
reverse3_3_in_ggg3(x1, x2, x3)  =  reverse3_3_in_ggg3(x1, x2, x3)
if_reverse3_3_in_1_ggg5(x1, x2, x3, x4, x5)  =  if_reverse3_3_in_1_ggg1(x5)
reverse3_3_out_ggg3(x1, x2, x3)  =  reverse3_3_out_ggg
reverse_2_out_gg2(x1, x2)  =  reverse_2_out_gg
palindrome_1_out_g1(x1)  =  palindrome_1_out_g
IF_REVERSE3_3_IN_1_GGG5(x1, x2, x3, x4, x5)  =  IF_REVERSE3_3_IN_1_GGG1(x5)
IF_PALINDROME_1_IN_1_G2(x1, x2)  =  IF_PALINDROME_1_IN_1_G1(x2)
REVERSE_2_IN_GG2(x1, x2)  =  REVERSE_2_IN_GG2(x1, x2)
REVERSE3_3_IN_GGG3(x1, x2, x3)  =  REVERSE3_3_IN_GGG3(x1, x2, x3)
IF_REVERSE_2_IN_1_GG3(x1, x2, x3)  =  IF_REVERSE_2_IN_1_GG1(x3)
PALINDROME_1_IN_G1(x1)  =  PALINDROME_1_IN_G1(x1)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph contains 1 SCC with 5 less nodes.

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

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

REVERSE3_3_IN_GGG3(._22(X, X1s), X2s, Ys) -> REVERSE3_3_IN_GGG3(X1s, ._22(X, X2s), Ys)

The TRS R consists of the following rules:

palindrome_1_in_g1(Xs) -> if_palindrome_1_in_1_g2(Xs, reverse_2_in_gg2(Xs, Xs))
reverse_2_in_gg2(X1s, X2s) -> if_reverse_2_in_1_gg3(X1s, X2s, reverse3_3_in_ggg3(X1s, []_0, X2s))
reverse3_3_in_ggg3(._22(X, X1s), X2s, Ys) -> if_reverse3_3_in_1_ggg5(X, X1s, X2s, Ys, reverse3_3_in_ggg3(X1s, ._22(X, X2s), Ys))
reverse3_3_in_ggg3([]_0, Xs, Xs) -> reverse3_3_out_ggg3([]_0, Xs, Xs)
if_reverse3_3_in_1_ggg5(X, X1s, X2s, Ys, reverse3_3_out_ggg3(X1s, ._22(X, X2s), Ys)) -> reverse3_3_out_ggg3(._22(X, X1s), X2s, Ys)
if_reverse_2_in_1_gg3(X1s, X2s, reverse3_3_out_ggg3(X1s, []_0, X2s)) -> reverse_2_out_gg2(X1s, X2s)
if_palindrome_1_in_1_g2(Xs, reverse_2_out_gg2(Xs, Xs)) -> palindrome_1_out_g1(Xs)

The argument filtering Pi contains the following mapping:
palindrome_1_in_g1(x1)  =  palindrome_1_in_g1(x1)
[]_0  =  []_0
._22(x1, x2)  =  ._22(x1, x2)
if_palindrome_1_in_1_g2(x1, x2)  =  if_palindrome_1_in_1_g1(x2)
reverse_2_in_gg2(x1, x2)  =  reverse_2_in_gg2(x1, x2)
if_reverse_2_in_1_gg3(x1, x2, x3)  =  if_reverse_2_in_1_gg1(x3)
reverse3_3_in_ggg3(x1, x2, x3)  =  reverse3_3_in_ggg3(x1, x2, x3)
if_reverse3_3_in_1_ggg5(x1, x2, x3, x4, x5)  =  if_reverse3_3_in_1_ggg1(x5)
reverse3_3_out_ggg3(x1, x2, x3)  =  reverse3_3_out_ggg
reverse_2_out_gg2(x1, x2)  =  reverse_2_out_gg
palindrome_1_out_g1(x1)  =  palindrome_1_out_g
REVERSE3_3_IN_GGG3(x1, x2, x3)  =  REVERSE3_3_IN_GGG3(x1, x2, 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:

REVERSE3_3_IN_GGG3(._22(X, X1s), X2s, Ys) -> REVERSE3_3_IN_GGG3(X1s, ._22(X, X2s), Ys)

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
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ PiDP
              ↳ UsableRulesProof
                ↳ PiDP
                  ↳ PiDPToQDPProof
QDP
                      ↳ QDPSizeChangeProof

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

REVERSE3_3_IN_GGG3(._22(X, X1s), X2s, Ys) -> REVERSE3_3_IN_GGG3(X1s, ._22(X, X2s), Ys)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {REVERSE3_3_IN_GGG3}.
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: