Left Termination of the query pattern palindrome_in_1(g) w.r.t. the given Prolog program could successfully be proven:



↳ Prolog
  ↳ PrologToPiTRSProof

Clauses:

palindrome(Xs) :- reverse(Xs, Xs).
reverse(X1s, X2s) :- reverse3(X1s, [], X2s).
reverse3(.(X, X1s), X2s, Ys) :- reverse3(X1s, .(X, X2s), Ys).
reverse3([], Xs, Xs).

Queries:

palindrome(g).

We use the technique of [30]. With regard to the inferred argument filtering the predicates were used in the following modes:
palindrome_in: (b)
reverse_in: (b,b)
reverse3_in: (b,b,b)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

palindrome_in_g(Xs) → U1_g(Xs, reverse_in_gg(Xs, Xs))
reverse_in_gg(X1s, X2s) → U2_gg(X1s, X2s, reverse3_in_ggg(X1s, [], X2s))
reverse3_in_ggg(.(X, X1s), X2s, Ys) → U3_ggg(X, X1s, X2s, Ys, reverse3_in_ggg(X1s, .(X, X2s), Ys))
reverse3_in_ggg([], Xs, Xs) → reverse3_out_ggg([], Xs, Xs)
U3_ggg(X, X1s, X2s, Ys, reverse3_out_ggg(X1s, .(X, X2s), Ys)) → reverse3_out_ggg(.(X, X1s), X2s, Ys)
U2_gg(X1s, X2s, reverse3_out_ggg(X1s, [], X2s)) → reverse_out_gg(X1s, X2s)
U1_g(Xs, reverse_out_gg(Xs, Xs)) → palindrome_out_g(Xs)

The argument filtering Pi contains the following mapping:
palindrome_in_g(x1)  =  palindrome_in_g(x1)
U1_g(x1, x2)  =  U1_g(x2)
reverse_in_gg(x1, x2)  =  reverse_in_gg(x1, x2)
U2_gg(x1, x2, x3)  =  U2_gg(x3)
reverse3_in_ggg(x1, x2, x3)  =  reverse3_in_ggg(x1, x2, x3)
.(x1, x2)  =  .(x1, x2)
U3_ggg(x1, x2, x3, x4, x5)  =  U3_ggg(x5)
[]  =  []
reverse3_out_ggg(x1, x2, x3)  =  reverse3_out_ggg
reverse_out_gg(x1, x2)  =  reverse_out_gg
palindrome_out_g(x1)  =  palindrome_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_in_g(Xs) → U1_g(Xs, reverse_in_gg(Xs, Xs))
reverse_in_gg(X1s, X2s) → U2_gg(X1s, X2s, reverse3_in_ggg(X1s, [], X2s))
reverse3_in_ggg(.(X, X1s), X2s, Ys) → U3_ggg(X, X1s, X2s, Ys, reverse3_in_ggg(X1s, .(X, X2s), Ys))
reverse3_in_ggg([], Xs, Xs) → reverse3_out_ggg([], Xs, Xs)
U3_ggg(X, X1s, X2s, Ys, reverse3_out_ggg(X1s, .(X, X2s), Ys)) → reverse3_out_ggg(.(X, X1s), X2s, Ys)
U2_gg(X1s, X2s, reverse3_out_ggg(X1s, [], X2s)) → reverse_out_gg(X1s, X2s)
U1_g(Xs, reverse_out_gg(Xs, Xs)) → palindrome_out_g(Xs)

The argument filtering Pi contains the following mapping:
palindrome_in_g(x1)  =  palindrome_in_g(x1)
U1_g(x1, x2)  =  U1_g(x2)
reverse_in_gg(x1, x2)  =  reverse_in_gg(x1, x2)
U2_gg(x1, x2, x3)  =  U2_gg(x3)
reverse3_in_ggg(x1, x2, x3)  =  reverse3_in_ggg(x1, x2, x3)
.(x1, x2)  =  .(x1, x2)
U3_ggg(x1, x2, x3, x4, x5)  =  U3_ggg(x5)
[]  =  []
reverse3_out_ggg(x1, x2, x3)  =  reverse3_out_ggg
reverse_out_gg(x1, x2)  =  reverse_out_gg
palindrome_out_g(x1)  =  palindrome_out_g


Using Dependency Pairs [1,30] we result in the following initial DP problem:
Pi DP problem:
The TRS P consists of the following rules:

PALINDROME_IN_G(Xs) → U1_G(Xs, reverse_in_gg(Xs, Xs))
PALINDROME_IN_G(Xs) → REVERSE_IN_GG(Xs, Xs)
REVERSE_IN_GG(X1s, X2s) → U2_GG(X1s, X2s, reverse3_in_ggg(X1s, [], X2s))
REVERSE_IN_GG(X1s, X2s) → REVERSE3_IN_GGG(X1s, [], X2s)
REVERSE3_IN_GGG(.(X, X1s), X2s, Ys) → U3_GGG(X, X1s, X2s, Ys, reverse3_in_ggg(X1s, .(X, X2s), Ys))
REVERSE3_IN_GGG(.(X, X1s), X2s, Ys) → REVERSE3_IN_GGG(X1s, .(X, X2s), Ys)

The TRS R consists of the following rules:

palindrome_in_g(Xs) → U1_g(Xs, reverse_in_gg(Xs, Xs))
reverse_in_gg(X1s, X2s) → U2_gg(X1s, X2s, reverse3_in_ggg(X1s, [], X2s))
reverse3_in_ggg(.(X, X1s), X2s, Ys) → U3_ggg(X, X1s, X2s, Ys, reverse3_in_ggg(X1s, .(X, X2s), Ys))
reverse3_in_ggg([], Xs, Xs) → reverse3_out_ggg([], Xs, Xs)
U3_ggg(X, X1s, X2s, Ys, reverse3_out_ggg(X1s, .(X, X2s), Ys)) → reverse3_out_ggg(.(X, X1s), X2s, Ys)
U2_gg(X1s, X2s, reverse3_out_ggg(X1s, [], X2s)) → reverse_out_gg(X1s, X2s)
U1_g(Xs, reverse_out_gg(Xs, Xs)) → palindrome_out_g(Xs)

The argument filtering Pi contains the following mapping:
palindrome_in_g(x1)  =  palindrome_in_g(x1)
U1_g(x1, x2)  =  U1_g(x2)
reverse_in_gg(x1, x2)  =  reverse_in_gg(x1, x2)
U2_gg(x1, x2, x3)  =  U2_gg(x3)
reverse3_in_ggg(x1, x2, x3)  =  reverse3_in_ggg(x1, x2, x3)
.(x1, x2)  =  .(x1, x2)
U3_ggg(x1, x2, x3, x4, x5)  =  U3_ggg(x5)
[]  =  []
reverse3_out_ggg(x1, x2, x3)  =  reverse3_out_ggg
reverse_out_gg(x1, x2)  =  reverse_out_gg
palindrome_out_g(x1)  =  palindrome_out_g
U3_GGG(x1, x2, x3, x4, x5)  =  U3_GGG(x5)
REVERSE3_IN_GGG(x1, x2, x3)  =  REVERSE3_IN_GGG(x1, x2, x3)
U1_G(x1, x2)  =  U1_G(x2)
PALINDROME_IN_G(x1)  =  PALINDROME_IN_G(x1)
U2_GG(x1, x2, x3)  =  U2_GG(x3)
REVERSE_IN_GG(x1, x2)  =  REVERSE_IN_GG(x1, x2)

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_IN_G(Xs) → U1_G(Xs, reverse_in_gg(Xs, Xs))
PALINDROME_IN_G(Xs) → REVERSE_IN_GG(Xs, Xs)
REVERSE_IN_GG(X1s, X2s) → U2_GG(X1s, X2s, reverse3_in_ggg(X1s, [], X2s))
REVERSE_IN_GG(X1s, X2s) → REVERSE3_IN_GGG(X1s, [], X2s)
REVERSE3_IN_GGG(.(X, X1s), X2s, Ys) → U3_GGG(X, X1s, X2s, Ys, reverse3_in_ggg(X1s, .(X, X2s), Ys))
REVERSE3_IN_GGG(.(X, X1s), X2s, Ys) → REVERSE3_IN_GGG(X1s, .(X, X2s), Ys)

The TRS R consists of the following rules:

palindrome_in_g(Xs) → U1_g(Xs, reverse_in_gg(Xs, Xs))
reverse_in_gg(X1s, X2s) → U2_gg(X1s, X2s, reverse3_in_ggg(X1s, [], X2s))
reverse3_in_ggg(.(X, X1s), X2s, Ys) → U3_ggg(X, X1s, X2s, Ys, reverse3_in_ggg(X1s, .(X, X2s), Ys))
reverse3_in_ggg([], Xs, Xs) → reverse3_out_ggg([], Xs, Xs)
U3_ggg(X, X1s, X2s, Ys, reverse3_out_ggg(X1s, .(X, X2s), Ys)) → reverse3_out_ggg(.(X, X1s), X2s, Ys)
U2_gg(X1s, X2s, reverse3_out_ggg(X1s, [], X2s)) → reverse_out_gg(X1s, X2s)
U1_g(Xs, reverse_out_gg(Xs, Xs)) → palindrome_out_g(Xs)

The argument filtering Pi contains the following mapping:
palindrome_in_g(x1)  =  palindrome_in_g(x1)
U1_g(x1, x2)  =  U1_g(x2)
reverse_in_gg(x1, x2)  =  reverse_in_gg(x1, x2)
U2_gg(x1, x2, x3)  =  U2_gg(x3)
reverse3_in_ggg(x1, x2, x3)  =  reverse3_in_ggg(x1, x2, x3)
.(x1, x2)  =  .(x1, x2)
U3_ggg(x1, x2, x3, x4, x5)  =  U3_ggg(x5)
[]  =  []
reverse3_out_ggg(x1, x2, x3)  =  reverse3_out_ggg
reverse_out_gg(x1, x2)  =  reverse_out_gg
palindrome_out_g(x1)  =  palindrome_out_g
U3_GGG(x1, x2, x3, x4, x5)  =  U3_GGG(x5)
REVERSE3_IN_GGG(x1, x2, x3)  =  REVERSE3_IN_GGG(x1, x2, x3)
U1_G(x1, x2)  =  U1_G(x2)
PALINDROME_IN_G(x1)  =  PALINDROME_IN_G(x1)
U2_GG(x1, x2, x3)  =  U2_GG(x3)
REVERSE_IN_GG(x1, x2)  =  REVERSE_IN_GG(x1, x2)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] 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_IN_GGG(.(X, X1s), X2s, Ys) → REVERSE3_IN_GGG(X1s, .(X, X2s), Ys)

The TRS R consists of the following rules:

palindrome_in_g(Xs) → U1_g(Xs, reverse_in_gg(Xs, Xs))
reverse_in_gg(X1s, X2s) → U2_gg(X1s, X2s, reverse3_in_ggg(X1s, [], X2s))
reverse3_in_ggg(.(X, X1s), X2s, Ys) → U3_ggg(X, X1s, X2s, Ys, reverse3_in_ggg(X1s, .(X, X2s), Ys))
reverse3_in_ggg([], Xs, Xs) → reverse3_out_ggg([], Xs, Xs)
U3_ggg(X, X1s, X2s, Ys, reverse3_out_ggg(X1s, .(X, X2s), Ys)) → reverse3_out_ggg(.(X, X1s), X2s, Ys)
U2_gg(X1s, X2s, reverse3_out_ggg(X1s, [], X2s)) → reverse_out_gg(X1s, X2s)
U1_g(Xs, reverse_out_gg(Xs, Xs)) → palindrome_out_g(Xs)

The argument filtering Pi contains the following mapping:
palindrome_in_g(x1)  =  palindrome_in_g(x1)
U1_g(x1, x2)  =  U1_g(x2)
reverse_in_gg(x1, x2)  =  reverse_in_gg(x1, x2)
U2_gg(x1, x2, x3)  =  U2_gg(x3)
reverse3_in_ggg(x1, x2, x3)  =  reverse3_in_ggg(x1, x2, x3)
.(x1, x2)  =  .(x1, x2)
U3_ggg(x1, x2, x3, x4, x5)  =  U3_ggg(x5)
[]  =  []
reverse3_out_ggg(x1, x2, x3)  =  reverse3_out_ggg
reverse_out_gg(x1, x2)  =  reverse_out_gg
palindrome_out_g(x1)  =  palindrome_out_g
REVERSE3_IN_GGG(x1, x2, x3)  =  REVERSE3_IN_GGG(x1, x2, x3)

We have to consider all (P,R,Pi)-chains
For (infinitary) constructor rewriting [30] 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_IN_GGG(.(X, X1s), X2s, Ys) → REVERSE3_IN_GGG(X1s, .(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 [30] into ordinary QDP problem [15] 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_IN_GGG(.(X, X1s), X2s, Ys) → REVERSE3_IN_GGG(X1s, .(X, X2s), Ys)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] 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: