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

0 Prolog
  ↳1 PrologToPiTRSProof (⇐)
    ↳2 PiTRS
    ↳3 DependencyPairsProof (⇔)
    ↳4 PiDP
    ↳5 DependencyGraphProof (⇔)
    ↳6 PiDP
    ↳7 UsableRulesProof (⇔)
    ↳8 PiDP
    ↳9 PiDPToQDPProof (⇔)
    ↳10 QDP
    ↳11 QDPSizeChangeProof (⇔)
    ↳12 TRUE
  ↳13 PrologToPiTRSProof (⇐)
    ↳14 PiTRS
    ↳15 DependencyPairsProof (⇔)
    ↳16 PiDP
    ↳17 DependencyGraphProof (⇔)
    ↳18 PiDP
    ↳19 UsableRulesProof (⇔)
    ↳20 PiDP
    ↳21 PiDPToQDPProof (⇔)
    ↳22 QDP

(0) Obligation:

Clauses:

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

Queries:

palindrome(g).

(1) PrologToPiTRSProof (SOUND transformation)

We use the technique of [LOPSTR]. With regard to the inferred argument filtering the predicates were used in the following modes:
palindrome_in: (b)
reverse_in: (b,b)
reverse_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, reverse_in_ggg(X1s, [], X2s))
reverse_in_ggg([], Xs, Xs) → reverse_out_ggg([], Xs, Xs)
reverse_in_ggg(.(X, X1s), X2s, Ys) → U3_ggg(X, X1s, X2s, Ys, reverse_in_ggg(X1s, .(X, X2s), Ys))
U3_ggg(X, X1s, X2s, Ys, reverse_out_ggg(X1s, .(X, X2s), Ys)) → reverse_out_ggg(.(X, X1s), X2s, Ys)
U2_gg(X1s, X2s, reverse_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)
reverse_in_ggg(x1, x2, x3)  =  reverse_in_ggg(x1, x2, x3)
[]  =  []
reverse_out_ggg(x1, x2, x3)  =  reverse_out_ggg
.(x1, x2)  =  .(x1, x2)
U3_ggg(x1, x2, x3, x4, x5)  =  U3_ggg(x5)
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

(2) Obligation:

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, reverse_in_ggg(X1s, [], X2s))
reverse_in_ggg([], Xs, Xs) → reverse_out_ggg([], Xs, Xs)
reverse_in_ggg(.(X, X1s), X2s, Ys) → U3_ggg(X, X1s, X2s, Ys, reverse_in_ggg(X1s, .(X, X2s), Ys))
U3_ggg(X, X1s, X2s, Ys, reverse_out_ggg(X1s, .(X, X2s), Ys)) → reverse_out_ggg(.(X, X1s), X2s, Ys)
U2_gg(X1s, X2s, reverse_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)
reverse_in_ggg(x1, x2, x3)  =  reverse_in_ggg(x1, x2, x3)
[]  =  []
reverse_out_ggg(x1, x2, x3)  =  reverse_out_ggg
.(x1, x2)  =  .(x1, x2)
U3_ggg(x1, x2, x3, x4, x5)  =  U3_ggg(x5)
reverse_out_gg(x1, x2)  =  reverse_out_gg
palindrome_out_g(x1)  =  palindrome_out_g

(3) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LOPSTR] 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, reverse_in_ggg(X1s, [], X2s))
REVERSE_IN_GG(X1s, X2s) → REVERSE_IN_GGG(X1s, [], X2s)
REVERSE_IN_GGG(.(X, X1s), X2s, Ys) → U3_GGG(X, X1s, X2s, Ys, reverse_in_ggg(X1s, .(X, X2s), Ys))
REVERSE_IN_GGG(.(X, X1s), X2s, Ys) → REVERSE_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, reverse_in_ggg(X1s, [], X2s))
reverse_in_ggg([], Xs, Xs) → reverse_out_ggg([], Xs, Xs)
reverse_in_ggg(.(X, X1s), X2s, Ys) → U3_ggg(X, X1s, X2s, Ys, reverse_in_ggg(X1s, .(X, X2s), Ys))
U3_ggg(X, X1s, X2s, Ys, reverse_out_ggg(X1s, .(X, X2s), Ys)) → reverse_out_ggg(.(X, X1s), X2s, Ys)
U2_gg(X1s, X2s, reverse_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)
reverse_in_ggg(x1, x2, x3)  =  reverse_in_ggg(x1, x2, x3)
[]  =  []
reverse_out_ggg(x1, x2, x3)  =  reverse_out_ggg
.(x1, x2)  =  .(x1, x2)
U3_ggg(x1, x2, x3, x4, x5)  =  U3_ggg(x5)
reverse_out_gg(x1, x2)  =  reverse_out_gg
palindrome_out_g(x1)  =  palindrome_out_g
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)
REVERSE_IN_GGG(x1, x2, x3)  =  REVERSE_IN_GGG(x1, x2, x3)
U3_GGG(x1, x2, x3, x4, x5)  =  U3_GGG(x5)

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

(4) Obligation:

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, reverse_in_ggg(X1s, [], X2s))
REVERSE_IN_GG(X1s, X2s) → REVERSE_IN_GGG(X1s, [], X2s)
REVERSE_IN_GGG(.(X, X1s), X2s, Ys) → U3_GGG(X, X1s, X2s, Ys, reverse_in_ggg(X1s, .(X, X2s), Ys))
REVERSE_IN_GGG(.(X, X1s), X2s, Ys) → REVERSE_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, reverse_in_ggg(X1s, [], X2s))
reverse_in_ggg([], Xs, Xs) → reverse_out_ggg([], Xs, Xs)
reverse_in_ggg(.(X, X1s), X2s, Ys) → U3_ggg(X, X1s, X2s, Ys, reverse_in_ggg(X1s, .(X, X2s), Ys))
U3_ggg(X, X1s, X2s, Ys, reverse_out_ggg(X1s, .(X, X2s), Ys)) → reverse_out_ggg(.(X, X1s), X2s, Ys)
U2_gg(X1s, X2s, reverse_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)
reverse_in_ggg(x1, x2, x3)  =  reverse_in_ggg(x1, x2, x3)
[]  =  []
reverse_out_ggg(x1, x2, x3)  =  reverse_out_ggg
.(x1, x2)  =  .(x1, x2)
U3_ggg(x1, x2, x3, x4, x5)  =  U3_ggg(x5)
reverse_out_gg(x1, x2)  =  reverse_out_gg
palindrome_out_g(x1)  =  palindrome_out_g
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)
REVERSE_IN_GGG(x1, x2, x3)  =  REVERSE_IN_GGG(x1, x2, x3)
U3_GGG(x1, x2, x3, x4, x5)  =  U3_GGG(x5)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 5 less nodes.

(6) Obligation:

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

REVERSE_IN_GGG(.(X, X1s), X2s, Ys) → REVERSE_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, reverse_in_ggg(X1s, [], X2s))
reverse_in_ggg([], Xs, Xs) → reverse_out_ggg([], Xs, Xs)
reverse_in_ggg(.(X, X1s), X2s, Ys) → U3_ggg(X, X1s, X2s, Ys, reverse_in_ggg(X1s, .(X, X2s), Ys))
U3_ggg(X, X1s, X2s, Ys, reverse_out_ggg(X1s, .(X, X2s), Ys)) → reverse_out_ggg(.(X, X1s), X2s, Ys)
U2_gg(X1s, X2s, reverse_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)
reverse_in_ggg(x1, x2, x3)  =  reverse_in_ggg(x1, x2, x3)
[]  =  []
reverse_out_ggg(x1, x2, x3)  =  reverse_out_ggg
.(x1, x2)  =  .(x1, x2)
U3_ggg(x1, x2, x3, x4, x5)  =  U3_ggg(x5)
reverse_out_gg(x1, x2)  =  reverse_out_gg
palindrome_out_g(x1)  =  palindrome_out_g
REVERSE_IN_GGG(x1, x2, x3)  =  REVERSE_IN_GGG(x1, x2, x3)

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

(7) UsableRulesProof (EQUIVALENT transformation)

For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.

(8) Obligation:

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

REVERSE_IN_GGG(.(X, X1s), X2s, Ys) → REVERSE_IN_GGG(X1s, .(X, X2s), Ys)

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

(9) PiDPToQDPProof (EQUIVALENT transformation)

Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.

(10) Obligation:

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

REVERSE_IN_GGG(.(X, X1s), X2s, Ys) → REVERSE_IN_GGG(X1s, .(X, X2s), Ys)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.

(11) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] 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:

  • REVERSE_IN_GGG(.(X, X1s), X2s, Ys) → REVERSE_IN_GGG(X1s, .(X, X2s), Ys)
    The graph contains the following edges 1 > 1, 3 >= 3

(12) TRUE

(13) PrologToPiTRSProof (SOUND transformation)

We use the technique of [LOPSTR]. With regard to the inferred argument filtering the predicates were used in the following modes:
palindrome_in: (b)
reverse_in: (b,b)
reverse_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, reverse_in_ggg(X1s, [], X2s))
reverse_in_ggg([], Xs, Xs) → reverse_out_ggg([], Xs, Xs)
reverse_in_ggg(.(X, X1s), X2s, Ys) → U3_ggg(X, X1s, X2s, Ys, reverse_in_ggg(X1s, .(X, X2s), Ys))
U3_ggg(X, X1s, X2s, Ys, reverse_out_ggg(X1s, .(X, X2s), Ys)) → reverse_out_ggg(.(X, X1s), X2s, Ys)
U2_gg(X1s, X2s, reverse_out_ggg(X1s, [], X2s)) → reverse_out_gg(X1s, X2s)
U1_g(Xs, reverse_out_gg(Xs, Xs)) → palindrome_out_g(Xs)

Pi is empty.

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(14) Obligation:

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, reverse_in_ggg(X1s, [], X2s))
reverse_in_ggg([], Xs, Xs) → reverse_out_ggg([], Xs, Xs)
reverse_in_ggg(.(X, X1s), X2s, Ys) → U3_ggg(X, X1s, X2s, Ys, reverse_in_ggg(X1s, .(X, X2s), Ys))
U3_ggg(X, X1s, X2s, Ys, reverse_out_ggg(X1s, .(X, X2s), Ys)) → reverse_out_ggg(.(X, X1s), X2s, Ys)
U2_gg(X1s, X2s, reverse_out_ggg(X1s, [], X2s)) → reverse_out_gg(X1s, X2s)
U1_g(Xs, reverse_out_gg(Xs, Xs)) → palindrome_out_g(Xs)

Pi is empty.

(15) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LOPSTR] 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, reverse_in_ggg(X1s, [], X2s))
REVERSE_IN_GG(X1s, X2s) → REVERSE_IN_GGG(X1s, [], X2s)
REVERSE_IN_GGG(.(X, X1s), X2s, Ys) → U3_GGG(X, X1s, X2s, Ys, reverse_in_ggg(X1s, .(X, X2s), Ys))
REVERSE_IN_GGG(.(X, X1s), X2s, Ys) → REVERSE_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, reverse_in_ggg(X1s, [], X2s))
reverse_in_ggg([], Xs, Xs) → reverse_out_ggg([], Xs, Xs)
reverse_in_ggg(.(X, X1s), X2s, Ys) → U3_ggg(X, X1s, X2s, Ys, reverse_in_ggg(X1s, .(X, X2s), Ys))
U3_ggg(X, X1s, X2s, Ys, reverse_out_ggg(X1s, .(X, X2s), Ys)) → reverse_out_ggg(.(X, X1s), X2s, Ys)
U2_gg(X1s, X2s, reverse_out_ggg(X1s, [], X2s)) → reverse_out_gg(X1s, X2s)
U1_g(Xs, reverse_out_gg(Xs, Xs)) → palindrome_out_g(Xs)

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

(16) Obligation:

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, reverse_in_ggg(X1s, [], X2s))
REVERSE_IN_GG(X1s, X2s) → REVERSE_IN_GGG(X1s, [], X2s)
REVERSE_IN_GGG(.(X, X1s), X2s, Ys) → U3_GGG(X, X1s, X2s, Ys, reverse_in_ggg(X1s, .(X, X2s), Ys))
REVERSE_IN_GGG(.(X, X1s), X2s, Ys) → REVERSE_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, reverse_in_ggg(X1s, [], X2s))
reverse_in_ggg([], Xs, Xs) → reverse_out_ggg([], Xs, Xs)
reverse_in_ggg(.(X, X1s), X2s, Ys) → U3_ggg(X, X1s, X2s, Ys, reverse_in_ggg(X1s, .(X, X2s), Ys))
U3_ggg(X, X1s, X2s, Ys, reverse_out_ggg(X1s, .(X, X2s), Ys)) → reverse_out_ggg(.(X, X1s), X2s, Ys)
U2_gg(X1s, X2s, reverse_out_ggg(X1s, [], X2s)) → reverse_out_gg(X1s, X2s)
U1_g(Xs, reverse_out_gg(Xs, Xs)) → palindrome_out_g(Xs)

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

(17) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 5 less nodes.

(18) Obligation:

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

REVERSE_IN_GGG(.(X, X1s), X2s, Ys) → REVERSE_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, reverse_in_ggg(X1s, [], X2s))
reverse_in_ggg([], Xs, Xs) → reverse_out_ggg([], Xs, Xs)
reverse_in_ggg(.(X, X1s), X2s, Ys) → U3_ggg(X, X1s, X2s, Ys, reverse_in_ggg(X1s, .(X, X2s), Ys))
U3_ggg(X, X1s, X2s, Ys, reverse_out_ggg(X1s, .(X, X2s), Ys)) → reverse_out_ggg(.(X, X1s), X2s, Ys)
U2_gg(X1s, X2s, reverse_out_ggg(X1s, [], X2s)) → reverse_out_gg(X1s, X2s)
U1_g(Xs, reverse_out_gg(Xs, Xs)) → palindrome_out_g(Xs)

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

(19) UsableRulesProof (EQUIVALENT transformation)

For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.

(20) Obligation:

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

REVERSE_IN_GGG(.(X, X1s), X2s, Ys) → REVERSE_IN_GGG(X1s, .(X, X2s), Ys)

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

(21) PiDPToQDPProof (EQUIVALENT transformation)

Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.

(22) Obligation:

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

REVERSE_IN_GGG(.(X, X1s), X2s, Ys) → REVERSE_IN_GGG(X1s, .(X, X2s), Ys)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.