YES Left Termination proof of ../tpdb/LP/SGST06/palindrome.pl
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(L) :- ','(halves(L, X1s, X2s, EvenOdd), ','(eq(EvenOdd, even), eq(X1s, X2s))).
palindrome(L) :- ','(halves(L, X1s, X2s, EvenOdd), ','(eq(EvenOdd, odd), last(X1s, X, X2s))).
halves([], [], [], even).
halves(.(X, []), .(X, []), [], odd).
halves(.(T, .(Y, Xs)), .(T, Ts), .(R, Rs), EvenOdd) :- ','(last(.(Y, Xs), R, Rests), halves(Rests, Ts, Rs, EvenOdd)).
last(.(T, []), T, []).
last(.(H, T), X, .(H, M)) :- last(T, X, M).
eq(X, X).

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)
halves_in: (b,f,f,f)
last_in: (b,f,f) (b,f,b)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

palindrome_in_g(L) → U1_g(L, halves_in_gaaa(L, X1s, X2s, EvenOdd))
halves_in_gaaa([], [], [], even) → halves_out_gaaa([], [], [], even)
halves_in_gaaa(.(X, []), .(X, []), [], odd) → halves_out_gaaa(.(X, []), .(X, []), [], odd)
halves_in_gaaa(.(T, .(Y, Xs)), .(T, Ts), .(R, Rs), EvenOdd) → U6_gaaa(T, Y, Xs, Ts, R, Rs, EvenOdd, last_in_gaa(.(Y, Xs), R, Rests))
last_in_gaa(.(T, []), T, []) → last_out_gaa(.(T, []), T, [])
last_in_gaa(.(H, T), X, .(H, M)) → U8_gaa(H, T, X, M, last_in_gaa(T, X, M))
U8_gaa(H, T, X, M, last_out_gaa(T, X, M)) → last_out_gaa(.(H, T), X, .(H, M))
U6_gaaa(T, Y, Xs, Ts, R, Rs, EvenOdd, last_out_gaa(.(Y, Xs), R, Rests)) → U7_gaaa(T, Y, Xs, Ts, R, Rs, EvenOdd, halves_in_gaaa(Rests, Ts, Rs, EvenOdd))
U7_gaaa(T, Y, Xs, Ts, R, Rs, EvenOdd, halves_out_gaaa(Rests, Ts, Rs, EvenOdd)) → halves_out_gaaa(.(T, .(Y, Xs)), .(T, Ts), .(R, Rs), EvenOdd)
U1_g(L, halves_out_gaaa(L, X1s, X2s, EvenOdd)) → U2_g(L, X1s, X2s, eq_in_gg(EvenOdd, even))
eq_in_gg(X, X) → eq_out_gg(X, X)
U2_g(L, X1s, X2s, eq_out_gg(EvenOdd, even)) → U3_g(L, eq_in_gg(X1s, X2s))
U3_g(L, eq_out_gg(X1s, X2s)) → palindrome_out_g(L)
U1_g(L, halves_out_gaaa(L, X1s, X2s, EvenOdd)) → U4_g(L, X1s, X2s, eq_in_gg(EvenOdd, odd))
U4_g(L, X1s, X2s, eq_out_gg(EvenOdd, odd)) → U5_g(L, last_in_gag(X1s, X, X2s))
last_in_gag(.(T, []), T, []) → last_out_gag(.(T, []), T, [])
last_in_gag(.(H, T), X, .(H, M)) → U8_gag(H, T, X, M, last_in_gag(T, X, M))
U8_gag(H, T, X, M, last_out_gag(T, X, M)) → last_out_gag(.(H, T), X, .(H, M))
U5_g(L, last_out_gag(X1s, X, X2s)) → palindrome_out_g(L)

The argument filtering Pi contains the following mapping:
palindrome_in_g(x1)  =  palindrome_in_g(x1)
U1_g(x1, x2)  =  U1_g(x2)
halves_in_gaaa(x1, x2, x3, x4)  =  halves_in_gaaa(x1)
[]  =  []
halves_out_gaaa(x1, x2, x3, x4)  =  halves_out_gaaa(x2, x3, x4)
.(x1, x2)  =  .(x1, x2)
U6_gaaa(x1, x2, x3, x4, x5, x6, x7, x8)  =  U6_gaaa(x1, x8)
last_in_gaa(x1, x2, x3)  =  last_in_gaa(x1)
last_out_gaa(x1, x2, x3)  =  last_out_gaa(x2, x3)
U8_gaa(x1, x2, x3, x4, x5)  =  U8_gaa(x1, x5)
U7_gaaa(x1, x2, x3, x4, x5, x6, x7, x8)  =  U7_gaaa(x1, x5, x8)
U2_g(x1, x2, x3, x4)  =  U2_g(x2, x3, x4)
eq_in_gg(x1, x2)  =  eq_in_gg(x1, x2)
eq_out_gg(x1, x2)  =  eq_out_gg
even  =  even
U3_g(x1, x2)  =  U3_g(x2)
palindrome_out_g(x1)  =  palindrome_out_g
U4_g(x1, x2, x3, x4)  =  U4_g(x2, x3, x4)
odd  =  odd
U5_g(x1, x2)  =  U5_g(x2)
last_in_gag(x1, x2, x3)  =  last_in_gag(x1, x3)
last_out_gag(x1, x2, x3)  =  last_out_gag(x2)
U8_gag(x1, x2, x3, x4, x5)  =  U8_gag(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:

palindrome_in_g(L) → U1_g(L, halves_in_gaaa(L, X1s, X2s, EvenOdd))
halves_in_gaaa([], [], [], even) → halves_out_gaaa([], [], [], even)
halves_in_gaaa(.(X, []), .(X, []), [], odd) → halves_out_gaaa(.(X, []), .(X, []), [], odd)
halves_in_gaaa(.(T, .(Y, Xs)), .(T, Ts), .(R, Rs), EvenOdd) → U6_gaaa(T, Y, Xs, Ts, R, Rs, EvenOdd, last_in_gaa(.(Y, Xs), R, Rests))
last_in_gaa(.(T, []), T, []) → last_out_gaa(.(T, []), T, [])
last_in_gaa(.(H, T), X, .(H, M)) → U8_gaa(H, T, X, M, last_in_gaa(T, X, M))
U8_gaa(H, T, X, M, last_out_gaa(T, X, M)) → last_out_gaa(.(H, T), X, .(H, M))
U6_gaaa(T, Y, Xs, Ts, R, Rs, EvenOdd, last_out_gaa(.(Y, Xs), R, Rests)) → U7_gaaa(T, Y, Xs, Ts, R, Rs, EvenOdd, halves_in_gaaa(Rests, Ts, Rs, EvenOdd))
U7_gaaa(T, Y, Xs, Ts, R, Rs, EvenOdd, halves_out_gaaa(Rests, Ts, Rs, EvenOdd)) → halves_out_gaaa(.(T, .(Y, Xs)), .(T, Ts), .(R, Rs), EvenOdd)
U1_g(L, halves_out_gaaa(L, X1s, X2s, EvenOdd)) → U2_g(L, X1s, X2s, eq_in_gg(EvenOdd, even))
eq_in_gg(X, X) → eq_out_gg(X, X)
U2_g(L, X1s, X2s, eq_out_gg(EvenOdd, even)) → U3_g(L, eq_in_gg(X1s, X2s))
U3_g(L, eq_out_gg(X1s, X2s)) → palindrome_out_g(L)
U1_g(L, halves_out_gaaa(L, X1s, X2s, EvenOdd)) → U4_g(L, X1s, X2s, eq_in_gg(EvenOdd, odd))
U4_g(L, X1s, X2s, eq_out_gg(EvenOdd, odd)) → U5_g(L, last_in_gag(X1s, X, X2s))
last_in_gag(.(T, []), T, []) → last_out_gag(.(T, []), T, [])
last_in_gag(.(H, T), X, .(H, M)) → U8_gag(H, T, X, M, last_in_gag(T, X, M))
U8_gag(H, T, X, M, last_out_gag(T, X, M)) → last_out_gag(.(H, T), X, .(H, M))
U5_g(L, last_out_gag(X1s, X, X2s)) → palindrome_out_g(L)

The argument filtering Pi contains the following mapping:
palindrome_in_g(x1)  =  palindrome_in_g(x1)
U1_g(x1, x2)  =  U1_g(x2)
halves_in_gaaa(x1, x2, x3, x4)  =  halves_in_gaaa(x1)
[]  =  []
halves_out_gaaa(x1, x2, x3, x4)  =  halves_out_gaaa(x2, x3, x4)
.(x1, x2)  =  .(x1, x2)
U6_gaaa(x1, x2, x3, x4, x5, x6, x7, x8)  =  U6_gaaa(x1, x8)
last_in_gaa(x1, x2, x3)  =  last_in_gaa(x1)
last_out_gaa(x1, x2, x3)  =  last_out_gaa(x2, x3)
U8_gaa(x1, x2, x3, x4, x5)  =  U8_gaa(x1, x5)
U7_gaaa(x1, x2, x3, x4, x5, x6, x7, x8)  =  U7_gaaa(x1, x5, x8)
U2_g(x1, x2, x3, x4)  =  U2_g(x2, x3, x4)
eq_in_gg(x1, x2)  =  eq_in_gg(x1, x2)
eq_out_gg(x1, x2)  =  eq_out_gg
even  =  even
U3_g(x1, x2)  =  U3_g(x2)
palindrome_out_g(x1)  =  palindrome_out_g
U4_g(x1, x2, x3, x4)  =  U4_g(x2, x3, x4)
odd  =  odd
U5_g(x1, x2)  =  U5_g(x2)
last_in_gag(x1, x2, x3)  =  last_in_gag(x1, x3)
last_out_gag(x1, x2, x3)  =  last_out_gag(x2)
U8_gag(x1, x2, x3, x4, x5)  =  U8_gag(x5)


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(L) → U1_G(L, halves_in_gaaa(L, X1s, X2s, EvenOdd))
PALINDROME_IN_G(L) → HALVES_IN_GAAA(L, X1s, X2s, EvenOdd)
HALVES_IN_GAAA(.(T, .(Y, Xs)), .(T, Ts), .(R, Rs), EvenOdd) → U6_GAAA(T, Y, Xs, Ts, R, Rs, EvenOdd, last_in_gaa(.(Y, Xs), R, Rests))
HALVES_IN_GAAA(.(T, .(Y, Xs)), .(T, Ts), .(R, Rs), EvenOdd) → LAST_IN_GAA(.(Y, Xs), R, Rests)
LAST_IN_GAA(.(H, T), X, .(H, M)) → U8_GAA(H, T, X, M, last_in_gaa(T, X, M))
LAST_IN_GAA(.(H, T), X, .(H, M)) → LAST_IN_GAA(T, X, M)
U6_GAAA(T, Y, Xs, Ts, R, Rs, EvenOdd, last_out_gaa(.(Y, Xs), R, Rests)) → U7_GAAA(T, Y, Xs, Ts, R, Rs, EvenOdd, halves_in_gaaa(Rests, Ts, Rs, EvenOdd))
U6_GAAA(T, Y, Xs, Ts, R, Rs, EvenOdd, last_out_gaa(.(Y, Xs), R, Rests)) → HALVES_IN_GAAA(Rests, Ts, Rs, EvenOdd)
U1_G(L, halves_out_gaaa(L, X1s, X2s, EvenOdd)) → U2_G(L, X1s, X2s, eq_in_gg(EvenOdd, even))
U1_G(L, halves_out_gaaa(L, X1s, X2s, EvenOdd)) → EQ_IN_GG(EvenOdd, even)
U2_G(L, X1s, X2s, eq_out_gg(EvenOdd, even)) → U3_G(L, eq_in_gg(X1s, X2s))
U2_G(L, X1s, X2s, eq_out_gg(EvenOdd, even)) → EQ_IN_GG(X1s, X2s)
U1_G(L, halves_out_gaaa(L, X1s, X2s, EvenOdd)) → U4_G(L, X1s, X2s, eq_in_gg(EvenOdd, odd))
U1_G(L, halves_out_gaaa(L, X1s, X2s, EvenOdd)) → EQ_IN_GG(EvenOdd, odd)
U4_G(L, X1s, X2s, eq_out_gg(EvenOdd, odd)) → U5_G(L, last_in_gag(X1s, X, X2s))
U4_G(L, X1s, X2s, eq_out_gg(EvenOdd, odd)) → LAST_IN_GAG(X1s, X, X2s)
LAST_IN_GAG(.(H, T), X, .(H, M)) → U8_GAG(H, T, X, M, last_in_gag(T, X, M))
LAST_IN_GAG(.(H, T), X, .(H, M)) → LAST_IN_GAG(T, X, M)

The TRS R consists of the following rules:

palindrome_in_g(L) → U1_g(L, halves_in_gaaa(L, X1s, X2s, EvenOdd))
halves_in_gaaa([], [], [], even) → halves_out_gaaa([], [], [], even)
halves_in_gaaa(.(X, []), .(X, []), [], odd) → halves_out_gaaa(.(X, []), .(X, []), [], odd)
halves_in_gaaa(.(T, .(Y, Xs)), .(T, Ts), .(R, Rs), EvenOdd) → U6_gaaa(T, Y, Xs, Ts, R, Rs, EvenOdd, last_in_gaa(.(Y, Xs), R, Rests))
last_in_gaa(.(T, []), T, []) → last_out_gaa(.(T, []), T, [])
last_in_gaa(.(H, T), X, .(H, M)) → U8_gaa(H, T, X, M, last_in_gaa(T, X, M))
U8_gaa(H, T, X, M, last_out_gaa(T, X, M)) → last_out_gaa(.(H, T), X, .(H, M))
U6_gaaa(T, Y, Xs, Ts, R, Rs, EvenOdd, last_out_gaa(.(Y, Xs), R, Rests)) → U7_gaaa(T, Y, Xs, Ts, R, Rs, EvenOdd, halves_in_gaaa(Rests, Ts, Rs, EvenOdd))
U7_gaaa(T, Y, Xs, Ts, R, Rs, EvenOdd, halves_out_gaaa(Rests, Ts, Rs, EvenOdd)) → halves_out_gaaa(.(T, .(Y, Xs)), .(T, Ts), .(R, Rs), EvenOdd)
U1_g(L, halves_out_gaaa(L, X1s, X2s, EvenOdd)) → U2_g(L, X1s, X2s, eq_in_gg(EvenOdd, even))
eq_in_gg(X, X) → eq_out_gg(X, X)
U2_g(L, X1s, X2s, eq_out_gg(EvenOdd, even)) → U3_g(L, eq_in_gg(X1s, X2s))
U3_g(L, eq_out_gg(X1s, X2s)) → palindrome_out_g(L)
U1_g(L, halves_out_gaaa(L, X1s, X2s, EvenOdd)) → U4_g(L, X1s, X2s, eq_in_gg(EvenOdd, odd))
U4_g(L, X1s, X2s, eq_out_gg(EvenOdd, odd)) → U5_g(L, last_in_gag(X1s, X, X2s))
last_in_gag(.(T, []), T, []) → last_out_gag(.(T, []), T, [])
last_in_gag(.(H, T), X, .(H, M)) → U8_gag(H, T, X, M, last_in_gag(T, X, M))
U8_gag(H, T, X, M, last_out_gag(T, X, M)) → last_out_gag(.(H, T), X, .(H, M))
U5_g(L, last_out_gag(X1s, X, X2s)) → palindrome_out_g(L)

The argument filtering Pi contains the following mapping:
palindrome_in_g(x1)  =  palindrome_in_g(x1)
U1_g(x1, x2)  =  U1_g(x2)
halves_in_gaaa(x1, x2, x3, x4)  =  halves_in_gaaa(x1)
[]  =  []
halves_out_gaaa(x1, x2, x3, x4)  =  halves_out_gaaa(x2, x3, x4)
.(x1, x2)  =  .(x1, x2)
U6_gaaa(x1, x2, x3, x4, x5, x6, x7, x8)  =  U6_gaaa(x1, x8)
last_in_gaa(x1, x2, x3)  =  last_in_gaa(x1)
last_out_gaa(x1, x2, x3)  =  last_out_gaa(x2, x3)
U8_gaa(x1, x2, x3, x4, x5)  =  U8_gaa(x1, x5)
U7_gaaa(x1, x2, x3, x4, x5, x6, x7, x8)  =  U7_gaaa(x1, x5, x8)
U2_g(x1, x2, x3, x4)  =  U2_g(x2, x3, x4)
eq_in_gg(x1, x2)  =  eq_in_gg(x1, x2)
eq_out_gg(x1, x2)  =  eq_out_gg
even  =  even
U3_g(x1, x2)  =  U3_g(x2)
palindrome_out_g(x1)  =  palindrome_out_g
U4_g(x1, x2, x3, x4)  =  U4_g(x2, x3, x4)
odd  =  odd
U5_g(x1, x2)  =  U5_g(x2)
last_in_gag(x1, x2, x3)  =  last_in_gag(x1, x3)
last_out_gag(x1, x2, x3)  =  last_out_gag(x2)
U8_gag(x1, x2, x3, x4, x5)  =  U8_gag(x5)
U7_GAAA(x1, x2, x3, x4, x5, x6, x7, x8)  =  U7_GAAA(x1, x5, x8)
HALVES_IN_GAAA(x1, x2, x3, x4)  =  HALVES_IN_GAAA(x1)
U1_G(x1, x2)  =  U1_G(x2)
U3_G(x1, x2)  =  U3_G(x2)
U8_GAG(x1, x2, x3, x4, x5)  =  U8_GAG(x5)
EQ_IN_GG(x1, x2)  =  EQ_IN_GG(x1, x2)
U8_GAA(x1, x2, x3, x4, x5)  =  U8_GAA(x1, x5)
LAST_IN_GAG(x1, x2, x3)  =  LAST_IN_GAG(x1, x3)
U6_GAAA(x1, x2, x3, x4, x5, x6, x7, x8)  =  U6_GAAA(x1, x8)
U4_G(x1, x2, x3, x4)  =  U4_G(x2, x3, x4)
PALINDROME_IN_G(x1)  =  PALINDROME_IN_G(x1)
U5_G(x1, x2)  =  U5_G(x2)
LAST_IN_GAA(x1, x2, x3)  =  LAST_IN_GAA(x1)
U2_G(x1, x2, x3, x4)  =  U2_G(x2, x3, x4)

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(L) → U1_G(L, halves_in_gaaa(L, X1s, X2s, EvenOdd))
PALINDROME_IN_G(L) → HALVES_IN_GAAA(L, X1s, X2s, EvenOdd)
HALVES_IN_GAAA(.(T, .(Y, Xs)), .(T, Ts), .(R, Rs), EvenOdd) → U6_GAAA(T, Y, Xs, Ts, R, Rs, EvenOdd, last_in_gaa(.(Y, Xs), R, Rests))
HALVES_IN_GAAA(.(T, .(Y, Xs)), .(T, Ts), .(R, Rs), EvenOdd) → LAST_IN_GAA(.(Y, Xs), R, Rests)
LAST_IN_GAA(.(H, T), X, .(H, M)) → U8_GAA(H, T, X, M, last_in_gaa(T, X, M))
LAST_IN_GAA(.(H, T), X, .(H, M)) → LAST_IN_GAA(T, X, M)
U6_GAAA(T, Y, Xs, Ts, R, Rs, EvenOdd, last_out_gaa(.(Y, Xs), R, Rests)) → U7_GAAA(T, Y, Xs, Ts, R, Rs, EvenOdd, halves_in_gaaa(Rests, Ts, Rs, EvenOdd))
U6_GAAA(T, Y, Xs, Ts, R, Rs, EvenOdd, last_out_gaa(.(Y, Xs), R, Rests)) → HALVES_IN_GAAA(Rests, Ts, Rs, EvenOdd)
U1_G(L, halves_out_gaaa(L, X1s, X2s, EvenOdd)) → U2_G(L, X1s, X2s, eq_in_gg(EvenOdd, even))
U1_G(L, halves_out_gaaa(L, X1s, X2s, EvenOdd)) → EQ_IN_GG(EvenOdd, even)
U2_G(L, X1s, X2s, eq_out_gg(EvenOdd, even)) → U3_G(L, eq_in_gg(X1s, X2s))
U2_G(L, X1s, X2s, eq_out_gg(EvenOdd, even)) → EQ_IN_GG(X1s, X2s)
U1_G(L, halves_out_gaaa(L, X1s, X2s, EvenOdd)) → U4_G(L, X1s, X2s, eq_in_gg(EvenOdd, odd))
U1_G(L, halves_out_gaaa(L, X1s, X2s, EvenOdd)) → EQ_IN_GG(EvenOdd, odd)
U4_G(L, X1s, X2s, eq_out_gg(EvenOdd, odd)) → U5_G(L, last_in_gag(X1s, X, X2s))
U4_G(L, X1s, X2s, eq_out_gg(EvenOdd, odd)) → LAST_IN_GAG(X1s, X, X2s)
LAST_IN_GAG(.(H, T), X, .(H, M)) → U8_GAG(H, T, X, M, last_in_gag(T, X, M))
LAST_IN_GAG(.(H, T), X, .(H, M)) → LAST_IN_GAG(T, X, M)

The TRS R consists of the following rules:

palindrome_in_g(L) → U1_g(L, halves_in_gaaa(L, X1s, X2s, EvenOdd))
halves_in_gaaa([], [], [], even) → halves_out_gaaa([], [], [], even)
halves_in_gaaa(.(X, []), .(X, []), [], odd) → halves_out_gaaa(.(X, []), .(X, []), [], odd)
halves_in_gaaa(.(T, .(Y, Xs)), .(T, Ts), .(R, Rs), EvenOdd) → U6_gaaa(T, Y, Xs, Ts, R, Rs, EvenOdd, last_in_gaa(.(Y, Xs), R, Rests))
last_in_gaa(.(T, []), T, []) → last_out_gaa(.(T, []), T, [])
last_in_gaa(.(H, T), X, .(H, M)) → U8_gaa(H, T, X, M, last_in_gaa(T, X, M))
U8_gaa(H, T, X, M, last_out_gaa(T, X, M)) → last_out_gaa(.(H, T), X, .(H, M))
U6_gaaa(T, Y, Xs, Ts, R, Rs, EvenOdd, last_out_gaa(.(Y, Xs), R, Rests)) → U7_gaaa(T, Y, Xs, Ts, R, Rs, EvenOdd, halves_in_gaaa(Rests, Ts, Rs, EvenOdd))
U7_gaaa(T, Y, Xs, Ts, R, Rs, EvenOdd, halves_out_gaaa(Rests, Ts, Rs, EvenOdd)) → halves_out_gaaa(.(T, .(Y, Xs)), .(T, Ts), .(R, Rs), EvenOdd)
U1_g(L, halves_out_gaaa(L, X1s, X2s, EvenOdd)) → U2_g(L, X1s, X2s, eq_in_gg(EvenOdd, even))
eq_in_gg(X, X) → eq_out_gg(X, X)
U2_g(L, X1s, X2s, eq_out_gg(EvenOdd, even)) → U3_g(L, eq_in_gg(X1s, X2s))
U3_g(L, eq_out_gg(X1s, X2s)) → palindrome_out_g(L)
U1_g(L, halves_out_gaaa(L, X1s, X2s, EvenOdd)) → U4_g(L, X1s, X2s, eq_in_gg(EvenOdd, odd))
U4_g(L, X1s, X2s, eq_out_gg(EvenOdd, odd)) → U5_g(L, last_in_gag(X1s, X, X2s))
last_in_gag(.(T, []), T, []) → last_out_gag(.(T, []), T, [])
last_in_gag(.(H, T), X, .(H, M)) → U8_gag(H, T, X, M, last_in_gag(T, X, M))
U8_gag(H, T, X, M, last_out_gag(T, X, M)) → last_out_gag(.(H, T), X, .(H, M))
U5_g(L, last_out_gag(X1s, X, X2s)) → palindrome_out_g(L)

The argument filtering Pi contains the following mapping:
palindrome_in_g(x1)  =  palindrome_in_g(x1)
U1_g(x1, x2)  =  U1_g(x2)
halves_in_gaaa(x1, x2, x3, x4)  =  halves_in_gaaa(x1)
[]  =  []
halves_out_gaaa(x1, x2, x3, x4)  =  halves_out_gaaa(x2, x3, x4)
.(x1, x2)  =  .(x1, x2)
U6_gaaa(x1, x2, x3, x4, x5, x6, x7, x8)  =  U6_gaaa(x1, x8)
last_in_gaa(x1, x2, x3)  =  last_in_gaa(x1)
last_out_gaa(x1, x2, x3)  =  last_out_gaa(x2, x3)
U8_gaa(x1, x2, x3, x4, x5)  =  U8_gaa(x1, x5)
U7_gaaa(x1, x2, x3, x4, x5, x6, x7, x8)  =  U7_gaaa(x1, x5, x8)
U2_g(x1, x2, x3, x4)  =  U2_g(x2, x3, x4)
eq_in_gg(x1, x2)  =  eq_in_gg(x1, x2)
eq_out_gg(x1, x2)  =  eq_out_gg
even  =  even
U3_g(x1, x2)  =  U3_g(x2)
palindrome_out_g(x1)  =  palindrome_out_g
U4_g(x1, x2, x3, x4)  =  U4_g(x2, x3, x4)
odd  =  odd
U5_g(x1, x2)  =  U5_g(x2)
last_in_gag(x1, x2, x3)  =  last_in_gag(x1, x3)
last_out_gag(x1, x2, x3)  =  last_out_gag(x2)
U8_gag(x1, x2, x3, x4, x5)  =  U8_gag(x5)
U7_GAAA(x1, x2, x3, x4, x5, x6, x7, x8)  =  U7_GAAA(x1, x5, x8)
HALVES_IN_GAAA(x1, x2, x3, x4)  =  HALVES_IN_GAAA(x1)
U1_G(x1, x2)  =  U1_G(x2)
U3_G(x1, x2)  =  U3_G(x2)
U8_GAG(x1, x2, x3, x4, x5)  =  U8_GAG(x5)
EQ_IN_GG(x1, x2)  =  EQ_IN_GG(x1, x2)
U8_GAA(x1, x2, x3, x4, x5)  =  U8_GAA(x1, x5)
LAST_IN_GAG(x1, x2, x3)  =  LAST_IN_GAG(x1, x3)
U6_GAAA(x1, x2, x3, x4, x5, x6, x7, x8)  =  U6_GAAA(x1, x8)
U4_G(x1, x2, x3, x4)  =  U4_G(x2, x3, x4)
PALINDROME_IN_G(x1)  =  PALINDROME_IN_G(x1)
U5_G(x1, x2)  =  U5_G(x2)
LAST_IN_GAA(x1, x2, x3)  =  LAST_IN_GAA(x1)
U2_G(x1, x2, x3, x4)  =  U2_G(x2, x3, x4)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 3 SCCs with 14 less nodes.

↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
                ↳ UsableRulesProof
              ↳ PiDP
              ↳ PiDP

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

LAST_IN_GAG(.(H, T), X, .(H, M)) → LAST_IN_GAG(T, X, M)

The TRS R consists of the following rules:

palindrome_in_g(L) → U1_g(L, halves_in_gaaa(L, X1s, X2s, EvenOdd))
halves_in_gaaa([], [], [], even) → halves_out_gaaa([], [], [], even)
halves_in_gaaa(.(X, []), .(X, []), [], odd) → halves_out_gaaa(.(X, []), .(X, []), [], odd)
halves_in_gaaa(.(T, .(Y, Xs)), .(T, Ts), .(R, Rs), EvenOdd) → U6_gaaa(T, Y, Xs, Ts, R, Rs, EvenOdd, last_in_gaa(.(Y, Xs), R, Rests))
last_in_gaa(.(T, []), T, []) → last_out_gaa(.(T, []), T, [])
last_in_gaa(.(H, T), X, .(H, M)) → U8_gaa(H, T, X, M, last_in_gaa(T, X, M))
U8_gaa(H, T, X, M, last_out_gaa(T, X, M)) → last_out_gaa(.(H, T), X, .(H, M))
U6_gaaa(T, Y, Xs, Ts, R, Rs, EvenOdd, last_out_gaa(.(Y, Xs), R, Rests)) → U7_gaaa(T, Y, Xs, Ts, R, Rs, EvenOdd, halves_in_gaaa(Rests, Ts, Rs, EvenOdd))
U7_gaaa(T, Y, Xs, Ts, R, Rs, EvenOdd, halves_out_gaaa(Rests, Ts, Rs, EvenOdd)) → halves_out_gaaa(.(T, .(Y, Xs)), .(T, Ts), .(R, Rs), EvenOdd)
U1_g(L, halves_out_gaaa(L, X1s, X2s, EvenOdd)) → U2_g(L, X1s, X2s, eq_in_gg(EvenOdd, even))
eq_in_gg(X, X) → eq_out_gg(X, X)
U2_g(L, X1s, X2s, eq_out_gg(EvenOdd, even)) → U3_g(L, eq_in_gg(X1s, X2s))
U3_g(L, eq_out_gg(X1s, X2s)) → palindrome_out_g(L)
U1_g(L, halves_out_gaaa(L, X1s, X2s, EvenOdd)) → U4_g(L, X1s, X2s, eq_in_gg(EvenOdd, odd))
U4_g(L, X1s, X2s, eq_out_gg(EvenOdd, odd)) → U5_g(L, last_in_gag(X1s, X, X2s))
last_in_gag(.(T, []), T, []) → last_out_gag(.(T, []), T, [])
last_in_gag(.(H, T), X, .(H, M)) → U8_gag(H, T, X, M, last_in_gag(T, X, M))
U8_gag(H, T, X, M, last_out_gag(T, X, M)) → last_out_gag(.(H, T), X, .(H, M))
U5_g(L, last_out_gag(X1s, X, X2s)) → palindrome_out_g(L)

The argument filtering Pi contains the following mapping:
palindrome_in_g(x1)  =  palindrome_in_g(x1)
U1_g(x1, x2)  =  U1_g(x2)
halves_in_gaaa(x1, x2, x3, x4)  =  halves_in_gaaa(x1)
[]  =  []
halves_out_gaaa(x1, x2, x3, x4)  =  halves_out_gaaa(x2, x3, x4)
.(x1, x2)  =  .(x1, x2)
U6_gaaa(x1, x2, x3, x4, x5, x6, x7, x8)  =  U6_gaaa(x1, x8)
last_in_gaa(x1, x2, x3)  =  last_in_gaa(x1)
last_out_gaa(x1, x2, x3)  =  last_out_gaa(x2, x3)
U8_gaa(x1, x2, x3, x4, x5)  =  U8_gaa(x1, x5)
U7_gaaa(x1, x2, x3, x4, x5, x6, x7, x8)  =  U7_gaaa(x1, x5, x8)
U2_g(x1, x2, x3, x4)  =  U2_g(x2, x3, x4)
eq_in_gg(x1, x2)  =  eq_in_gg(x1, x2)
eq_out_gg(x1, x2)  =  eq_out_gg
even  =  even
U3_g(x1, x2)  =  U3_g(x2)
palindrome_out_g(x1)  =  palindrome_out_g
U4_g(x1, x2, x3, x4)  =  U4_g(x2, x3, x4)
odd  =  odd
U5_g(x1, x2)  =  U5_g(x2)
last_in_gag(x1, x2, x3)  =  last_in_gag(x1, x3)
last_out_gag(x1, x2, x3)  =  last_out_gag(x2)
U8_gag(x1, x2, x3, x4, x5)  =  U8_gag(x5)
LAST_IN_GAG(x1, x2, x3)  =  LAST_IN_GAG(x1, 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
            ↳ AND
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
              ↳ PiDP
              ↳ PiDP

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

LAST_IN_GAG(.(H, T), X, .(H, M)) → LAST_IN_GAG(T, X, M)

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

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
            ↳ AND
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ QDPSizeChangeProof
              ↳ PiDP
              ↳ PiDP

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

LAST_IN_GAG(.(H, T), .(H, M)) → LAST_IN_GAG(T, M)

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: 



↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
              ↳ PiDP

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

LAST_IN_GAA(.(H, T), X, .(H, M)) → LAST_IN_GAA(T, X, M)

The TRS R consists of the following rules:

palindrome_in_g(L) → U1_g(L, halves_in_gaaa(L, X1s, X2s, EvenOdd))
halves_in_gaaa([], [], [], even) → halves_out_gaaa([], [], [], even)
halves_in_gaaa(.(X, []), .(X, []), [], odd) → halves_out_gaaa(.(X, []), .(X, []), [], odd)
halves_in_gaaa(.(T, .(Y, Xs)), .(T, Ts), .(R, Rs), EvenOdd) → U6_gaaa(T, Y, Xs, Ts, R, Rs, EvenOdd, last_in_gaa(.(Y, Xs), R, Rests))
last_in_gaa(.(T, []), T, []) → last_out_gaa(.(T, []), T, [])
last_in_gaa(.(H, T), X, .(H, M)) → U8_gaa(H, T, X, M, last_in_gaa(T, X, M))
U8_gaa(H, T, X, M, last_out_gaa(T, X, M)) → last_out_gaa(.(H, T), X, .(H, M))
U6_gaaa(T, Y, Xs, Ts, R, Rs, EvenOdd, last_out_gaa(.(Y, Xs), R, Rests)) → U7_gaaa(T, Y, Xs, Ts, R, Rs, EvenOdd, halves_in_gaaa(Rests, Ts, Rs, EvenOdd))
U7_gaaa(T, Y, Xs, Ts, R, Rs, EvenOdd, halves_out_gaaa(Rests, Ts, Rs, EvenOdd)) → halves_out_gaaa(.(T, .(Y, Xs)), .(T, Ts), .(R, Rs), EvenOdd)
U1_g(L, halves_out_gaaa(L, X1s, X2s, EvenOdd)) → U2_g(L, X1s, X2s, eq_in_gg(EvenOdd, even))
eq_in_gg(X, X) → eq_out_gg(X, X)
U2_g(L, X1s, X2s, eq_out_gg(EvenOdd, even)) → U3_g(L, eq_in_gg(X1s, X2s))
U3_g(L, eq_out_gg(X1s, X2s)) → palindrome_out_g(L)
U1_g(L, halves_out_gaaa(L, X1s, X2s, EvenOdd)) → U4_g(L, X1s, X2s, eq_in_gg(EvenOdd, odd))
U4_g(L, X1s, X2s, eq_out_gg(EvenOdd, odd)) → U5_g(L, last_in_gag(X1s, X, X2s))
last_in_gag(.(T, []), T, []) → last_out_gag(.(T, []), T, [])
last_in_gag(.(H, T), X, .(H, M)) → U8_gag(H, T, X, M, last_in_gag(T, X, M))
U8_gag(H, T, X, M, last_out_gag(T, X, M)) → last_out_gag(.(H, T), X, .(H, M))
U5_g(L, last_out_gag(X1s, X, X2s)) → palindrome_out_g(L)

The argument filtering Pi contains the following mapping:
palindrome_in_g(x1)  =  palindrome_in_g(x1)
U1_g(x1, x2)  =  U1_g(x2)
halves_in_gaaa(x1, x2, x3, x4)  =  halves_in_gaaa(x1)
[]  =  []
halves_out_gaaa(x1, x2, x3, x4)  =  halves_out_gaaa(x2, x3, x4)
.(x1, x2)  =  .(x1, x2)
U6_gaaa(x1, x2, x3, x4, x5, x6, x7, x8)  =  U6_gaaa(x1, x8)
last_in_gaa(x1, x2, x3)  =  last_in_gaa(x1)
last_out_gaa(x1, x2, x3)  =  last_out_gaa(x2, x3)
U8_gaa(x1, x2, x3, x4, x5)  =  U8_gaa(x1, x5)
U7_gaaa(x1, x2, x3, x4, x5, x6, x7, x8)  =  U7_gaaa(x1, x5, x8)
U2_g(x1, x2, x3, x4)  =  U2_g(x2, x3, x4)
eq_in_gg(x1, x2)  =  eq_in_gg(x1, x2)
eq_out_gg(x1, x2)  =  eq_out_gg
even  =  even
U3_g(x1, x2)  =  U3_g(x2)
palindrome_out_g(x1)  =  palindrome_out_g
U4_g(x1, x2, x3, x4)  =  U4_g(x2, x3, x4)
odd  =  odd
U5_g(x1, x2)  =  U5_g(x2)
last_in_gag(x1, x2, x3)  =  last_in_gag(x1, x3)
last_out_gag(x1, x2, x3)  =  last_out_gag(x2)
U8_gag(x1, x2, x3, x4, x5)  =  U8_gag(x5)
LAST_IN_GAA(x1, x2, x3)  =  LAST_IN_GAA(x1)

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
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
              ↳ PiDP

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

LAST_IN_GAA(.(H, T), X, .(H, M)) → LAST_IN_GAA(T, X, M)

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

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
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ QDPSizeChangeProof
              ↳ PiDP

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

LAST_IN_GAA(.(H, T)) → LAST_IN_GAA(T)

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: 



↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof

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

U6_GAAA(T, Y, Xs, Ts, R, Rs, EvenOdd, last_out_gaa(.(Y, Xs), R, Rests)) → HALVES_IN_GAAA(Rests, Ts, Rs, EvenOdd)
HALVES_IN_GAAA(.(T, .(Y, Xs)), .(T, Ts), .(R, Rs), EvenOdd) → U6_GAAA(T, Y, Xs, Ts, R, Rs, EvenOdd, last_in_gaa(.(Y, Xs), R, Rests))

The TRS R consists of the following rules:

palindrome_in_g(L) → U1_g(L, halves_in_gaaa(L, X1s, X2s, EvenOdd))
halves_in_gaaa([], [], [], even) → halves_out_gaaa([], [], [], even)
halves_in_gaaa(.(X, []), .(X, []), [], odd) → halves_out_gaaa(.(X, []), .(X, []), [], odd)
halves_in_gaaa(.(T, .(Y, Xs)), .(T, Ts), .(R, Rs), EvenOdd) → U6_gaaa(T, Y, Xs, Ts, R, Rs, EvenOdd, last_in_gaa(.(Y, Xs), R, Rests))
last_in_gaa(.(T, []), T, []) → last_out_gaa(.(T, []), T, [])
last_in_gaa(.(H, T), X, .(H, M)) → U8_gaa(H, T, X, M, last_in_gaa(T, X, M))
U8_gaa(H, T, X, M, last_out_gaa(T, X, M)) → last_out_gaa(.(H, T), X, .(H, M))
U6_gaaa(T, Y, Xs, Ts, R, Rs, EvenOdd, last_out_gaa(.(Y, Xs), R, Rests)) → U7_gaaa(T, Y, Xs, Ts, R, Rs, EvenOdd, halves_in_gaaa(Rests, Ts, Rs, EvenOdd))
U7_gaaa(T, Y, Xs, Ts, R, Rs, EvenOdd, halves_out_gaaa(Rests, Ts, Rs, EvenOdd)) → halves_out_gaaa(.(T, .(Y, Xs)), .(T, Ts), .(R, Rs), EvenOdd)
U1_g(L, halves_out_gaaa(L, X1s, X2s, EvenOdd)) → U2_g(L, X1s, X2s, eq_in_gg(EvenOdd, even))
eq_in_gg(X, X) → eq_out_gg(X, X)
U2_g(L, X1s, X2s, eq_out_gg(EvenOdd, even)) → U3_g(L, eq_in_gg(X1s, X2s))
U3_g(L, eq_out_gg(X1s, X2s)) → palindrome_out_g(L)
U1_g(L, halves_out_gaaa(L, X1s, X2s, EvenOdd)) → U4_g(L, X1s, X2s, eq_in_gg(EvenOdd, odd))
U4_g(L, X1s, X2s, eq_out_gg(EvenOdd, odd)) → U5_g(L, last_in_gag(X1s, X, X2s))
last_in_gag(.(T, []), T, []) → last_out_gag(.(T, []), T, [])
last_in_gag(.(H, T), X, .(H, M)) → U8_gag(H, T, X, M, last_in_gag(T, X, M))
U8_gag(H, T, X, M, last_out_gag(T, X, M)) → last_out_gag(.(H, T), X, .(H, M))
U5_g(L, last_out_gag(X1s, X, X2s)) → palindrome_out_g(L)

The argument filtering Pi contains the following mapping:
palindrome_in_g(x1)  =  palindrome_in_g(x1)
U1_g(x1, x2)  =  U1_g(x2)
halves_in_gaaa(x1, x2, x3, x4)  =  halves_in_gaaa(x1)
[]  =  []
halves_out_gaaa(x1, x2, x3, x4)  =  halves_out_gaaa(x2, x3, x4)
.(x1, x2)  =  .(x1, x2)
U6_gaaa(x1, x2, x3, x4, x5, x6, x7, x8)  =  U6_gaaa(x1, x8)
last_in_gaa(x1, x2, x3)  =  last_in_gaa(x1)
last_out_gaa(x1, x2, x3)  =  last_out_gaa(x2, x3)
U8_gaa(x1, x2, x3, x4, x5)  =  U8_gaa(x1, x5)
U7_gaaa(x1, x2, x3, x4, x5, x6, x7, x8)  =  U7_gaaa(x1, x5, x8)
U2_g(x1, x2, x3, x4)  =  U2_g(x2, x3, x4)
eq_in_gg(x1, x2)  =  eq_in_gg(x1, x2)
eq_out_gg(x1, x2)  =  eq_out_gg
even  =  even
U3_g(x1, x2)  =  U3_g(x2)
palindrome_out_g(x1)  =  palindrome_out_g
U4_g(x1, x2, x3, x4)  =  U4_g(x2, x3, x4)
odd  =  odd
U5_g(x1, x2)  =  U5_g(x2)
last_in_gag(x1, x2, x3)  =  last_in_gag(x1, x3)
last_out_gag(x1, x2, x3)  =  last_out_gag(x2)
U8_gag(x1, x2, x3, x4, x5)  =  U8_gag(x5)
HALVES_IN_GAAA(x1, x2, x3, x4)  =  HALVES_IN_GAAA(x1)
U6_GAAA(x1, x2, x3, x4, x5, x6, x7, x8)  =  U6_GAAA(x1, x8)

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
            ↳ AND
              ↳ PiDP
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof

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

U6_GAAA(T, Y, Xs, Ts, R, Rs, EvenOdd, last_out_gaa(.(Y, Xs), R, Rests)) → HALVES_IN_GAAA(Rests, Ts, Rs, EvenOdd)
HALVES_IN_GAAA(.(T, .(Y, Xs)), .(T, Ts), .(R, Rs), EvenOdd) → U6_GAAA(T, Y, Xs, Ts, R, Rs, EvenOdd, last_in_gaa(.(Y, Xs), R, Rests))

The TRS R consists of the following rules:

last_in_gaa(.(T, []), T, []) → last_out_gaa(.(T, []), T, [])
last_in_gaa(.(H, T), X, .(H, M)) → U8_gaa(H, T, X, M, last_in_gaa(T, X, M))
U8_gaa(H, T, X, M, last_out_gaa(T, X, M)) → last_out_gaa(.(H, T), X, .(H, M))

The argument filtering Pi contains the following mapping:
[]  =  []
.(x1, x2)  =  .(x1, x2)
last_in_gaa(x1, x2, x3)  =  last_in_gaa(x1)
last_out_gaa(x1, x2, x3)  =  last_out_gaa(x2, x3)
U8_gaa(x1, x2, x3, x4, x5)  =  U8_gaa(x1, x5)
HALVES_IN_GAAA(x1, x2, x3, x4)  =  HALVES_IN_GAAA(x1)
U6_GAAA(x1, x2, x3, x4, x5, x6, x7, x8)  =  U6_GAAA(x1, x8)

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
            ↳ AND
              ↳ PiDP
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ RuleRemovalProof

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

HALVES_IN_GAAA(.(T, .(Y, Xs))) → U6_GAAA(T, last_in_gaa(.(Y, Xs)))
U6_GAAA(T, last_out_gaa(R, Rests)) → HALVES_IN_GAAA(Rests)

The TRS R consists of the following rules:

last_in_gaa(.(T, [])) → last_out_gaa(T, [])
last_in_gaa(.(H, T)) → U8_gaa(H, last_in_gaa(T))
U8_gaa(H, last_out_gaa(X, M)) → last_out_gaa(X, .(H, M))

The set Q consists of the following terms:

last_in_gaa(x0)
U8_gaa(x0, x1)

We have to consider all (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

HALVES_IN_GAAA(.(T, .(Y, Xs))) → U6_GAAA(T, last_in_gaa(.(Y, Xs)))
U6_GAAA(T, last_out_gaa(R, Rests)) → HALVES_IN_GAAA(Rests)

Strictly oriented rules of the TRS R:

last_in_gaa(.(T, [])) → last_out_gaa(T, [])
last_in_gaa(.(H, T)) → U8_gaa(H, last_in_gaa(T))
U8_gaa(H, last_out_gaa(X, M)) → last_out_gaa(X, .(H, M))

Used ordering: POLO with Polynomial interpretation [25]:

POL(.(x1, x2)) = 2 + 2·x1 + 2·x2   
POL(HALVES_IN_GAAA(x1)) = 1 + 2·x1   
POL(U6_GAAA(x1, x2)) = 2 + 2·x1 + 2·x2   
POL(U8_gaa(x1, x2)) = 1 + 2·x1 + 2·x2   
POL([]) = 1   
POL(last_in_gaa(x1)) = 2·x1   
POL(last_out_gaa(x1, x2)) = 2 + x1 + x2   



↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ RuleRemovalProof
                          ↳ QDP
                            ↳ PisEmptyProof

Q DP problem:
P is empty.
R is empty.
The set Q consists of the following terms:

last_in_gaa(x0)
U8_gaa(x0, x1)

We have to consider all (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.