(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).
Query: palindrome(g)
(1) PrologToPiTRSProof (SOUND transformation)
We use the technique of [TOCL09]. 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) YES