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].Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:
palindrome_in(Xs) → U1(Xs, reverse_in(Xs, Xs))
reverse_in(X1s, X2s) → U2(X1s, X2s, reverse3_in(X1s, [], X2s))
reverse3_in([], Xs, Xs) → reverse3_out([], Xs, Xs)
reverse3_in(.(X, X1s), X2s, Ys) → U3(X, X1s, X2s, Ys, reverse3_in(X1s, .(X, X2s), Ys))
U3(X, X1s, X2s, Ys, reverse3_out(X1s, .(X, X2s), Ys)) → reverse3_out(.(X, X1s), X2s, Ys)
U2(X1s, X2s, reverse3_out(X1s, [], X2s)) → reverse_out(X1s, X2s)
U1(Xs, reverse_out(Xs, Xs)) → palindrome_out(Xs)
The argument filtering Pi contains the following mapping:
palindrome_in(x1) = palindrome_in(x1)
U1(x1, x2) = U1(x2)
reverse_in(x1, x2) = reverse_in(x1, x2)
U2(x1, x2, x3) = U2(x3)
reverse3_in(x1, x2, x3) = reverse3_in(x1, x2, x3)
[] = []
reverse3_out(x1, x2, x3) = reverse3_out
.(x1, x2) = .(x1, x2)
U3(x1, x2, x3, x4, x5) = U3(x5)
reverse_out(x1, x2) = reverse_out
palindrome_out(x1) = palindrome_out
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(Xs) → U1(Xs, reverse_in(Xs, Xs))
reverse_in(X1s, X2s) → U2(X1s, X2s, reverse3_in(X1s, [], X2s))
reverse3_in([], Xs, Xs) → reverse3_out([], Xs, Xs)
reverse3_in(.(X, X1s), X2s, Ys) → U3(X, X1s, X2s, Ys, reverse3_in(X1s, .(X, X2s), Ys))
U3(X, X1s, X2s, Ys, reverse3_out(X1s, .(X, X2s), Ys)) → reverse3_out(.(X, X1s), X2s, Ys)
U2(X1s, X2s, reverse3_out(X1s, [], X2s)) → reverse_out(X1s, X2s)
U1(Xs, reverse_out(Xs, Xs)) → palindrome_out(Xs)
The argument filtering Pi contains the following mapping:
palindrome_in(x1) = palindrome_in(x1)
U1(x1, x2) = U1(x2)
reverse_in(x1, x2) = reverse_in(x1, x2)
U2(x1, x2, x3) = U2(x3)
reverse3_in(x1, x2, x3) = reverse3_in(x1, x2, x3)
[] = []
reverse3_out(x1, x2, x3) = reverse3_out
.(x1, x2) = .(x1, x2)
U3(x1, x2, x3, x4, x5) = U3(x5)
reverse_out(x1, x2) = reverse_out
palindrome_out(x1) = palindrome_out
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(Xs) → U11(Xs, reverse_in(Xs, Xs))
PALINDROME_IN(Xs) → REVERSE_IN(Xs, Xs)
REVERSE_IN(X1s, X2s) → U21(X1s, X2s, reverse3_in(X1s, [], X2s))
REVERSE_IN(X1s, X2s) → REVERSE3_IN(X1s, [], X2s)
REVERSE3_IN(.(X, X1s), X2s, Ys) → U31(X, X1s, X2s, Ys, reverse3_in(X1s, .(X, X2s), Ys))
REVERSE3_IN(.(X, X1s), X2s, Ys) → REVERSE3_IN(X1s, .(X, X2s), Ys)
The TRS R consists of the following rules:
palindrome_in(Xs) → U1(Xs, reverse_in(Xs, Xs))
reverse_in(X1s, X2s) → U2(X1s, X2s, reverse3_in(X1s, [], X2s))
reverse3_in([], Xs, Xs) → reverse3_out([], Xs, Xs)
reverse3_in(.(X, X1s), X2s, Ys) → U3(X, X1s, X2s, Ys, reverse3_in(X1s, .(X, X2s), Ys))
U3(X, X1s, X2s, Ys, reverse3_out(X1s, .(X, X2s), Ys)) → reverse3_out(.(X, X1s), X2s, Ys)
U2(X1s, X2s, reverse3_out(X1s, [], X2s)) → reverse_out(X1s, X2s)
U1(Xs, reverse_out(Xs, Xs)) → palindrome_out(Xs)
The argument filtering Pi contains the following mapping:
palindrome_in(x1) = palindrome_in(x1)
U1(x1, x2) = U1(x2)
reverse_in(x1, x2) = reverse_in(x1, x2)
U2(x1, x2, x3) = U2(x3)
reverse3_in(x1, x2, x3) = reverse3_in(x1, x2, x3)
[] = []
reverse3_out(x1, x2, x3) = reverse3_out
.(x1, x2) = .(x1, x2)
U3(x1, x2, x3, x4, x5) = U3(x5)
reverse_out(x1, x2) = reverse_out
palindrome_out(x1) = palindrome_out
REVERSE_IN(x1, x2) = REVERSE_IN(x1, x2)
U31(x1, x2, x3, x4, x5) = U31(x5)
U11(x1, x2) = U11(x2)
REVERSE3_IN(x1, x2, x3) = REVERSE3_IN(x1, x2, x3)
U21(x1, x2, x3) = U21(x3)
PALINDROME_IN(x1) = PALINDROME_IN(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_IN(Xs) → U11(Xs, reverse_in(Xs, Xs))
PALINDROME_IN(Xs) → REVERSE_IN(Xs, Xs)
REVERSE_IN(X1s, X2s) → U21(X1s, X2s, reverse3_in(X1s, [], X2s))
REVERSE_IN(X1s, X2s) → REVERSE3_IN(X1s, [], X2s)
REVERSE3_IN(.(X, X1s), X2s, Ys) → U31(X, X1s, X2s, Ys, reverse3_in(X1s, .(X, X2s), Ys))
REVERSE3_IN(.(X, X1s), X2s, Ys) → REVERSE3_IN(X1s, .(X, X2s), Ys)
The TRS R consists of the following rules:
palindrome_in(Xs) → U1(Xs, reverse_in(Xs, Xs))
reverse_in(X1s, X2s) → U2(X1s, X2s, reverse3_in(X1s, [], X2s))
reverse3_in([], Xs, Xs) → reverse3_out([], Xs, Xs)
reverse3_in(.(X, X1s), X2s, Ys) → U3(X, X1s, X2s, Ys, reverse3_in(X1s, .(X, X2s), Ys))
U3(X, X1s, X2s, Ys, reverse3_out(X1s, .(X, X2s), Ys)) → reverse3_out(.(X, X1s), X2s, Ys)
U2(X1s, X2s, reverse3_out(X1s, [], X2s)) → reverse_out(X1s, X2s)
U1(Xs, reverse_out(Xs, Xs)) → palindrome_out(Xs)
The argument filtering Pi contains the following mapping:
palindrome_in(x1) = palindrome_in(x1)
U1(x1, x2) = U1(x2)
reverse_in(x1, x2) = reverse_in(x1, x2)
U2(x1, x2, x3) = U2(x3)
reverse3_in(x1, x2, x3) = reverse3_in(x1, x2, x3)
[] = []
reverse3_out(x1, x2, x3) = reverse3_out
.(x1, x2) = .(x1, x2)
U3(x1, x2, x3, x4, x5) = U3(x5)
reverse_out(x1, x2) = reverse_out
palindrome_out(x1) = palindrome_out
REVERSE_IN(x1, x2) = REVERSE_IN(x1, x2)
U31(x1, x2, x3, x4, x5) = U31(x5)
U11(x1, x2) = U11(x2)
REVERSE3_IN(x1, x2, x3) = REVERSE3_IN(x1, x2, x3)
U21(x1, x2, x3) = U21(x3)
PALINDROME_IN(x1) = PALINDROME_IN(x1)
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(.(X, X1s), X2s, Ys) → REVERSE3_IN(X1s, .(X, X2s), Ys)
The TRS R consists of the following rules:
palindrome_in(Xs) → U1(Xs, reverse_in(Xs, Xs))
reverse_in(X1s, X2s) → U2(X1s, X2s, reverse3_in(X1s, [], X2s))
reverse3_in([], Xs, Xs) → reverse3_out([], Xs, Xs)
reverse3_in(.(X, X1s), X2s, Ys) → U3(X, X1s, X2s, Ys, reverse3_in(X1s, .(X, X2s), Ys))
U3(X, X1s, X2s, Ys, reverse3_out(X1s, .(X, X2s), Ys)) → reverse3_out(.(X, X1s), X2s, Ys)
U2(X1s, X2s, reverse3_out(X1s, [], X2s)) → reverse_out(X1s, X2s)
U1(Xs, reverse_out(Xs, Xs)) → palindrome_out(Xs)
The argument filtering Pi contains the following mapping:
palindrome_in(x1) = palindrome_in(x1)
U1(x1, x2) = U1(x2)
reverse_in(x1, x2) = reverse_in(x1, x2)
U2(x1, x2, x3) = U2(x3)
reverse3_in(x1, x2, x3) = reverse3_in(x1, x2, x3)
[] = []
reverse3_out(x1, x2, x3) = reverse3_out
.(x1, x2) = .(x1, x2)
U3(x1, x2, x3, x4, x5) = U3(x5)
reverse_out(x1, x2) = reverse_out
palindrome_out(x1) = palindrome_out
REVERSE3_IN(x1, x2, x3) = REVERSE3_IN(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(.(X, X1s), X2s, Ys) → REVERSE3_IN(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(.(X, X1s), X2s, Ys) → REVERSE3_IN(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:
- REVERSE3_IN(.(X, X1s), X2s, Ys) → REVERSE3_IN(X1s, .(X, X2s), Ys)
The graph contains the following edges 1 > 1, 3 >= 3