Left Termination of the query pattern
select_in_3(g, a, a)
w.r.t. the given Prolog program could not be shown:
↳ Prolog
↳ PrologToPiTRSProof
↳ PrologToPiTRSProof
Clauses:
select(X, .(X, Xs), Xs).
select(X, .(Y, Xs), .(Y, Zs)) :- select(X, Xs, Zs).
Queries:
select(g,a,a).
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:
select_in(X, .(Y, Xs), .(Y, Zs)) → U1(X, Y, Xs, Zs, select_in(X, Xs, Zs))
select_in(X, .(X, Xs), Xs) → select_out(X, .(X, Xs), Xs)
U1(X, Y, Xs, Zs, select_out(X, Xs, Zs)) → select_out(X, .(Y, Xs), .(Y, Zs))
The argument filtering Pi contains the following mapping:
select_in(x1, x2, x3) = select_in(x1)
.(x1, x2) = .(x1, x2)
U1(x1, x2, x3, x4, x5) = U1(x5)
select_out(x1, x2, x3) = select_out
Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog
↳ Prolog
↳ PrologToPiTRSProof
↳ PiTRS
↳ DependencyPairsProof
↳ PrologToPiTRSProof
Pi-finite rewrite system:
The TRS R consists of the following rules:
select_in(X, .(Y, Xs), .(Y, Zs)) → U1(X, Y, Xs, Zs, select_in(X, Xs, Zs))
select_in(X, .(X, Xs), Xs) → select_out(X, .(X, Xs), Xs)
U1(X, Y, Xs, Zs, select_out(X, Xs, Zs)) → select_out(X, .(Y, Xs), .(Y, Zs))
The argument filtering Pi contains the following mapping:
select_in(x1, x2, x3) = select_in(x1)
.(x1, x2) = .(x1, x2)
U1(x1, x2, x3, x4, x5) = U1(x5)
select_out(x1, x2, x3) = select_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:
SELECT_IN(X, .(Y, Xs), .(Y, Zs)) → U11(X, Y, Xs, Zs, select_in(X, Xs, Zs))
SELECT_IN(X, .(Y, Xs), .(Y, Zs)) → SELECT_IN(X, Xs, Zs)
The TRS R consists of the following rules:
select_in(X, .(Y, Xs), .(Y, Zs)) → U1(X, Y, Xs, Zs, select_in(X, Xs, Zs))
select_in(X, .(X, Xs), Xs) → select_out(X, .(X, Xs), Xs)
U1(X, Y, Xs, Zs, select_out(X, Xs, Zs)) → select_out(X, .(Y, Xs), .(Y, Zs))
The argument filtering Pi contains the following mapping:
select_in(x1, x2, x3) = select_in(x1)
.(x1, x2) = .(x1, x2)
U1(x1, x2, x3, x4, x5) = U1(x5)
select_out(x1, x2, x3) = select_out
SELECT_IN(x1, x2, x3) = SELECT_IN(x1)
U11(x1, x2, x3, x4, x5) = U11(x5)
We have to consider all (P,R,Pi)-chains
↳ Prolog
↳ PrologToPiTRSProof
↳ PiTRS
↳ DependencyPairsProof
↳ PiDP
↳ DependencyGraphProof
↳ PrologToPiTRSProof
Pi DP problem:
The TRS P consists of the following rules:
SELECT_IN(X, .(Y, Xs), .(Y, Zs)) → U11(X, Y, Xs, Zs, select_in(X, Xs, Zs))
SELECT_IN(X, .(Y, Xs), .(Y, Zs)) → SELECT_IN(X, Xs, Zs)
The TRS R consists of the following rules:
select_in(X, .(Y, Xs), .(Y, Zs)) → U1(X, Y, Xs, Zs, select_in(X, Xs, Zs))
select_in(X, .(X, Xs), Xs) → select_out(X, .(X, Xs), Xs)
U1(X, Y, Xs, Zs, select_out(X, Xs, Zs)) → select_out(X, .(Y, Xs), .(Y, Zs))
The argument filtering Pi contains the following mapping:
select_in(x1, x2, x3) = select_in(x1)
.(x1, x2) = .(x1, x2)
U1(x1, x2, x3, x4, x5) = U1(x5)
select_out(x1, x2, x3) = select_out
SELECT_IN(x1, x2, x3) = SELECT_IN(x1)
U11(x1, x2, x3, x4, x5) = U11(x5)
We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 1 SCC with 1 less node.
↳ Prolog
↳ PrologToPiTRSProof
↳ PiTRS
↳ DependencyPairsProof
↳ PiDP
↳ DependencyGraphProof
↳ PiDP
↳ UsableRulesProof
↳ PrologToPiTRSProof
Pi DP problem:
The TRS P consists of the following rules:
SELECT_IN(X, .(Y, Xs), .(Y, Zs)) → SELECT_IN(X, Xs, Zs)
The TRS R consists of the following rules:
select_in(X, .(Y, Xs), .(Y, Zs)) → U1(X, Y, Xs, Zs, select_in(X, Xs, Zs))
select_in(X, .(X, Xs), Xs) → select_out(X, .(X, Xs), Xs)
U1(X, Y, Xs, Zs, select_out(X, Xs, Zs)) → select_out(X, .(Y, Xs), .(Y, Zs))
The argument filtering Pi contains the following mapping:
select_in(x1, x2, x3) = select_in(x1)
.(x1, x2) = .(x1, x2)
U1(x1, x2, x3, x4, x5) = U1(x5)
select_out(x1, x2, x3) = select_out
SELECT_IN(x1, x2, x3) = SELECT_IN(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
↳ PiDP
↳ UsableRulesProof
↳ PiDP
↳ PiDPToQDPProof
↳ PrologToPiTRSProof
Pi DP problem:
The TRS P consists of the following rules:
SELECT_IN(X, .(Y, Xs), .(Y, Zs)) → SELECT_IN(X, Xs, Zs)
R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2) = .(x1, x2)
SELECT_IN(x1, x2, x3) = SELECT_IN(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
↳ PiDP
↳ UsableRulesProof
↳ PiDP
↳ PiDPToQDPProof
↳ QDP
↳ NonTerminationProof
↳ PrologToPiTRSProof
Q DP problem:
The TRS P consists of the following rules:
SELECT_IN(X) → SELECT_IN(X)
R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.
The TRS P consists of the following rules:
SELECT_IN(X) → SELECT_IN(X)
The TRS R consists of the following rules:none
s = SELECT_IN(X) evaluates to t =SELECT_IN(X)
Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
- Semiunifier: [ ]
- Matcher: [ ]
Rewriting sequence
The DP semiunifies directly so there is only one rewrite step from SELECT_IN(X) to SELECT_IN(X).
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:
select_in(X, .(Y, Xs), .(Y, Zs)) → U1(X, Y, Xs, Zs, select_in(X, Xs, Zs))
select_in(X, .(X, Xs), Xs) → select_out(X, .(X, Xs), Xs)
U1(X, Y, Xs, Zs, select_out(X, Xs, Zs)) → select_out(X, .(Y, Xs), .(Y, Zs))
The argument filtering Pi contains the following mapping:
select_in(x1, x2, x3) = select_in(x1)
.(x1, x2) = .(x1, x2)
U1(x1, x2, x3, x4, x5) = U1(x1, x5)
select_out(x1, x2, x3) = select_out(x1)
Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog
↳ Prolog
↳ PrologToPiTRSProof
↳ PrologToPiTRSProof
↳ PiTRS
↳ DependencyPairsProof
Pi-finite rewrite system:
The TRS R consists of the following rules:
select_in(X, .(Y, Xs), .(Y, Zs)) → U1(X, Y, Xs, Zs, select_in(X, Xs, Zs))
select_in(X, .(X, Xs), Xs) → select_out(X, .(X, Xs), Xs)
U1(X, Y, Xs, Zs, select_out(X, Xs, Zs)) → select_out(X, .(Y, Xs), .(Y, Zs))
The argument filtering Pi contains the following mapping:
select_in(x1, x2, x3) = select_in(x1)
.(x1, x2) = .(x1, x2)
U1(x1, x2, x3, x4, x5) = U1(x1, x5)
select_out(x1, x2, x3) = select_out(x1)
Using Dependency Pairs [1,30] we result in the following initial DP problem:
Pi DP problem:
The TRS P consists of the following rules:
SELECT_IN(X, .(Y, Xs), .(Y, Zs)) → U11(X, Y, Xs, Zs, select_in(X, Xs, Zs))
SELECT_IN(X, .(Y, Xs), .(Y, Zs)) → SELECT_IN(X, Xs, Zs)
The TRS R consists of the following rules:
select_in(X, .(Y, Xs), .(Y, Zs)) → U1(X, Y, Xs, Zs, select_in(X, Xs, Zs))
select_in(X, .(X, Xs), Xs) → select_out(X, .(X, Xs), Xs)
U1(X, Y, Xs, Zs, select_out(X, Xs, Zs)) → select_out(X, .(Y, Xs), .(Y, Zs))
The argument filtering Pi contains the following mapping:
select_in(x1, x2, x3) = select_in(x1)
.(x1, x2) = .(x1, x2)
U1(x1, x2, x3, x4, x5) = U1(x1, x5)
select_out(x1, x2, x3) = select_out(x1)
SELECT_IN(x1, x2, x3) = SELECT_IN(x1)
U11(x1, x2, x3, x4, x5) = U11(x1, x5)
We have to consider all (P,R,Pi)-chains
↳ Prolog
↳ PrologToPiTRSProof
↳ PrologToPiTRSProof
↳ PiTRS
↳ DependencyPairsProof
↳ PiDP
↳ DependencyGraphProof
Pi DP problem:
The TRS P consists of the following rules:
SELECT_IN(X, .(Y, Xs), .(Y, Zs)) → U11(X, Y, Xs, Zs, select_in(X, Xs, Zs))
SELECT_IN(X, .(Y, Xs), .(Y, Zs)) → SELECT_IN(X, Xs, Zs)
The TRS R consists of the following rules:
select_in(X, .(Y, Xs), .(Y, Zs)) → U1(X, Y, Xs, Zs, select_in(X, Xs, Zs))
select_in(X, .(X, Xs), Xs) → select_out(X, .(X, Xs), Xs)
U1(X, Y, Xs, Zs, select_out(X, Xs, Zs)) → select_out(X, .(Y, Xs), .(Y, Zs))
The argument filtering Pi contains the following mapping:
select_in(x1, x2, x3) = select_in(x1)
.(x1, x2) = .(x1, x2)
U1(x1, x2, x3, x4, x5) = U1(x1, x5)
select_out(x1, x2, x3) = select_out(x1)
SELECT_IN(x1, x2, x3) = SELECT_IN(x1)
U11(x1, x2, x3, x4, x5) = U11(x1, x5)
We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 1 SCC with 1 less node.
↳ Prolog
↳ PrologToPiTRSProof
↳ PrologToPiTRSProof
↳ PiTRS
↳ DependencyPairsProof
↳ PiDP
↳ DependencyGraphProof
↳ PiDP
↳ UsableRulesProof
Pi DP problem:
The TRS P consists of the following rules:
SELECT_IN(X, .(Y, Xs), .(Y, Zs)) → SELECT_IN(X, Xs, Zs)
The TRS R consists of the following rules:
select_in(X, .(Y, Xs), .(Y, Zs)) → U1(X, Y, Xs, Zs, select_in(X, Xs, Zs))
select_in(X, .(X, Xs), Xs) → select_out(X, .(X, Xs), Xs)
U1(X, Y, Xs, Zs, select_out(X, Xs, Zs)) → select_out(X, .(Y, Xs), .(Y, Zs))
The argument filtering Pi contains the following mapping:
select_in(x1, x2, x3) = select_in(x1)
.(x1, x2) = .(x1, x2)
U1(x1, x2, x3, x4, x5) = U1(x1, x5)
select_out(x1, x2, x3) = select_out(x1)
SELECT_IN(x1, x2, x3) = SELECT_IN(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
↳ PrologToPiTRSProof
↳ PiTRS
↳ DependencyPairsProof
↳ PiDP
↳ DependencyGraphProof
↳ PiDP
↳ UsableRulesProof
↳ PiDP
↳ PiDPToQDPProof
Pi DP problem:
The TRS P consists of the following rules:
SELECT_IN(X, .(Y, Xs), .(Y, Zs)) → SELECT_IN(X, Xs, Zs)
R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2) = .(x1, x2)
SELECT_IN(x1, x2, x3) = SELECT_IN(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
↳ PrologToPiTRSProof
↳ PiTRS
↳ DependencyPairsProof
↳ PiDP
↳ DependencyGraphProof
↳ PiDP
↳ UsableRulesProof
↳ PiDP
↳ PiDPToQDPProof
↳ QDP
↳ NonTerminationProof
Q DP problem:
The TRS P consists of the following rules:
SELECT_IN(X) → SELECT_IN(X)
R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.
The TRS P consists of the following rules:
SELECT_IN(X) → SELECT_IN(X)
The TRS R consists of the following rules:none
s = SELECT_IN(X) evaluates to t =SELECT_IN(X)
Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
- Semiunifier: [ ]
- Matcher: [ ]
Rewriting sequence
The DP semiunifies directly so there is only one rewrite step from SELECT_IN(X) to SELECT_IN(X).