Left Termination of the query pattern
append_in_3(a, a, a)
w.r.t. the given Prolog program could not be shown:
↳ Prolog
↳ PrologToPiTRSProof
↳ PrologToPiTRSProof
Clauses:
append([], L, L).
append(.(H, L1), L2, .(H, L3)) :- append(L1, L2, L3).
append1([], L, L).
append1(.(H, L1), L2, .(H, L3)) :- append1(L1, L2, L3).
Queries:
append(a,a,a).
We use the technique of [30]. With regard to the inferred argument filtering the predicates were used in the following modes:
append_in: (f,f,f)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:
append_in_aaa([], L, L) → append_out_aaa([], L, L)
append_in_aaa(.(H, L1), L2, .(H, L3)) → U1_aaa(H, L1, L2, L3, append_in_aaa(L1, L2, L3))
U1_aaa(H, L1, L2, L3, append_out_aaa(L1, L2, L3)) → append_out_aaa(.(H, L1), L2, .(H, L3))
The argument filtering Pi contains the following mapping:
append_in_aaa(x1, x2, x3) = append_in_aaa
append_out_aaa(x1, x2, x3) = append_out_aaa
U1_aaa(x1, x2, x3, x4, x5) = U1_aaa(x5)
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:
append_in_aaa([], L, L) → append_out_aaa([], L, L)
append_in_aaa(.(H, L1), L2, .(H, L3)) → U1_aaa(H, L1, L2, L3, append_in_aaa(L1, L2, L3))
U1_aaa(H, L1, L2, L3, append_out_aaa(L1, L2, L3)) → append_out_aaa(.(H, L1), L2, .(H, L3))
The argument filtering Pi contains the following mapping:
append_in_aaa(x1, x2, x3) = append_in_aaa
append_out_aaa(x1, x2, x3) = append_out_aaa
U1_aaa(x1, x2, x3, x4, x5) = U1_aaa(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:
APPEND_IN_AAA(.(H, L1), L2, .(H, L3)) → U1_AAA(H, L1, L2, L3, append_in_aaa(L1, L2, L3))
APPEND_IN_AAA(.(H, L1), L2, .(H, L3)) → APPEND_IN_AAA(L1, L2, L3)
The TRS R consists of the following rules:
append_in_aaa([], L, L) → append_out_aaa([], L, L)
append_in_aaa(.(H, L1), L2, .(H, L3)) → U1_aaa(H, L1, L2, L3, append_in_aaa(L1, L2, L3))
U1_aaa(H, L1, L2, L3, append_out_aaa(L1, L2, L3)) → append_out_aaa(.(H, L1), L2, .(H, L3))
The argument filtering Pi contains the following mapping:
append_in_aaa(x1, x2, x3) = append_in_aaa
append_out_aaa(x1, x2, x3) = append_out_aaa
U1_aaa(x1, x2, x3, x4, x5) = U1_aaa(x5)
U1_AAA(x1, x2, x3, x4, x5) = U1_AAA(x5)
APPEND_IN_AAA(x1, x2, x3) = APPEND_IN_AAA
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:
APPEND_IN_AAA(.(H, L1), L2, .(H, L3)) → U1_AAA(H, L1, L2, L3, append_in_aaa(L1, L2, L3))
APPEND_IN_AAA(.(H, L1), L2, .(H, L3)) → APPEND_IN_AAA(L1, L2, L3)
The TRS R consists of the following rules:
append_in_aaa([], L, L) → append_out_aaa([], L, L)
append_in_aaa(.(H, L1), L2, .(H, L3)) → U1_aaa(H, L1, L2, L3, append_in_aaa(L1, L2, L3))
U1_aaa(H, L1, L2, L3, append_out_aaa(L1, L2, L3)) → append_out_aaa(.(H, L1), L2, .(H, L3))
The argument filtering Pi contains the following mapping:
append_in_aaa(x1, x2, x3) = append_in_aaa
append_out_aaa(x1, x2, x3) = append_out_aaa
U1_aaa(x1, x2, x3, x4, x5) = U1_aaa(x5)
U1_AAA(x1, x2, x3, x4, x5) = U1_AAA(x5)
APPEND_IN_AAA(x1, x2, x3) = APPEND_IN_AAA
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:
APPEND_IN_AAA(.(H, L1), L2, .(H, L3)) → APPEND_IN_AAA(L1, L2, L3)
The TRS R consists of the following rules:
append_in_aaa([], L, L) → append_out_aaa([], L, L)
append_in_aaa(.(H, L1), L2, .(H, L3)) → U1_aaa(H, L1, L2, L3, append_in_aaa(L1, L2, L3))
U1_aaa(H, L1, L2, L3, append_out_aaa(L1, L2, L3)) → append_out_aaa(.(H, L1), L2, .(H, L3))
The argument filtering Pi contains the following mapping:
append_in_aaa(x1, x2, x3) = append_in_aaa
append_out_aaa(x1, x2, x3) = append_out_aaa
U1_aaa(x1, x2, x3, x4, x5) = U1_aaa(x5)
APPEND_IN_AAA(x1, x2, x3) = APPEND_IN_AAA
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:
APPEND_IN_AAA(.(H, L1), L2, .(H, L3)) → APPEND_IN_AAA(L1, L2, L3)
R is empty.
The argument filtering Pi contains the following mapping:
APPEND_IN_AAA(x1, x2, x3) = APPEND_IN_AAA
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:
APPEND_IN_AAA → APPEND_IN_AAA
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:
APPEND_IN_AAA → APPEND_IN_AAA
The TRS R consists of the following rules:none
s = APPEND_IN_AAA evaluates to t =APPEND_IN_AAA
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 APPEND_IN_AAA to APPEND_IN_AAA.
We use the technique of [30]. With regard to the inferred argument filtering the predicates were used in the following modes:
append_in: (f,f,f)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:
append_in_aaa([], L, L) → append_out_aaa([], L, L)
append_in_aaa(.(H, L1), L2, .(H, L3)) → U1_aaa(H, L1, L2, L3, append_in_aaa(L1, L2, L3))
U1_aaa(H, L1, L2, L3, append_out_aaa(L1, L2, L3)) → append_out_aaa(.(H, L1), L2, .(H, L3))
The argument filtering Pi contains the following mapping:
append_in_aaa(x1, x2, x3) = append_in_aaa
append_out_aaa(x1, x2, x3) = append_out_aaa
U1_aaa(x1, x2, x3, x4, x5) = U1_aaa(x5)
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:
append_in_aaa([], L, L) → append_out_aaa([], L, L)
append_in_aaa(.(H, L1), L2, .(H, L3)) → U1_aaa(H, L1, L2, L3, append_in_aaa(L1, L2, L3))
U1_aaa(H, L1, L2, L3, append_out_aaa(L1, L2, L3)) → append_out_aaa(.(H, L1), L2, .(H, L3))
The argument filtering Pi contains the following mapping:
append_in_aaa(x1, x2, x3) = append_in_aaa
append_out_aaa(x1, x2, x3) = append_out_aaa
U1_aaa(x1, x2, x3, x4, x5) = U1_aaa(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:
APPEND_IN_AAA(.(H, L1), L2, .(H, L3)) → U1_AAA(H, L1, L2, L3, append_in_aaa(L1, L2, L3))
APPEND_IN_AAA(.(H, L1), L2, .(H, L3)) → APPEND_IN_AAA(L1, L2, L3)
The TRS R consists of the following rules:
append_in_aaa([], L, L) → append_out_aaa([], L, L)
append_in_aaa(.(H, L1), L2, .(H, L3)) → U1_aaa(H, L1, L2, L3, append_in_aaa(L1, L2, L3))
U1_aaa(H, L1, L2, L3, append_out_aaa(L1, L2, L3)) → append_out_aaa(.(H, L1), L2, .(H, L3))
The argument filtering Pi contains the following mapping:
append_in_aaa(x1, x2, x3) = append_in_aaa
append_out_aaa(x1, x2, x3) = append_out_aaa
U1_aaa(x1, x2, x3, x4, x5) = U1_aaa(x5)
U1_AAA(x1, x2, x3, x4, x5) = U1_AAA(x5)
APPEND_IN_AAA(x1, x2, x3) = APPEND_IN_AAA
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:
APPEND_IN_AAA(.(H, L1), L2, .(H, L3)) → U1_AAA(H, L1, L2, L3, append_in_aaa(L1, L2, L3))
APPEND_IN_AAA(.(H, L1), L2, .(H, L3)) → APPEND_IN_AAA(L1, L2, L3)
The TRS R consists of the following rules:
append_in_aaa([], L, L) → append_out_aaa([], L, L)
append_in_aaa(.(H, L1), L2, .(H, L3)) → U1_aaa(H, L1, L2, L3, append_in_aaa(L1, L2, L3))
U1_aaa(H, L1, L2, L3, append_out_aaa(L1, L2, L3)) → append_out_aaa(.(H, L1), L2, .(H, L3))
The argument filtering Pi contains the following mapping:
append_in_aaa(x1, x2, x3) = append_in_aaa
append_out_aaa(x1, x2, x3) = append_out_aaa
U1_aaa(x1, x2, x3, x4, x5) = U1_aaa(x5)
U1_AAA(x1, x2, x3, x4, x5) = U1_AAA(x5)
APPEND_IN_AAA(x1, x2, x3) = APPEND_IN_AAA
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:
APPEND_IN_AAA(.(H, L1), L2, .(H, L3)) → APPEND_IN_AAA(L1, L2, L3)
The TRS R consists of the following rules:
append_in_aaa([], L, L) → append_out_aaa([], L, L)
append_in_aaa(.(H, L1), L2, .(H, L3)) → U1_aaa(H, L1, L2, L3, append_in_aaa(L1, L2, L3))
U1_aaa(H, L1, L2, L3, append_out_aaa(L1, L2, L3)) → append_out_aaa(.(H, L1), L2, .(H, L3))
The argument filtering Pi contains the following mapping:
append_in_aaa(x1, x2, x3) = append_in_aaa
append_out_aaa(x1, x2, x3) = append_out_aaa
U1_aaa(x1, x2, x3, x4, x5) = U1_aaa(x5)
APPEND_IN_AAA(x1, x2, x3) = APPEND_IN_AAA
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:
APPEND_IN_AAA(.(H, L1), L2, .(H, L3)) → APPEND_IN_AAA(L1, L2, L3)
R is empty.
The argument filtering Pi contains the following mapping:
APPEND_IN_AAA(x1, x2, x3) = APPEND_IN_AAA
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:
APPEND_IN_AAA → APPEND_IN_AAA
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:
APPEND_IN_AAA → APPEND_IN_AAA
The TRS R consists of the following rules:none
s = APPEND_IN_AAA evaluates to t =APPEND_IN_AAA
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 APPEND_IN_AAA to APPEND_IN_AAA.