(0) Obligation:
Clauses:
p(X) :- ','(l(X), ','(!, q(X))).
q(.(A, [])).
r(1).
l([]).
l(.(H, T)) :- ','(r(H), l(T)).
Queries:
p(a).
(1) PrologToPrologProblemTransformerProof (SOUND transformation)
Built Prolog problem from termination graph.
(2) Obligation:
Clauses:
l83([]).
l83(.(1, T46)) :- l83(T46).
p1(.(1, [])).
p1(.(1, .(1, .(1, .(1, .(1, .(1, .(1, T38)))))))) :- l83(T38).
p1(.(1, .(1, .(1, .(1, .(1, .(1, .(1, T38)))))))) :- l83(T38).
Queries:
p1(a).
(3) PrologToPiTRSProof (SOUND transformation)
We use the technique of [LOPSTR]. With regard to the inferred argument filtering the predicates were used in the following modes:
p1_in: (f)
l83_in: (f)
Transforming
Prolog into the following
Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:
p1_in_a(.(1, [])) → p1_out_a(.(1, []))
p1_in_a(.(1, .(1, .(1, .(1, .(1, .(1, .(1, T38)))))))) → U2_a(T38, l83_in_a(T38))
l83_in_a([]) → l83_out_a([])
l83_in_a(.(1, T46)) → U1_a(T46, l83_in_a(T46))
U1_a(T46, l83_out_a(T46)) → l83_out_a(.(1, T46))
U2_a(T38, l83_out_a(T38)) → p1_out_a(.(1, .(1, .(1, .(1, .(1, .(1, .(1, T38))))))))
The argument filtering Pi contains the following mapping:
p1_in_a(
x1) =
p1_in_a
p1_out_a(
x1) =
p1_out_a(
x1)
U2_a(
x1,
x2) =
U2_a(
x2)
l83_in_a(
x1) =
l83_in_a
l83_out_a(
x1) =
l83_out_a(
x1)
U1_a(
x1,
x2) =
U1_a(
x2)
.(
x1,
x2) =
.(
x1,
x2)
1 =
1
Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog
(4) Obligation:
Pi-finite rewrite system:
The TRS R consists of the following rules:
p1_in_a(.(1, [])) → p1_out_a(.(1, []))
p1_in_a(.(1, .(1, .(1, .(1, .(1, .(1, .(1, T38)))))))) → U2_a(T38, l83_in_a(T38))
l83_in_a([]) → l83_out_a([])
l83_in_a(.(1, T46)) → U1_a(T46, l83_in_a(T46))
U1_a(T46, l83_out_a(T46)) → l83_out_a(.(1, T46))
U2_a(T38, l83_out_a(T38)) → p1_out_a(.(1, .(1, .(1, .(1, .(1, .(1, .(1, T38))))))))
The argument filtering Pi contains the following mapping:
p1_in_a(
x1) =
p1_in_a
p1_out_a(
x1) =
p1_out_a(
x1)
U2_a(
x1,
x2) =
U2_a(
x2)
l83_in_a(
x1) =
l83_in_a
l83_out_a(
x1) =
l83_out_a(
x1)
U1_a(
x1,
x2) =
U1_a(
x2)
.(
x1,
x2) =
.(
x1,
x2)
1 =
1
(5) 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:
P1_IN_A(.(1, .(1, .(1, .(1, .(1, .(1, .(1, T38)))))))) → U2_A(T38, l83_in_a(T38))
P1_IN_A(.(1, .(1, .(1, .(1, .(1, .(1, .(1, T38)))))))) → L83_IN_A(T38)
L83_IN_A(.(1, T46)) → U1_A(T46, l83_in_a(T46))
L83_IN_A(.(1, T46)) → L83_IN_A(T46)
The TRS R consists of the following rules:
p1_in_a(.(1, [])) → p1_out_a(.(1, []))
p1_in_a(.(1, .(1, .(1, .(1, .(1, .(1, .(1, T38)))))))) → U2_a(T38, l83_in_a(T38))
l83_in_a([]) → l83_out_a([])
l83_in_a(.(1, T46)) → U1_a(T46, l83_in_a(T46))
U1_a(T46, l83_out_a(T46)) → l83_out_a(.(1, T46))
U2_a(T38, l83_out_a(T38)) → p1_out_a(.(1, .(1, .(1, .(1, .(1, .(1, .(1, T38))))))))
The argument filtering Pi contains the following mapping:
p1_in_a(
x1) =
p1_in_a
p1_out_a(
x1) =
p1_out_a(
x1)
U2_a(
x1,
x2) =
U2_a(
x2)
l83_in_a(
x1) =
l83_in_a
l83_out_a(
x1) =
l83_out_a(
x1)
U1_a(
x1,
x2) =
U1_a(
x2)
.(
x1,
x2) =
.(
x1,
x2)
1 =
1
P1_IN_A(
x1) =
P1_IN_A
U2_A(
x1,
x2) =
U2_A(
x2)
L83_IN_A(
x1) =
L83_IN_A
U1_A(
x1,
x2) =
U1_A(
x2)
We have to consider all (P,R,Pi)-chains
(6) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
P1_IN_A(.(1, .(1, .(1, .(1, .(1, .(1, .(1, T38)))))))) → U2_A(T38, l83_in_a(T38))
P1_IN_A(.(1, .(1, .(1, .(1, .(1, .(1, .(1, T38)))))))) → L83_IN_A(T38)
L83_IN_A(.(1, T46)) → U1_A(T46, l83_in_a(T46))
L83_IN_A(.(1, T46)) → L83_IN_A(T46)
The TRS R consists of the following rules:
p1_in_a(.(1, [])) → p1_out_a(.(1, []))
p1_in_a(.(1, .(1, .(1, .(1, .(1, .(1, .(1, T38)))))))) → U2_a(T38, l83_in_a(T38))
l83_in_a([]) → l83_out_a([])
l83_in_a(.(1, T46)) → U1_a(T46, l83_in_a(T46))
U1_a(T46, l83_out_a(T46)) → l83_out_a(.(1, T46))
U2_a(T38, l83_out_a(T38)) → p1_out_a(.(1, .(1, .(1, .(1, .(1, .(1, .(1, T38))))))))
The argument filtering Pi contains the following mapping:
p1_in_a(
x1) =
p1_in_a
p1_out_a(
x1) =
p1_out_a(
x1)
U2_a(
x1,
x2) =
U2_a(
x2)
l83_in_a(
x1) =
l83_in_a
l83_out_a(
x1) =
l83_out_a(
x1)
U1_a(
x1,
x2) =
U1_a(
x2)
.(
x1,
x2) =
.(
x1,
x2)
1 =
1
P1_IN_A(
x1) =
P1_IN_A
U2_A(
x1,
x2) =
U2_A(
x2)
L83_IN_A(
x1) =
L83_IN_A
U1_A(
x1,
x2) =
U1_A(
x2)
We have to consider all (P,R,Pi)-chains
(7) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 3 less nodes.
(8) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
L83_IN_A(.(1, T46)) → L83_IN_A(T46)
The TRS R consists of the following rules:
p1_in_a(.(1, [])) → p1_out_a(.(1, []))
p1_in_a(.(1, .(1, .(1, .(1, .(1, .(1, .(1, T38)))))))) → U2_a(T38, l83_in_a(T38))
l83_in_a([]) → l83_out_a([])
l83_in_a(.(1, T46)) → U1_a(T46, l83_in_a(T46))
U1_a(T46, l83_out_a(T46)) → l83_out_a(.(1, T46))
U2_a(T38, l83_out_a(T38)) → p1_out_a(.(1, .(1, .(1, .(1, .(1, .(1, .(1, T38))))))))
The argument filtering Pi contains the following mapping:
p1_in_a(
x1) =
p1_in_a
p1_out_a(
x1) =
p1_out_a(
x1)
U2_a(
x1,
x2) =
U2_a(
x2)
l83_in_a(
x1) =
l83_in_a
l83_out_a(
x1) =
l83_out_a(
x1)
U1_a(
x1,
x2) =
U1_a(
x2)
.(
x1,
x2) =
.(
x1,
x2)
1 =
1
L83_IN_A(
x1) =
L83_IN_A
We have to consider all (P,R,Pi)-chains
(9) UsableRulesProof (EQUIVALENT transformation)
For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.
(10) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
L83_IN_A(.(1, T46)) → L83_IN_A(T46)
R is empty.
The argument filtering Pi contains the following mapping:
.(
x1,
x2) =
.(
x1,
x2)
1 =
1
L83_IN_A(
x1) =
L83_IN_A
We have to consider all (P,R,Pi)-chains
(11) PiDPToQDPProof (SOUND transformation)
Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.
(12) Obligation:
Q DP problem:
The TRS P consists of the following rules:
L83_IN_A → L83_IN_A
R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
(13) NonTerminationProof (EQUIVALENT transformation)
We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.
s =
L83_IN_A evaluates to t =
L83_IN_AThus s starts an infinite chain as s semiunifies with t with the following substitutions:
- Matcher: [ ]
- Semiunifier: [ ]
Rewriting sequenceThe DP semiunifies directly so there is only one rewrite step from L83_IN_A to L83_IN_A.
(14) FALSE
(15) PrologToPiTRSProof (SOUND transformation)
We use the technique of [LOPSTR]. With regard to the inferred argument filtering the predicates were used in the following modes:
p1_in: (f)
l83_in: (f)
Transforming
Prolog into the following
Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:
p1_in_a(.(1, [])) → p1_out_a(.(1, []))
p1_in_a(.(1, .(1, .(1, .(1, .(1, .(1, .(1, T38)))))))) → U2_a(T38, l83_in_a(T38))
l83_in_a([]) → l83_out_a([])
l83_in_a(.(1, T46)) → U1_a(T46, l83_in_a(T46))
U1_a(T46, l83_out_a(T46)) → l83_out_a(.(1, T46))
U2_a(T38, l83_out_a(T38)) → p1_out_a(.(1, .(1, .(1, .(1, .(1, .(1, .(1, T38))))))))
The argument filtering Pi contains the following mapping:
p1_in_a(
x1) =
p1_in_a
p1_out_a(
x1) =
p1_out_a(
x1)
U2_a(
x1,
x2) =
U2_a(
x2)
l83_in_a(
x1) =
l83_in_a
l83_out_a(
x1) =
l83_out_a(
x1)
U1_a(
x1,
x2) =
U1_a(
x2)
.(
x1,
x2) =
.(
x1,
x2)
1 =
1
Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog
(16) Obligation:
Pi-finite rewrite system:
The TRS R consists of the following rules:
p1_in_a(.(1, [])) → p1_out_a(.(1, []))
p1_in_a(.(1, .(1, .(1, .(1, .(1, .(1, .(1, T38)))))))) → U2_a(T38, l83_in_a(T38))
l83_in_a([]) → l83_out_a([])
l83_in_a(.(1, T46)) → U1_a(T46, l83_in_a(T46))
U1_a(T46, l83_out_a(T46)) → l83_out_a(.(1, T46))
U2_a(T38, l83_out_a(T38)) → p1_out_a(.(1, .(1, .(1, .(1, .(1, .(1, .(1, T38))))))))
The argument filtering Pi contains the following mapping:
p1_in_a(
x1) =
p1_in_a
p1_out_a(
x1) =
p1_out_a(
x1)
U2_a(
x1,
x2) =
U2_a(
x2)
l83_in_a(
x1) =
l83_in_a
l83_out_a(
x1) =
l83_out_a(
x1)
U1_a(
x1,
x2) =
U1_a(
x2)
.(
x1,
x2) =
.(
x1,
x2)
1 =
1
(17) 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:
P1_IN_A(.(1, .(1, .(1, .(1, .(1, .(1, .(1, T38)))))))) → U2_A(T38, l83_in_a(T38))
P1_IN_A(.(1, .(1, .(1, .(1, .(1, .(1, .(1, T38)))))))) → L83_IN_A(T38)
L83_IN_A(.(1, T46)) → U1_A(T46, l83_in_a(T46))
L83_IN_A(.(1, T46)) → L83_IN_A(T46)
The TRS R consists of the following rules:
p1_in_a(.(1, [])) → p1_out_a(.(1, []))
p1_in_a(.(1, .(1, .(1, .(1, .(1, .(1, .(1, T38)))))))) → U2_a(T38, l83_in_a(T38))
l83_in_a([]) → l83_out_a([])
l83_in_a(.(1, T46)) → U1_a(T46, l83_in_a(T46))
U1_a(T46, l83_out_a(T46)) → l83_out_a(.(1, T46))
U2_a(T38, l83_out_a(T38)) → p1_out_a(.(1, .(1, .(1, .(1, .(1, .(1, .(1, T38))))))))
The argument filtering Pi contains the following mapping:
p1_in_a(
x1) =
p1_in_a
p1_out_a(
x1) =
p1_out_a(
x1)
U2_a(
x1,
x2) =
U2_a(
x2)
l83_in_a(
x1) =
l83_in_a
l83_out_a(
x1) =
l83_out_a(
x1)
U1_a(
x1,
x2) =
U1_a(
x2)
.(
x1,
x2) =
.(
x1,
x2)
1 =
1
P1_IN_A(
x1) =
P1_IN_A
U2_A(
x1,
x2) =
U2_A(
x2)
L83_IN_A(
x1) =
L83_IN_A
U1_A(
x1,
x2) =
U1_A(
x2)
We have to consider all (P,R,Pi)-chains
(18) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
P1_IN_A(.(1, .(1, .(1, .(1, .(1, .(1, .(1, T38)))))))) → U2_A(T38, l83_in_a(T38))
P1_IN_A(.(1, .(1, .(1, .(1, .(1, .(1, .(1, T38)))))))) → L83_IN_A(T38)
L83_IN_A(.(1, T46)) → U1_A(T46, l83_in_a(T46))
L83_IN_A(.(1, T46)) → L83_IN_A(T46)
The TRS R consists of the following rules:
p1_in_a(.(1, [])) → p1_out_a(.(1, []))
p1_in_a(.(1, .(1, .(1, .(1, .(1, .(1, .(1, T38)))))))) → U2_a(T38, l83_in_a(T38))
l83_in_a([]) → l83_out_a([])
l83_in_a(.(1, T46)) → U1_a(T46, l83_in_a(T46))
U1_a(T46, l83_out_a(T46)) → l83_out_a(.(1, T46))
U2_a(T38, l83_out_a(T38)) → p1_out_a(.(1, .(1, .(1, .(1, .(1, .(1, .(1, T38))))))))
The argument filtering Pi contains the following mapping:
p1_in_a(
x1) =
p1_in_a
p1_out_a(
x1) =
p1_out_a(
x1)
U2_a(
x1,
x2) =
U2_a(
x2)
l83_in_a(
x1) =
l83_in_a
l83_out_a(
x1) =
l83_out_a(
x1)
U1_a(
x1,
x2) =
U1_a(
x2)
.(
x1,
x2) =
.(
x1,
x2)
1 =
1
P1_IN_A(
x1) =
P1_IN_A
U2_A(
x1,
x2) =
U2_A(
x2)
L83_IN_A(
x1) =
L83_IN_A
U1_A(
x1,
x2) =
U1_A(
x2)
We have to consider all (P,R,Pi)-chains
(19) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 3 less nodes.
(20) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
L83_IN_A(.(1, T46)) → L83_IN_A(T46)
The TRS R consists of the following rules:
p1_in_a(.(1, [])) → p1_out_a(.(1, []))
p1_in_a(.(1, .(1, .(1, .(1, .(1, .(1, .(1, T38)))))))) → U2_a(T38, l83_in_a(T38))
l83_in_a([]) → l83_out_a([])
l83_in_a(.(1, T46)) → U1_a(T46, l83_in_a(T46))
U1_a(T46, l83_out_a(T46)) → l83_out_a(.(1, T46))
U2_a(T38, l83_out_a(T38)) → p1_out_a(.(1, .(1, .(1, .(1, .(1, .(1, .(1, T38))))))))
The argument filtering Pi contains the following mapping:
p1_in_a(
x1) =
p1_in_a
p1_out_a(
x1) =
p1_out_a(
x1)
U2_a(
x1,
x2) =
U2_a(
x2)
l83_in_a(
x1) =
l83_in_a
l83_out_a(
x1) =
l83_out_a(
x1)
U1_a(
x1,
x2) =
U1_a(
x2)
.(
x1,
x2) =
.(
x1,
x2)
1 =
1
L83_IN_A(
x1) =
L83_IN_A
We have to consider all (P,R,Pi)-chains
(21) UsableRulesProof (EQUIVALENT transformation)
For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.
(22) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
L83_IN_A(.(1, T46)) → L83_IN_A(T46)
R is empty.
The argument filtering Pi contains the following mapping:
.(
x1,
x2) =
.(
x1,
x2)
1 =
1
L83_IN_A(
x1) =
L83_IN_A
We have to consider all (P,R,Pi)-chains
(23) PiDPToQDPProof (SOUND transformation)
Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.
(24) Obligation:
Q DP problem:
The TRS P consists of the following rules:
L83_IN_A → L83_IN_A
R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
(25) NonTerminationProof (EQUIVALENT transformation)
We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.
s =
L83_IN_A evaluates to t =
L83_IN_AThus s starts an infinite chain as s semiunifies with t with the following substitutions:
- Matcher: [ ]
- Semiunifier: [ ]
Rewriting sequenceThe DP semiunifies directly so there is only one rewrite step from L83_IN_A to L83_IN_A.
(26) FALSE