(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_AL83_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_A

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
  • Matcher: [ ]
  • Semiunifier: [ ]




Rewriting sequence

The 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_AL83_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_A

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
  • Matcher: [ ]
  • Semiunifier: [ ]




Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from L83_IN_A to L83_IN_A.



(26) FALSE