Left Termination of the query pattern p_in_1(a) w.r.t. the given Prolog program could not be shown:

0 Prolog
↳1 PrologToPrologProblemTransformerProof (⇐)
↳2 Prolog
    ↳3 PrologToPiTRSProof (⇐)
        ↳4 PiTRS
        ↳5 DependencyPairsProof (⇔)
        ↳6 PiDP
        ↳7 DependencyGraphProof (⇔)
        ↳8 AND
            ↳9 PiDP
                ↳10 UsableRulesProof (⇔)
                ↳11 PiDP
                ↳12 PiDPToQDPProof (⇔)
                ↳13 QDP
                ↳14 QDPSizeChangeProof (⇔)
                ↳15 YES
            ↳16 PiDP
                ↳17 UsableRulesProof (⇔)
                ↳18 PiDP
                ↳19 PiDPToQDPProof (⇐)
                ↳20 QDP
                ↳21 NonTerminationProof (⇔)
                ↳22 NO
    ↳23 PrologToPiTRSProof (⇐)
        ↳24 PiTRS
        ↳25 DependencyPairsProof (⇔)
        ↳26 PiDP
        ↳27 DependencyGraphProof (⇔)
        ↳28 AND
            ↳29 PiDP
                ↳30 UsableRulesProof (⇔)
                ↳31 PiDP
                ↳32 PiDPToQDPProof (⇔)
                ↳33 QDP
                ↳34 QDPSizeChangeProof (⇔)
                ↳35 YES
            ↳36 PiDP
                ↳37 UsableRulesProof (⇔)
                ↳38 PiDP
                ↳39 PiDPToQDPProof (⇐)
                ↳40 QDP
                ↳41 NonTerminationProof (⇔)
                ↳42 NO

(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:

l16([]).
l16(.(1, T23)) :- l16(T23).
p1(.(1, T13)) :- l16(T13).
p1(.(1, [])) :- l16([]).

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)
l16_in: (f) (b)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

p1_in_a(.(1, T13)) → U2_a(T13, l16_in_a(T13))
l16_in_a([]) → l16_out_a([])
l16_in_a(.(1, T23)) → U1_a(T23, l16_in_a(T23))
U1_a(T23, l16_out_a(T23)) → l16_out_a(.(1, T23))
U2_a(T13, l16_out_a(T13)) → p1_out_a(.(1, T13))
p1_in_a(.(1, [])) → U3_a(l16_in_g([]))
l16_in_g([]) → l16_out_g([])
l16_in_g(.(1, T23)) → U1_g(T23, l16_in_g(T23))
U1_g(T23, l16_out_g(T23)) → l16_out_g(.(1, T23))
U3_a(l16_out_g([])) → p1_out_a(.(1, []))

The argument filtering Pi contains the following mapping:
p1_in_a(x1)  =  p1_in_a
U2_a(x1, x2)  =  U2_a(x2)
l16_in_a(x1)  =  l16_in_a
l16_out_a(x1)  =  l16_out_a(x1)
U1_a(x1, x2)  =  U1_a(x2)
p1_out_a(x1)  =  p1_out_a(x1)
U3_a(x1)  =  U3_a(x1)
l16_in_g(x1)  =  l16_in_g(x1)
[]  =  []
l16_out_g(x1)  =  l16_out_g
.(x1, x2)  =  .(x1, x2)
1  =  1
U1_g(x1, x2)  =  U1_g(x2)

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, T13)) → U2_a(T13, l16_in_a(T13))
l16_in_a([]) → l16_out_a([])
l16_in_a(.(1, T23)) → U1_a(T23, l16_in_a(T23))
U1_a(T23, l16_out_a(T23)) → l16_out_a(.(1, T23))
U2_a(T13, l16_out_a(T13)) → p1_out_a(.(1, T13))
p1_in_a(.(1, [])) → U3_a(l16_in_g([]))
l16_in_g([]) → l16_out_g([])
l16_in_g(.(1, T23)) → U1_g(T23, l16_in_g(T23))
U1_g(T23, l16_out_g(T23)) → l16_out_g(.(1, T23))
U3_a(l16_out_g([])) → p1_out_a(.(1, []))

The argument filtering Pi contains the following mapping:
p1_in_a(x1)  =  p1_in_a
U2_a(x1, x2)  =  U2_a(x2)
l16_in_a(x1)  =  l16_in_a
l16_out_a(x1)  =  l16_out_a(x1)
U1_a(x1, x2)  =  U1_a(x2)
p1_out_a(x1)  =  p1_out_a(x1)
U3_a(x1)  =  U3_a(x1)
l16_in_g(x1)  =  l16_in_g(x1)
[]  =  []
l16_out_g(x1)  =  l16_out_g
.(x1, x2)  =  .(x1, x2)
1  =  1
U1_g(x1, x2)  =  U1_g(x2)

(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, T13)) → U2_A(T13, l16_in_a(T13))
P1_IN_A(.(1, T13)) → L16_IN_A(T13)
L16_IN_A(.(1, T23)) → U1_A(T23, l16_in_a(T23))
L16_IN_A(.(1, T23)) → L16_IN_A(T23)
P1_IN_A(.(1, [])) → U3_A(l16_in_g([]))
P1_IN_A(.(1, [])) → L16_IN_G([])
L16_IN_G(.(1, T23)) → U1_G(T23, l16_in_g(T23))
L16_IN_G(.(1, T23)) → L16_IN_G(T23)

The TRS R consists of the following rules:

p1_in_a(.(1, T13)) → U2_a(T13, l16_in_a(T13))
l16_in_a([]) → l16_out_a([])
l16_in_a(.(1, T23)) → U1_a(T23, l16_in_a(T23))
U1_a(T23, l16_out_a(T23)) → l16_out_a(.(1, T23))
U2_a(T13, l16_out_a(T13)) → p1_out_a(.(1, T13))
p1_in_a(.(1, [])) → U3_a(l16_in_g([]))
l16_in_g([]) → l16_out_g([])
l16_in_g(.(1, T23)) → U1_g(T23, l16_in_g(T23))
U1_g(T23, l16_out_g(T23)) → l16_out_g(.(1, T23))
U3_a(l16_out_g([])) → p1_out_a(.(1, []))

The argument filtering Pi contains the following mapping:
p1_in_a(x1)  =  p1_in_a
U2_a(x1, x2)  =  U2_a(x2)
l16_in_a(x1)  =  l16_in_a
l16_out_a(x1)  =  l16_out_a(x1)
U1_a(x1, x2)  =  U1_a(x2)
p1_out_a(x1)  =  p1_out_a(x1)
U3_a(x1)  =  U3_a(x1)
l16_in_g(x1)  =  l16_in_g(x1)
[]  =  []
l16_out_g(x1)  =  l16_out_g
.(x1, x2)  =  .(x1, x2)
1  =  1
U1_g(x1, x2)  =  U1_g(x2)
P1_IN_A(x1)  =  P1_IN_A
U2_A(x1, x2)  =  U2_A(x2)
L16_IN_A(x1)  =  L16_IN_A
U1_A(x1, x2)  =  U1_A(x2)
U3_A(x1)  =  U3_A(x1)
L16_IN_G(x1)  =  L16_IN_G(x1)
U1_G(x1, x2)  =  U1_G(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, T13)) → U2_A(T13, l16_in_a(T13))
P1_IN_A(.(1, T13)) → L16_IN_A(T13)
L16_IN_A(.(1, T23)) → U1_A(T23, l16_in_a(T23))
L16_IN_A(.(1, T23)) → L16_IN_A(T23)
P1_IN_A(.(1, [])) → U3_A(l16_in_g([]))
P1_IN_A(.(1, [])) → L16_IN_G([])
L16_IN_G(.(1, T23)) → U1_G(T23, l16_in_g(T23))
L16_IN_G(.(1, T23)) → L16_IN_G(T23)

The TRS R consists of the following rules:

p1_in_a(.(1, T13)) → U2_a(T13, l16_in_a(T13))
l16_in_a([]) → l16_out_a([])
l16_in_a(.(1, T23)) → U1_a(T23, l16_in_a(T23))
U1_a(T23, l16_out_a(T23)) → l16_out_a(.(1, T23))
U2_a(T13, l16_out_a(T13)) → p1_out_a(.(1, T13))
p1_in_a(.(1, [])) → U3_a(l16_in_g([]))
l16_in_g([]) → l16_out_g([])
l16_in_g(.(1, T23)) → U1_g(T23, l16_in_g(T23))
U1_g(T23, l16_out_g(T23)) → l16_out_g(.(1, T23))
U3_a(l16_out_g([])) → p1_out_a(.(1, []))

The argument filtering Pi contains the following mapping:
p1_in_a(x1)  =  p1_in_a
U2_a(x1, x2)  =  U2_a(x2)
l16_in_a(x1)  =  l16_in_a
l16_out_a(x1)  =  l16_out_a(x1)
U1_a(x1, x2)  =  U1_a(x2)
p1_out_a(x1)  =  p1_out_a(x1)
U3_a(x1)  =  U3_a(x1)
l16_in_g(x1)  =  l16_in_g(x1)
[]  =  []
l16_out_g(x1)  =  l16_out_g
.(x1, x2)  =  .(x1, x2)
1  =  1
U1_g(x1, x2)  =  U1_g(x2)
P1_IN_A(x1)  =  P1_IN_A
U2_A(x1, x2)  =  U2_A(x2)
L16_IN_A(x1)  =  L16_IN_A
U1_A(x1, x2)  =  U1_A(x2)
U3_A(x1)  =  U3_A(x1)
L16_IN_G(x1)  =  L16_IN_G(x1)
U1_G(x1, x2)  =  U1_G(x2)

We have to consider all (P,R,Pi)-chains

(7) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 2 SCCs with 6 less nodes.

(8) Complex Obligation (AND)

(9) Obligation:

Pi DP problem:
The TRS P consists of the following rules:

L16_IN_G(.(1, T23)) → L16_IN_G(T23)

The TRS R consists of the following rules:

p1_in_a(.(1, T13)) → U2_a(T13, l16_in_a(T13))
l16_in_a([]) → l16_out_a([])
l16_in_a(.(1, T23)) → U1_a(T23, l16_in_a(T23))
U1_a(T23, l16_out_a(T23)) → l16_out_a(.(1, T23))
U2_a(T13, l16_out_a(T13)) → p1_out_a(.(1, T13))
p1_in_a(.(1, [])) → U3_a(l16_in_g([]))
l16_in_g([]) → l16_out_g([])
l16_in_g(.(1, T23)) → U1_g(T23, l16_in_g(T23))
U1_g(T23, l16_out_g(T23)) → l16_out_g(.(1, T23))
U3_a(l16_out_g([])) → p1_out_a(.(1, []))

The argument filtering Pi contains the following mapping:
p1_in_a(x1)  =  p1_in_a
U2_a(x1, x2)  =  U2_a(x2)
l16_in_a(x1)  =  l16_in_a
l16_out_a(x1)  =  l16_out_a(x1)
U1_a(x1, x2)  =  U1_a(x2)
p1_out_a(x1)  =  p1_out_a(x1)
U3_a(x1)  =  U3_a(x1)
l16_in_g(x1)  =  l16_in_g(x1)
[]  =  []
l16_out_g(x1)  =  l16_out_g
.(x1, x2)  =  .(x1, x2)
1  =  1
U1_g(x1, x2)  =  U1_g(x2)
L16_IN_G(x1)  =  L16_IN_G(x1)

We have to consider all (P,R,Pi)-chains

(10) UsableRulesProof (EQUIVALENT transformation)

For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.

(11) Obligation:

Pi DP problem:
The TRS P consists of the following rules:

L16_IN_G(.(1, T23)) → L16_IN_G(T23)

R is empty.
Pi is empty.
We have to consider all (P,R,Pi)-chains

(12) PiDPToQDPProof (EQUIVALENT transformation)

Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.

(13) Obligation:

Q DP problem:
The TRS P consists of the following rules:

L16_IN_G(.(1, T23)) → L16_IN_G(T23)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.

(14) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] 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:

  • L16_IN_G(.(1, T23)) → L16_IN_G(T23)
    The graph contains the following edges 1 > 1

(15) YES

(16) Obligation:

Pi DP problem:
The TRS P consists of the following rules:

L16_IN_A(.(1, T23)) → L16_IN_A(T23)

The TRS R consists of the following rules:

p1_in_a(.(1, T13)) → U2_a(T13, l16_in_a(T13))
l16_in_a([]) → l16_out_a([])
l16_in_a(.(1, T23)) → U1_a(T23, l16_in_a(T23))
U1_a(T23, l16_out_a(T23)) → l16_out_a(.(1, T23))
U2_a(T13, l16_out_a(T13)) → p1_out_a(.(1, T13))
p1_in_a(.(1, [])) → U3_a(l16_in_g([]))
l16_in_g([]) → l16_out_g([])
l16_in_g(.(1, T23)) → U1_g(T23, l16_in_g(T23))
U1_g(T23, l16_out_g(T23)) → l16_out_g(.(1, T23))
U3_a(l16_out_g([])) → p1_out_a(.(1, []))

The argument filtering Pi contains the following mapping:
p1_in_a(x1)  =  p1_in_a
U2_a(x1, x2)  =  U2_a(x2)
l16_in_a(x1)  =  l16_in_a
l16_out_a(x1)  =  l16_out_a(x1)
U1_a(x1, x2)  =  U1_a(x2)
p1_out_a(x1)  =  p1_out_a(x1)
U3_a(x1)  =  U3_a(x1)
l16_in_g(x1)  =  l16_in_g(x1)
[]  =  []
l16_out_g(x1)  =  l16_out_g
.(x1, x2)  =  .(x1, x2)
1  =  1
U1_g(x1, x2)  =  U1_g(x2)
L16_IN_A(x1)  =  L16_IN_A

We have to consider all (P,R,Pi)-chains

(17) UsableRulesProof (EQUIVALENT transformation)

For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.

(18) Obligation:

Pi DP problem:
The TRS P consists of the following rules:

L16_IN_A(.(1, T23)) → L16_IN_A(T23)

R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x1, x2)
1  =  1
L16_IN_A(x1)  =  L16_IN_A

We have to consider all (P,R,Pi)-chains

(19) PiDPToQDPProof (SOUND transformation)

Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.

(20) Obligation:

Q DP problem:
The TRS P consists of the following rules:

L16_IN_A → L16_IN_A

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.

(21) 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 = L16_IN_A evaluates to t =L16_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 L16_IN_A to L16_IN_A.



(22) NO

(23) 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)
l16_in: (f) (b)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

p1_in_a(.(1, T13)) → U2_a(T13, l16_in_a(T13))
l16_in_a([]) → l16_out_a([])
l16_in_a(.(1, T23)) → U1_a(T23, l16_in_a(T23))
U1_a(T23, l16_out_a(T23)) → l16_out_a(.(1, T23))
U2_a(T13, l16_out_a(T13)) → p1_out_a(.(1, T13))
p1_in_a(.(1, [])) → U3_a(l16_in_g([]))
l16_in_g([]) → l16_out_g([])
l16_in_g(.(1, T23)) → U1_g(T23, l16_in_g(T23))
U1_g(T23, l16_out_g(T23)) → l16_out_g(.(1, T23))
U3_a(l16_out_g([])) → p1_out_a(.(1, []))

The argument filtering Pi contains the following mapping:
p1_in_a(x1)  =  p1_in_a
U2_a(x1, x2)  =  U2_a(x2)
l16_in_a(x1)  =  l16_in_a
l16_out_a(x1)  =  l16_out_a(x1)
U1_a(x1, x2)  =  U1_a(x2)
p1_out_a(x1)  =  p1_out_a(x1)
U3_a(x1)  =  U3_a(x1)
l16_in_g(x1)  =  l16_in_g(x1)
[]  =  []
l16_out_g(x1)  =  l16_out_g(x1)
.(x1, x2)  =  .(x1, x2)
1  =  1
U1_g(x1, x2)  =  U1_g(x1, x2)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(24) Obligation:

Pi-finite rewrite system:
The TRS R consists of the following rules:

p1_in_a(.(1, T13)) → U2_a(T13, l16_in_a(T13))
l16_in_a([]) → l16_out_a([])
l16_in_a(.(1, T23)) → U1_a(T23, l16_in_a(T23))
U1_a(T23, l16_out_a(T23)) → l16_out_a(.(1, T23))
U2_a(T13, l16_out_a(T13)) → p1_out_a(.(1, T13))
p1_in_a(.(1, [])) → U3_a(l16_in_g([]))
l16_in_g([]) → l16_out_g([])
l16_in_g(.(1, T23)) → U1_g(T23, l16_in_g(T23))
U1_g(T23, l16_out_g(T23)) → l16_out_g(.(1, T23))
U3_a(l16_out_g([])) → p1_out_a(.(1, []))

The argument filtering Pi contains the following mapping:
p1_in_a(x1)  =  p1_in_a
U2_a(x1, x2)  =  U2_a(x2)
l16_in_a(x1)  =  l16_in_a
l16_out_a(x1)  =  l16_out_a(x1)
U1_a(x1, x2)  =  U1_a(x2)
p1_out_a(x1)  =  p1_out_a(x1)
U3_a(x1)  =  U3_a(x1)
l16_in_g(x1)  =  l16_in_g(x1)
[]  =  []
l16_out_g(x1)  =  l16_out_g(x1)
.(x1, x2)  =  .(x1, x2)
1  =  1
U1_g(x1, x2)  =  U1_g(x1, x2)

(25) 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, T13)) → U2_A(T13, l16_in_a(T13))
P1_IN_A(.(1, T13)) → L16_IN_A(T13)
L16_IN_A(.(1, T23)) → U1_A(T23, l16_in_a(T23))
L16_IN_A(.(1, T23)) → L16_IN_A(T23)
P1_IN_A(.(1, [])) → U3_A(l16_in_g([]))
P1_IN_A(.(1, [])) → L16_IN_G([])
L16_IN_G(.(1, T23)) → U1_G(T23, l16_in_g(T23))
L16_IN_G(.(1, T23)) → L16_IN_G(T23)

The TRS R consists of the following rules:

p1_in_a(.(1, T13)) → U2_a(T13, l16_in_a(T13))
l16_in_a([]) → l16_out_a([])
l16_in_a(.(1, T23)) → U1_a(T23, l16_in_a(T23))
U1_a(T23, l16_out_a(T23)) → l16_out_a(.(1, T23))
U2_a(T13, l16_out_a(T13)) → p1_out_a(.(1, T13))
p1_in_a(.(1, [])) → U3_a(l16_in_g([]))
l16_in_g([]) → l16_out_g([])
l16_in_g(.(1, T23)) → U1_g(T23, l16_in_g(T23))
U1_g(T23, l16_out_g(T23)) → l16_out_g(.(1, T23))
U3_a(l16_out_g([])) → p1_out_a(.(1, []))

The argument filtering Pi contains the following mapping:
p1_in_a(x1)  =  p1_in_a
U2_a(x1, x2)  =  U2_a(x2)
l16_in_a(x1)  =  l16_in_a
l16_out_a(x1)  =  l16_out_a(x1)
U1_a(x1, x2)  =  U1_a(x2)
p1_out_a(x1)  =  p1_out_a(x1)
U3_a(x1)  =  U3_a(x1)
l16_in_g(x1)  =  l16_in_g(x1)
[]  =  []
l16_out_g(x1)  =  l16_out_g(x1)
.(x1, x2)  =  .(x1, x2)
1  =  1
U1_g(x1, x2)  =  U1_g(x1, x2)
P1_IN_A(x1)  =  P1_IN_A
U2_A(x1, x2)  =  U2_A(x2)
L16_IN_A(x1)  =  L16_IN_A
U1_A(x1, x2)  =  U1_A(x2)
U3_A(x1)  =  U3_A(x1)
L16_IN_G(x1)  =  L16_IN_G(x1)
U1_G(x1, x2)  =  U1_G(x1, x2)

We have to consider all (P,R,Pi)-chains

(26) Obligation:

Pi DP problem:
The TRS P consists of the following rules:

P1_IN_A(.(1, T13)) → U2_A(T13, l16_in_a(T13))
P1_IN_A(.(1, T13)) → L16_IN_A(T13)
L16_IN_A(.(1, T23)) → U1_A(T23, l16_in_a(T23))
L16_IN_A(.(1, T23)) → L16_IN_A(T23)
P1_IN_A(.(1, [])) → U3_A(l16_in_g([]))
P1_IN_A(.(1, [])) → L16_IN_G([])
L16_IN_G(.(1, T23)) → U1_G(T23, l16_in_g(T23))
L16_IN_G(.(1, T23)) → L16_IN_G(T23)

The TRS R consists of the following rules:

p1_in_a(.(1, T13)) → U2_a(T13, l16_in_a(T13))
l16_in_a([]) → l16_out_a([])
l16_in_a(.(1, T23)) → U1_a(T23, l16_in_a(T23))
U1_a(T23, l16_out_a(T23)) → l16_out_a(.(1, T23))
U2_a(T13, l16_out_a(T13)) → p1_out_a(.(1, T13))
p1_in_a(.(1, [])) → U3_a(l16_in_g([]))
l16_in_g([]) → l16_out_g([])
l16_in_g(.(1, T23)) → U1_g(T23, l16_in_g(T23))
U1_g(T23, l16_out_g(T23)) → l16_out_g(.(1, T23))
U3_a(l16_out_g([])) → p1_out_a(.(1, []))

The argument filtering Pi contains the following mapping:
p1_in_a(x1)  =  p1_in_a
U2_a(x1, x2)  =  U2_a(x2)
l16_in_a(x1)  =  l16_in_a
l16_out_a(x1)  =  l16_out_a(x1)
U1_a(x1, x2)  =  U1_a(x2)
p1_out_a(x1)  =  p1_out_a(x1)
U3_a(x1)  =  U3_a(x1)
l16_in_g(x1)  =  l16_in_g(x1)
[]  =  []
l16_out_g(x1)  =  l16_out_g(x1)
.(x1, x2)  =  .(x1, x2)
1  =  1
U1_g(x1, x2)  =  U1_g(x1, x2)
P1_IN_A(x1)  =  P1_IN_A
U2_A(x1, x2)  =  U2_A(x2)
L16_IN_A(x1)  =  L16_IN_A
U1_A(x1, x2)  =  U1_A(x2)
U3_A(x1)  =  U3_A(x1)
L16_IN_G(x1)  =  L16_IN_G(x1)
U1_G(x1, x2)  =  U1_G(x1, x2)

We have to consider all (P,R,Pi)-chains

(27) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 2 SCCs with 6 less nodes.

(28) Complex Obligation (AND)

(29) Obligation:

Pi DP problem:
The TRS P consists of the following rules:

L16_IN_G(.(1, T23)) → L16_IN_G(T23)

The TRS R consists of the following rules:

p1_in_a(.(1, T13)) → U2_a(T13, l16_in_a(T13))
l16_in_a([]) → l16_out_a([])
l16_in_a(.(1, T23)) → U1_a(T23, l16_in_a(T23))
U1_a(T23, l16_out_a(T23)) → l16_out_a(.(1, T23))
U2_a(T13, l16_out_a(T13)) → p1_out_a(.(1, T13))
p1_in_a(.(1, [])) → U3_a(l16_in_g([]))
l16_in_g([]) → l16_out_g([])
l16_in_g(.(1, T23)) → U1_g(T23, l16_in_g(T23))
U1_g(T23, l16_out_g(T23)) → l16_out_g(.(1, T23))
U3_a(l16_out_g([])) → p1_out_a(.(1, []))

The argument filtering Pi contains the following mapping:
p1_in_a(x1)  =  p1_in_a
U2_a(x1, x2)  =  U2_a(x2)
l16_in_a(x1)  =  l16_in_a
l16_out_a(x1)  =  l16_out_a(x1)
U1_a(x1, x2)  =  U1_a(x2)
p1_out_a(x1)  =  p1_out_a(x1)
U3_a(x1)  =  U3_a(x1)
l16_in_g(x1)  =  l16_in_g(x1)
[]  =  []
l16_out_g(x1)  =  l16_out_g(x1)
.(x1, x2)  =  .(x1, x2)
1  =  1
U1_g(x1, x2)  =  U1_g(x1, x2)
L16_IN_G(x1)  =  L16_IN_G(x1)

We have to consider all (P,R,Pi)-chains

(30) UsableRulesProof (EQUIVALENT transformation)

For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.

(31) Obligation:

Pi DP problem:
The TRS P consists of the following rules:

L16_IN_G(.(1, T23)) → L16_IN_G(T23)

R is empty.
Pi is empty.
We have to consider all (P,R,Pi)-chains

(32) PiDPToQDPProof (EQUIVALENT transformation)

Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.

(33) Obligation:

Q DP problem:
The TRS P consists of the following rules:

L16_IN_G(.(1, T23)) → L16_IN_G(T23)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.

(34) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] 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:

  • L16_IN_G(.(1, T23)) → L16_IN_G(T23)
    The graph contains the following edges 1 > 1

(35) YES

(36) Obligation:

Pi DP problem:
The TRS P consists of the following rules:

L16_IN_A(.(1, T23)) → L16_IN_A(T23)

The TRS R consists of the following rules:

p1_in_a(.(1, T13)) → U2_a(T13, l16_in_a(T13))
l16_in_a([]) → l16_out_a([])
l16_in_a(.(1, T23)) → U1_a(T23, l16_in_a(T23))
U1_a(T23, l16_out_a(T23)) → l16_out_a(.(1, T23))
U2_a(T13, l16_out_a(T13)) → p1_out_a(.(1, T13))
p1_in_a(.(1, [])) → U3_a(l16_in_g([]))
l16_in_g([]) → l16_out_g([])
l16_in_g(.(1, T23)) → U1_g(T23, l16_in_g(T23))
U1_g(T23, l16_out_g(T23)) → l16_out_g(.(1, T23))
U3_a(l16_out_g([])) → p1_out_a(.(1, []))

The argument filtering Pi contains the following mapping:
p1_in_a(x1)  =  p1_in_a
U2_a(x1, x2)  =  U2_a(x2)
l16_in_a(x1)  =  l16_in_a
l16_out_a(x1)  =  l16_out_a(x1)
U1_a(x1, x2)  =  U1_a(x2)
p1_out_a(x1)  =  p1_out_a(x1)
U3_a(x1)  =  U3_a(x1)
l16_in_g(x1)  =  l16_in_g(x1)
[]  =  []
l16_out_g(x1)  =  l16_out_g(x1)
.(x1, x2)  =  .(x1, x2)
1  =  1
U1_g(x1, x2)  =  U1_g(x1, x2)
L16_IN_A(x1)  =  L16_IN_A

We have to consider all (P,R,Pi)-chains

(37) UsableRulesProof (EQUIVALENT transformation)

For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.

(38) Obligation:

Pi DP problem:
The TRS P consists of the following rules:

L16_IN_A(.(1, T23)) → L16_IN_A(T23)

R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x1, x2)
1  =  1
L16_IN_A(x1)  =  L16_IN_A

We have to consider all (P,R,Pi)-chains

(39) PiDPToQDPProof (SOUND transformation)

Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.

(40) Obligation:

Q DP problem:
The TRS P consists of the following rules:

L16_IN_A → L16_IN_A

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.

(41) 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 = L16_IN_A evaluates to t =L16_IN_A

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 L16_IN_A to L16_IN_A.



(42) NO