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

0 Prolog
↳1 CutEliminatorProof (⇐)
↳2 Prolog
  ↳3 PrologToPiTRSProof (⇐)
    ↳4 PiTRS
    ↳5 DependencyPairsProof (⇔)
    ↳6 PiDP
    ↳7 DependencyGraphProof (⇔)
    ↳8 PiDP
    ↳9 UsableRulesProof (⇔)
    ↳10 PiDP
    ↳11 PiDPToQDPProof (⇐)
    ↳12 QDP
    ↳13 Rewriting (⇔)
    ↳14 QDP
    ↳15 UsableRulesProof (⇔)
    ↳16 QDP
    ↳17 QReductionProof (⇔)
    ↳18 QDP
    ↳19 Instantiation (⇔)
    ↳20 QDP
    ↳21 NonTerminationProof (⇔)
    ↳22 FALSE
  ↳23 PrologToPiTRSProof (⇐)
    ↳24 PiTRS
    ↳25 DependencyPairsProof (⇔)
    ↳26 PiDP
    ↳27 DependencyGraphProof (⇔)
    ↳28 PiDP
    ↳29 UsableRulesProof (⇔)
    ↳30 PiDP
    ↳31 PiDPToQDPProof (⇐)
    ↳32 QDP
    ↳33 Rewriting (⇔)
    ↳34 QDP
    ↳35 UsableRulesProof (⇔)
    ↳36 QDP
    ↳37 QReductionProof (⇔)
    ↳38 QDP
    ↳39 Instantiation (⇔)
    ↳40 QDP
    ↳41 NonTerminationProof (⇔)
    ↳42 FALSE

(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) CutEliminatorProof (SOUND transformation)

Eliminated all cuts by simply ignoring them[PROLOG].

(2) Obligation:

Clauses:

p(X) :- ','(l(X), q(X)).
q(.(A, [])).
r(1).
l([]).
l(.(H, T)) :- ','(r(H), l(T)).

Queries:

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

p_in_a(X) → U1_a(X, l_in_a(X))
l_in_a([]) → l_out_a([])
l_in_a(.(H, T)) → U3_a(H, T, r_in_a(H))
r_in_a(1) → r_out_a(1)
U3_a(H, T, r_out_a(H)) → U4_a(H, T, l_in_a(T))
U4_a(H, T, l_out_a(T)) → l_out_a(.(H, T))
U1_a(X, l_out_a(X)) → U2_a(X, q_in_g(X))
q_in_g(.(A, [])) → q_out_g(.(A, []))
U2_a(X, q_out_g(X)) → p_out_a(X)

The argument filtering Pi contains the following mapping:
p_in_a(x1)  =  p_in_a
U1_a(x1, x2)  =  U1_a(x2)
l_in_a(x1)  =  l_in_a
l_out_a(x1)  =  l_out_a(x1)
U3_a(x1, x2, x3)  =  U3_a(x3)
r_in_a(x1)  =  r_in_a
r_out_a(x1)  =  r_out_a(x1)
U4_a(x1, x2, x3)  =  U4_a(x1, x3)
U2_a(x1, x2)  =  U2_a(x1, x2)
q_in_g(x1)  =  q_in_g(x1)
.(x1, x2)  =  .(x1, x2)
[]  =  []
q_out_g(x1)  =  q_out_g
p_out_a(x1)  =  p_out_a(x1)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(4) Obligation:

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

p_in_a(X) → U1_a(X, l_in_a(X))
l_in_a([]) → l_out_a([])
l_in_a(.(H, T)) → U3_a(H, T, r_in_a(H))
r_in_a(1) → r_out_a(1)
U3_a(H, T, r_out_a(H)) → U4_a(H, T, l_in_a(T))
U4_a(H, T, l_out_a(T)) → l_out_a(.(H, T))
U1_a(X, l_out_a(X)) → U2_a(X, q_in_g(X))
q_in_g(.(A, [])) → q_out_g(.(A, []))
U2_a(X, q_out_g(X)) → p_out_a(X)

The argument filtering Pi contains the following mapping:
p_in_a(x1)  =  p_in_a
U1_a(x1, x2)  =  U1_a(x2)
l_in_a(x1)  =  l_in_a
l_out_a(x1)  =  l_out_a(x1)
U3_a(x1, x2, x3)  =  U3_a(x3)
r_in_a(x1)  =  r_in_a
r_out_a(x1)  =  r_out_a(x1)
U4_a(x1, x2, x3)  =  U4_a(x1, x3)
U2_a(x1, x2)  =  U2_a(x1, x2)
q_in_g(x1)  =  q_in_g(x1)
.(x1, x2)  =  .(x1, x2)
[]  =  []
q_out_g(x1)  =  q_out_g
p_out_a(x1)  =  p_out_a(x1)

(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:

P_IN_A(X) → U1_A(X, l_in_a(X))
P_IN_A(X) → L_IN_A(X)
L_IN_A(.(H, T)) → U3_A(H, T, r_in_a(H))
L_IN_A(.(H, T)) → R_IN_A(H)
U3_A(H, T, r_out_a(H)) → U4_A(H, T, l_in_a(T))
U3_A(H, T, r_out_a(H)) → L_IN_A(T)
U1_A(X, l_out_a(X)) → U2_A(X, q_in_g(X))
U1_A(X, l_out_a(X)) → Q_IN_G(X)

The TRS R consists of the following rules:

p_in_a(X) → U1_a(X, l_in_a(X))
l_in_a([]) → l_out_a([])
l_in_a(.(H, T)) → U3_a(H, T, r_in_a(H))
r_in_a(1) → r_out_a(1)
U3_a(H, T, r_out_a(H)) → U4_a(H, T, l_in_a(T))
U4_a(H, T, l_out_a(T)) → l_out_a(.(H, T))
U1_a(X, l_out_a(X)) → U2_a(X, q_in_g(X))
q_in_g(.(A, [])) → q_out_g(.(A, []))
U2_a(X, q_out_g(X)) → p_out_a(X)

The argument filtering Pi contains the following mapping:
p_in_a(x1)  =  p_in_a
U1_a(x1, x2)  =  U1_a(x2)
l_in_a(x1)  =  l_in_a
l_out_a(x1)  =  l_out_a(x1)
U3_a(x1, x2, x3)  =  U3_a(x3)
r_in_a(x1)  =  r_in_a
r_out_a(x1)  =  r_out_a(x1)
U4_a(x1, x2, x3)  =  U4_a(x1, x3)
U2_a(x1, x2)  =  U2_a(x1, x2)
q_in_g(x1)  =  q_in_g(x1)
.(x1, x2)  =  .(x1, x2)
[]  =  []
q_out_g(x1)  =  q_out_g
p_out_a(x1)  =  p_out_a(x1)
P_IN_A(x1)  =  P_IN_A
U1_A(x1, x2)  =  U1_A(x2)
L_IN_A(x1)  =  L_IN_A
U3_A(x1, x2, x3)  =  U3_A(x3)
R_IN_A(x1)  =  R_IN_A
U4_A(x1, x2, x3)  =  U4_A(x1, x3)
U2_A(x1, x2)  =  U2_A(x1, x2)
Q_IN_G(x1)  =  Q_IN_G(x1)

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

(6) Obligation:

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

P_IN_A(X) → U1_A(X, l_in_a(X))
P_IN_A(X) → L_IN_A(X)
L_IN_A(.(H, T)) → U3_A(H, T, r_in_a(H))
L_IN_A(.(H, T)) → R_IN_A(H)
U3_A(H, T, r_out_a(H)) → U4_A(H, T, l_in_a(T))
U3_A(H, T, r_out_a(H)) → L_IN_A(T)
U1_A(X, l_out_a(X)) → U2_A(X, q_in_g(X))
U1_A(X, l_out_a(X)) → Q_IN_G(X)

The TRS R consists of the following rules:

p_in_a(X) → U1_a(X, l_in_a(X))
l_in_a([]) → l_out_a([])
l_in_a(.(H, T)) → U3_a(H, T, r_in_a(H))
r_in_a(1) → r_out_a(1)
U3_a(H, T, r_out_a(H)) → U4_a(H, T, l_in_a(T))
U4_a(H, T, l_out_a(T)) → l_out_a(.(H, T))
U1_a(X, l_out_a(X)) → U2_a(X, q_in_g(X))
q_in_g(.(A, [])) → q_out_g(.(A, []))
U2_a(X, q_out_g(X)) → p_out_a(X)

The argument filtering Pi contains the following mapping:
p_in_a(x1)  =  p_in_a
U1_a(x1, x2)  =  U1_a(x2)
l_in_a(x1)  =  l_in_a
l_out_a(x1)  =  l_out_a(x1)
U3_a(x1, x2, x3)  =  U3_a(x3)
r_in_a(x1)  =  r_in_a
r_out_a(x1)  =  r_out_a(x1)
U4_a(x1, x2, x3)  =  U4_a(x1, x3)
U2_a(x1, x2)  =  U2_a(x1, x2)
q_in_g(x1)  =  q_in_g(x1)
.(x1, x2)  =  .(x1, x2)
[]  =  []
q_out_g(x1)  =  q_out_g
p_out_a(x1)  =  p_out_a(x1)
P_IN_A(x1)  =  P_IN_A
U1_A(x1, x2)  =  U1_A(x2)
L_IN_A(x1)  =  L_IN_A
U3_A(x1, x2, x3)  =  U3_A(x3)
R_IN_A(x1)  =  R_IN_A
U4_A(x1, x2, x3)  =  U4_A(x1, x3)
U2_A(x1, x2)  =  U2_A(x1, x2)
Q_IN_G(x1)  =  Q_IN_G(x1)

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

(7) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 6 less nodes.

(8) Obligation:

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

U3_A(H, T, r_out_a(H)) → L_IN_A(T)
L_IN_A(.(H, T)) → U3_A(H, T, r_in_a(H))

The TRS R consists of the following rules:

p_in_a(X) → U1_a(X, l_in_a(X))
l_in_a([]) → l_out_a([])
l_in_a(.(H, T)) → U3_a(H, T, r_in_a(H))
r_in_a(1) → r_out_a(1)
U3_a(H, T, r_out_a(H)) → U4_a(H, T, l_in_a(T))
U4_a(H, T, l_out_a(T)) → l_out_a(.(H, T))
U1_a(X, l_out_a(X)) → U2_a(X, q_in_g(X))
q_in_g(.(A, [])) → q_out_g(.(A, []))
U2_a(X, q_out_g(X)) → p_out_a(X)

The argument filtering Pi contains the following mapping:
p_in_a(x1)  =  p_in_a
U1_a(x1, x2)  =  U1_a(x2)
l_in_a(x1)  =  l_in_a
l_out_a(x1)  =  l_out_a(x1)
U3_a(x1, x2, x3)  =  U3_a(x3)
r_in_a(x1)  =  r_in_a
r_out_a(x1)  =  r_out_a(x1)
U4_a(x1, x2, x3)  =  U4_a(x1, x3)
U2_a(x1, x2)  =  U2_a(x1, x2)
q_in_g(x1)  =  q_in_g(x1)
.(x1, x2)  =  .(x1, x2)
[]  =  []
q_out_g(x1)  =  q_out_g
p_out_a(x1)  =  p_out_a(x1)
L_IN_A(x1)  =  L_IN_A
U3_A(x1, x2, x3)  =  U3_A(x3)

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:

U3_A(H, T, r_out_a(H)) → L_IN_A(T)
L_IN_A(.(H, T)) → U3_A(H, T, r_in_a(H))

The TRS R consists of the following rules:

r_in_a(1) → r_out_a(1)

The argument filtering Pi contains the following mapping:
r_in_a(x1)  =  r_in_a
r_out_a(x1)  =  r_out_a(x1)
.(x1, x2)  =  .(x1, x2)
L_IN_A(x1)  =  L_IN_A
U3_A(x1, x2, x3)  =  U3_A(x3)

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:

U3_A(r_out_a(H)) → L_IN_A
L_IN_A → U3_A(r_in_a)

The TRS R consists of the following rules:

r_in_a → r_out_a(1)

The set Q consists of the following terms:

r_in_a

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

(13) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule L_IN_A → U3_A(r_in_a) at position [0] we obtained the following new rules [LPAR04]:

L_IN_A → U3_A(r_out_a(1))

(14) Obligation:

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

U3_A(r_out_a(H)) → L_IN_A
L_IN_A → U3_A(r_out_a(1))

The TRS R consists of the following rules:

r_in_a → r_out_a(1)

The set Q consists of the following terms:

r_in_a

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

(15) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(16) Obligation:

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

U3_A(r_out_a(H)) → L_IN_A
L_IN_A → U3_A(r_out_a(1))

R is empty.
The set Q consists of the following terms:

r_in_a

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

(17) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

r_in_a

(18) Obligation:

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

U3_A(r_out_a(H)) → L_IN_A
L_IN_A → U3_A(r_out_a(1))

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

(19) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule U3_A(r_out_a(H)) → L_IN_A we obtained the following new rules [LPAR04]:

U3_A(r_out_a(1)) → L_IN_A

(20) Obligation:

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

L_IN_A → U3_A(r_out_a(1))
U3_A(r_out_a(1)) → L_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 narrowing to the left:

s = U3_A(r_out_a(1)) evaluates to t =U3_A(r_out_a(1))

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



Rewriting sequence

U3_A(r_out_a(1)) → L_IN_A
with rule U3_A(r_out_a(1)) → L_IN_A at position [] and matcher [ ]

L_IN_A → U3_A(r_out_a(1))
with rule L_IN_A → U3_A(r_out_a(1))

Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence


All these steps are and every following step will be a correct step w.r.t to Q.



(22) FALSE

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

p_in_a(X) → U1_a(X, l_in_a(X))
l_in_a([]) → l_out_a([])
l_in_a(.(H, T)) → U3_a(H, T, r_in_a(H))
r_in_a(1) → r_out_a(1)
U3_a(H, T, r_out_a(H)) → U4_a(H, T, l_in_a(T))
U4_a(H, T, l_out_a(T)) → l_out_a(.(H, T))
U1_a(X, l_out_a(X)) → U2_a(X, q_in_g(X))
q_in_g(.(A, [])) → q_out_g(.(A, []))
U2_a(X, q_out_g(X)) → p_out_a(X)

The argument filtering Pi contains the following mapping:
p_in_a(x1)  =  p_in_a
U1_a(x1, x2)  =  U1_a(x2)
l_in_a(x1)  =  l_in_a
l_out_a(x1)  =  l_out_a(x1)
U3_a(x1, x2, x3)  =  U3_a(x3)
r_in_a(x1)  =  r_in_a
r_out_a(x1)  =  r_out_a(x1)
U4_a(x1, x2, x3)  =  U4_a(x1, x3)
U2_a(x1, x2)  =  U2_a(x1, x2)
q_in_g(x1)  =  q_in_g(x1)
.(x1, x2)  =  .(x1, x2)
[]  =  []
q_out_g(x1)  =  q_out_g(x1)
p_out_a(x1)  =  p_out_a(x1)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(24) Obligation:

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

p_in_a(X) → U1_a(X, l_in_a(X))
l_in_a([]) → l_out_a([])
l_in_a(.(H, T)) → U3_a(H, T, r_in_a(H))
r_in_a(1) → r_out_a(1)
U3_a(H, T, r_out_a(H)) → U4_a(H, T, l_in_a(T))
U4_a(H, T, l_out_a(T)) → l_out_a(.(H, T))
U1_a(X, l_out_a(X)) → U2_a(X, q_in_g(X))
q_in_g(.(A, [])) → q_out_g(.(A, []))
U2_a(X, q_out_g(X)) → p_out_a(X)

The argument filtering Pi contains the following mapping:
p_in_a(x1)  =  p_in_a
U1_a(x1, x2)  =  U1_a(x2)
l_in_a(x1)  =  l_in_a
l_out_a(x1)  =  l_out_a(x1)
U3_a(x1, x2, x3)  =  U3_a(x3)
r_in_a(x1)  =  r_in_a
r_out_a(x1)  =  r_out_a(x1)
U4_a(x1, x2, x3)  =  U4_a(x1, x3)
U2_a(x1, x2)  =  U2_a(x1, x2)
q_in_g(x1)  =  q_in_g(x1)
.(x1, x2)  =  .(x1, x2)
[]  =  []
q_out_g(x1)  =  q_out_g(x1)
p_out_a(x1)  =  p_out_a(x1)

(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:

P_IN_A(X) → U1_A(X, l_in_a(X))
P_IN_A(X) → L_IN_A(X)
L_IN_A(.(H, T)) → U3_A(H, T, r_in_a(H))
L_IN_A(.(H, T)) → R_IN_A(H)
U3_A(H, T, r_out_a(H)) → U4_A(H, T, l_in_a(T))
U3_A(H, T, r_out_a(H)) → L_IN_A(T)
U1_A(X, l_out_a(X)) → U2_A(X, q_in_g(X))
U1_A(X, l_out_a(X)) → Q_IN_G(X)

The TRS R consists of the following rules:

p_in_a(X) → U1_a(X, l_in_a(X))
l_in_a([]) → l_out_a([])
l_in_a(.(H, T)) → U3_a(H, T, r_in_a(H))
r_in_a(1) → r_out_a(1)
U3_a(H, T, r_out_a(H)) → U4_a(H, T, l_in_a(T))
U4_a(H, T, l_out_a(T)) → l_out_a(.(H, T))
U1_a(X, l_out_a(X)) → U2_a(X, q_in_g(X))
q_in_g(.(A, [])) → q_out_g(.(A, []))
U2_a(X, q_out_g(X)) → p_out_a(X)

The argument filtering Pi contains the following mapping:
p_in_a(x1)  =  p_in_a
U1_a(x1, x2)  =  U1_a(x2)
l_in_a(x1)  =  l_in_a
l_out_a(x1)  =  l_out_a(x1)
U3_a(x1, x2, x3)  =  U3_a(x3)
r_in_a(x1)  =  r_in_a
r_out_a(x1)  =  r_out_a(x1)
U4_a(x1, x2, x3)  =  U4_a(x1, x3)
U2_a(x1, x2)  =  U2_a(x1, x2)
q_in_g(x1)  =  q_in_g(x1)
.(x1, x2)  =  .(x1, x2)
[]  =  []
q_out_g(x1)  =  q_out_g(x1)
p_out_a(x1)  =  p_out_a(x1)
P_IN_A(x1)  =  P_IN_A
U1_A(x1, x2)  =  U1_A(x2)
L_IN_A(x1)  =  L_IN_A
U3_A(x1, x2, x3)  =  U3_A(x3)
R_IN_A(x1)  =  R_IN_A
U4_A(x1, x2, x3)  =  U4_A(x1, x3)
U2_A(x1, x2)  =  U2_A(x1, x2)
Q_IN_G(x1)  =  Q_IN_G(x1)

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

(26) Obligation:

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

P_IN_A(X) → U1_A(X, l_in_a(X))
P_IN_A(X) → L_IN_A(X)
L_IN_A(.(H, T)) → U3_A(H, T, r_in_a(H))
L_IN_A(.(H, T)) → R_IN_A(H)
U3_A(H, T, r_out_a(H)) → U4_A(H, T, l_in_a(T))
U3_A(H, T, r_out_a(H)) → L_IN_A(T)
U1_A(X, l_out_a(X)) → U2_A(X, q_in_g(X))
U1_A(X, l_out_a(X)) → Q_IN_G(X)

The TRS R consists of the following rules:

p_in_a(X) → U1_a(X, l_in_a(X))
l_in_a([]) → l_out_a([])
l_in_a(.(H, T)) → U3_a(H, T, r_in_a(H))
r_in_a(1) → r_out_a(1)
U3_a(H, T, r_out_a(H)) → U4_a(H, T, l_in_a(T))
U4_a(H, T, l_out_a(T)) → l_out_a(.(H, T))
U1_a(X, l_out_a(X)) → U2_a(X, q_in_g(X))
q_in_g(.(A, [])) → q_out_g(.(A, []))
U2_a(X, q_out_g(X)) → p_out_a(X)

The argument filtering Pi contains the following mapping:
p_in_a(x1)  =  p_in_a
U1_a(x1, x2)  =  U1_a(x2)
l_in_a(x1)  =  l_in_a
l_out_a(x1)  =  l_out_a(x1)
U3_a(x1, x2, x3)  =  U3_a(x3)
r_in_a(x1)  =  r_in_a
r_out_a(x1)  =  r_out_a(x1)
U4_a(x1, x2, x3)  =  U4_a(x1, x3)
U2_a(x1, x2)  =  U2_a(x1, x2)
q_in_g(x1)  =  q_in_g(x1)
.(x1, x2)  =  .(x1, x2)
[]  =  []
q_out_g(x1)  =  q_out_g(x1)
p_out_a(x1)  =  p_out_a(x1)
P_IN_A(x1)  =  P_IN_A
U1_A(x1, x2)  =  U1_A(x2)
L_IN_A(x1)  =  L_IN_A
U3_A(x1, x2, x3)  =  U3_A(x3)
R_IN_A(x1)  =  R_IN_A
U4_A(x1, x2, x3)  =  U4_A(x1, x3)
U2_A(x1, x2)  =  U2_A(x1, x2)
Q_IN_G(x1)  =  Q_IN_G(x1)

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

(27) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 6 less nodes.

(28) Obligation:

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

U3_A(H, T, r_out_a(H)) → L_IN_A(T)
L_IN_A(.(H, T)) → U3_A(H, T, r_in_a(H))

The TRS R consists of the following rules:

p_in_a(X) → U1_a(X, l_in_a(X))
l_in_a([]) → l_out_a([])
l_in_a(.(H, T)) → U3_a(H, T, r_in_a(H))
r_in_a(1) → r_out_a(1)
U3_a(H, T, r_out_a(H)) → U4_a(H, T, l_in_a(T))
U4_a(H, T, l_out_a(T)) → l_out_a(.(H, T))
U1_a(X, l_out_a(X)) → U2_a(X, q_in_g(X))
q_in_g(.(A, [])) → q_out_g(.(A, []))
U2_a(X, q_out_g(X)) → p_out_a(X)

The argument filtering Pi contains the following mapping:
p_in_a(x1)  =  p_in_a
U1_a(x1, x2)  =  U1_a(x2)
l_in_a(x1)  =  l_in_a
l_out_a(x1)  =  l_out_a(x1)
U3_a(x1, x2, x3)  =  U3_a(x3)
r_in_a(x1)  =  r_in_a
r_out_a(x1)  =  r_out_a(x1)
U4_a(x1, x2, x3)  =  U4_a(x1, x3)
U2_a(x1, x2)  =  U2_a(x1, x2)
q_in_g(x1)  =  q_in_g(x1)
.(x1, x2)  =  .(x1, x2)
[]  =  []
q_out_g(x1)  =  q_out_g(x1)
p_out_a(x1)  =  p_out_a(x1)
L_IN_A(x1)  =  L_IN_A
U3_A(x1, x2, x3)  =  U3_A(x3)

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

(29) UsableRulesProof (EQUIVALENT transformation)

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

(30) Obligation:

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

U3_A(H, T, r_out_a(H)) → L_IN_A(T)
L_IN_A(.(H, T)) → U3_A(H, T, r_in_a(H))

The TRS R consists of the following rules:

r_in_a(1) → r_out_a(1)

The argument filtering Pi contains the following mapping:
r_in_a(x1)  =  r_in_a
r_out_a(x1)  =  r_out_a(x1)
.(x1, x2)  =  .(x1, x2)
L_IN_A(x1)  =  L_IN_A
U3_A(x1, x2, x3)  =  U3_A(x3)

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

(31) PiDPToQDPProof (SOUND transformation)

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

(32) Obligation:

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

U3_A(r_out_a(H)) → L_IN_A
L_IN_A → U3_A(r_in_a)

The TRS R consists of the following rules:

r_in_a → r_out_a(1)

The set Q consists of the following terms:

r_in_a

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

(33) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule L_IN_A → U3_A(r_in_a) at position [0] we obtained the following new rules [LPAR04]:

L_IN_A → U3_A(r_out_a(1))

(34) Obligation:

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

U3_A(r_out_a(H)) → L_IN_A
L_IN_A → U3_A(r_out_a(1))

The TRS R consists of the following rules:

r_in_a → r_out_a(1)

The set Q consists of the following terms:

r_in_a

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

(35) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(36) Obligation:

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

U3_A(r_out_a(H)) → L_IN_A
L_IN_A → U3_A(r_out_a(1))

R is empty.
The set Q consists of the following terms:

r_in_a

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

(37) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

r_in_a

(38) Obligation:

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

U3_A(r_out_a(H)) → L_IN_A
L_IN_A → U3_A(r_out_a(1))

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

(39) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule U3_A(r_out_a(H)) → L_IN_A we obtained the following new rules [LPAR04]:

U3_A(r_out_a(1)) → L_IN_A

(40) Obligation:

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

L_IN_A → U3_A(r_out_a(1))
U3_A(r_out_a(1)) → L_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 narrowing to the left:

s = U3_A(r_out_a(1)) evaluates to t =U3_A(r_out_a(1))

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



Rewriting sequence

U3_A(r_out_a(1)) → L_IN_A
with rule U3_A(r_out_a(1)) → L_IN_A at position [] and matcher [ ]

L_IN_A → U3_A(r_out_a(1))
with rule L_IN_A → U3_A(r_out_a(1))

Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence


All these steps are and every following step will be a correct step w.r.t to Q.



(42) FALSE