Left Termination of the query pattern num_in_1(g) w.r.t. the given Prolog program could not be shown:

(0) Obligation:

Clauses:

num(0) :- !.
num(X) :- ','(p(X, Y), num(Y)).
p(0, 0).
p(s(X), X).

Queries:

num(g).

(1) CutEliminatorProof (SOUND transformation)

Eliminated all cuts by simply ignoring them[PROLOG].

(2) Obligation:

Clauses:

num(0) :- true.
num(X) :- ','(p(X, Y), num(Y)).
p(0, 0).
p(s(X), X).

Queries:

num(g).

(3) UndefinedPredicateHandlerProof (SOUND transformation)

Added facts for all undefined predicates [PROLOG].

(4) Obligation:

Clauses:

num(0) :- true.
num(X) :- ','(p(X, Y), num(Y)).
p(0, 0).
p(s(X), X).
true.

Queries:

num(g).

(5) PrologToPiTRSProof (SOUND transformation)

We use the technique of [LOPSTR]. With regard to the inferred argument filtering the predicates were used in the following modes:
num_in: (b)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

num_in_g(0) → U1_g(true_in_)
true_in_ → true_out_
U1_g(true_out_) → num_out_g(0)
num_in_g(X) → U2_g(X, p_in_ga(X, Y))
p_in_ga(0, 0) → p_out_ga(0, 0)
p_in_ga(s(X), X) → p_out_ga(s(X), X)
U2_g(X, p_out_ga(X, Y)) → U3_g(X, num_in_g(Y))
U3_g(X, num_out_g(Y)) → num_out_g(X)

The argument filtering Pi contains the following mapping:
num_in_g(x1)  =  num_in_g(x1)
0  =  0
U1_g(x1)  =  U1_g(x1)
true_in_  =  true_in_
true_out_  =  true_out_
num_out_g(x1)  =  num_out_g(x1)
U2_g(x1, x2)  =  U2_g(x1, x2)
p_in_ga(x1, x2)  =  p_in_ga(x1)
p_out_ga(x1, x2)  =  p_out_ga(x1, x2)
s(x1)  =  s(x1)
U3_g(x1, x2)  =  U3_g(x1, x2)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(6) Obligation:

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

num_in_g(0) → U1_g(true_in_)
true_in_ → true_out_
U1_g(true_out_) → num_out_g(0)
num_in_g(X) → U2_g(X, p_in_ga(X, Y))
p_in_ga(0, 0) → p_out_ga(0, 0)
p_in_ga(s(X), X) → p_out_ga(s(X), X)
U2_g(X, p_out_ga(X, Y)) → U3_g(X, num_in_g(Y))
U3_g(X, num_out_g(Y)) → num_out_g(X)

The argument filtering Pi contains the following mapping:
num_in_g(x1)  =  num_in_g(x1)
0  =  0
U1_g(x1)  =  U1_g(x1)
true_in_  =  true_in_
true_out_  =  true_out_
num_out_g(x1)  =  num_out_g(x1)
U2_g(x1, x2)  =  U2_g(x1, x2)
p_in_ga(x1, x2)  =  p_in_ga(x1)
p_out_ga(x1, x2)  =  p_out_ga(x1, x2)
s(x1)  =  s(x1)
U3_g(x1, x2)  =  U3_g(x1, x2)

(7) 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:

NUM_IN_G(0) → U1_G(true_in_)
NUM_IN_G(0) → TRUE_IN_
NUM_IN_G(X) → U2_G(X, p_in_ga(X, Y))
NUM_IN_G(X) → P_IN_GA(X, Y)
U2_G(X, p_out_ga(X, Y)) → U3_G(X, num_in_g(Y))
U2_G(X, p_out_ga(X, Y)) → NUM_IN_G(Y)

The TRS R consists of the following rules:

num_in_g(0) → U1_g(true_in_)
true_in_ → true_out_
U1_g(true_out_) → num_out_g(0)
num_in_g(X) → U2_g(X, p_in_ga(X, Y))
p_in_ga(0, 0) → p_out_ga(0, 0)
p_in_ga(s(X), X) → p_out_ga(s(X), X)
U2_g(X, p_out_ga(X, Y)) → U3_g(X, num_in_g(Y))
U3_g(X, num_out_g(Y)) → num_out_g(X)

The argument filtering Pi contains the following mapping:
num_in_g(x1)  =  num_in_g(x1)
0  =  0
U1_g(x1)  =  U1_g(x1)
true_in_  =  true_in_
true_out_  =  true_out_
num_out_g(x1)  =  num_out_g(x1)
U2_g(x1, x2)  =  U2_g(x1, x2)
p_in_ga(x1, x2)  =  p_in_ga(x1)
p_out_ga(x1, x2)  =  p_out_ga(x1, x2)
s(x1)  =  s(x1)
U3_g(x1, x2)  =  U3_g(x1, x2)
NUM_IN_G(x1)  =  NUM_IN_G(x1)
U1_G(x1)  =  U1_G(x1)
TRUE_IN_  =  TRUE_IN_
U2_G(x1, x2)  =  U2_G(x1, x2)
P_IN_GA(x1, x2)  =  P_IN_GA(x1)
U3_G(x1, x2)  =  U3_G(x1, x2)

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

(8) Obligation:

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

NUM_IN_G(0) → U1_G(true_in_)
NUM_IN_G(0) → TRUE_IN_
NUM_IN_G(X) → U2_G(X, p_in_ga(X, Y))
NUM_IN_G(X) → P_IN_GA(X, Y)
U2_G(X, p_out_ga(X, Y)) → U3_G(X, num_in_g(Y))
U2_G(X, p_out_ga(X, Y)) → NUM_IN_G(Y)

The TRS R consists of the following rules:

num_in_g(0) → U1_g(true_in_)
true_in_ → true_out_
U1_g(true_out_) → num_out_g(0)
num_in_g(X) → U2_g(X, p_in_ga(X, Y))
p_in_ga(0, 0) → p_out_ga(0, 0)
p_in_ga(s(X), X) → p_out_ga(s(X), X)
U2_g(X, p_out_ga(X, Y)) → U3_g(X, num_in_g(Y))
U3_g(X, num_out_g(Y)) → num_out_g(X)

The argument filtering Pi contains the following mapping:
num_in_g(x1)  =  num_in_g(x1)
0  =  0
U1_g(x1)  =  U1_g(x1)
true_in_  =  true_in_
true_out_  =  true_out_
num_out_g(x1)  =  num_out_g(x1)
U2_g(x1, x2)  =  U2_g(x1, x2)
p_in_ga(x1, x2)  =  p_in_ga(x1)
p_out_ga(x1, x2)  =  p_out_ga(x1, x2)
s(x1)  =  s(x1)
U3_g(x1, x2)  =  U3_g(x1, x2)
NUM_IN_G(x1)  =  NUM_IN_G(x1)
U1_G(x1)  =  U1_G(x1)
TRUE_IN_  =  TRUE_IN_
U2_G(x1, x2)  =  U2_G(x1, x2)
P_IN_GA(x1, x2)  =  P_IN_GA(x1)
U3_G(x1, x2)  =  U3_G(x1, x2)

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

(9) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 4 less nodes.

(10) Obligation:

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

NUM_IN_G(X) → U2_G(X, p_in_ga(X, Y))
U2_G(X, p_out_ga(X, Y)) → NUM_IN_G(Y)

The TRS R consists of the following rules:

num_in_g(0) → U1_g(true_in_)
true_in_ → true_out_
U1_g(true_out_) → num_out_g(0)
num_in_g(X) → U2_g(X, p_in_ga(X, Y))
p_in_ga(0, 0) → p_out_ga(0, 0)
p_in_ga(s(X), X) → p_out_ga(s(X), X)
U2_g(X, p_out_ga(X, Y)) → U3_g(X, num_in_g(Y))
U3_g(X, num_out_g(Y)) → num_out_g(X)

The argument filtering Pi contains the following mapping:
num_in_g(x1)  =  num_in_g(x1)
0  =  0
U1_g(x1)  =  U1_g(x1)
true_in_  =  true_in_
true_out_  =  true_out_
num_out_g(x1)  =  num_out_g(x1)
U2_g(x1, x2)  =  U2_g(x1, x2)
p_in_ga(x1, x2)  =  p_in_ga(x1)
p_out_ga(x1, x2)  =  p_out_ga(x1, x2)
s(x1)  =  s(x1)
U3_g(x1, x2)  =  U3_g(x1, x2)
NUM_IN_G(x1)  =  NUM_IN_G(x1)
U2_G(x1, x2)  =  U2_G(x1, x2)

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

(11) UsableRulesProof (EQUIVALENT transformation)

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

(12) Obligation:

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

NUM_IN_G(X) → U2_G(X, p_in_ga(X, Y))
U2_G(X, p_out_ga(X, Y)) → NUM_IN_G(Y)

The TRS R consists of the following rules:

p_in_ga(0, 0) → p_out_ga(0, 0)
p_in_ga(s(X), X) → p_out_ga(s(X), X)

The argument filtering Pi contains the following mapping:
0  =  0
p_in_ga(x1, x2)  =  p_in_ga(x1)
p_out_ga(x1, x2)  =  p_out_ga(x1, x2)
s(x1)  =  s(x1)
NUM_IN_G(x1)  =  NUM_IN_G(x1)
U2_G(x1, x2)  =  U2_G(x1, x2)

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

(13) PiDPToQDPProof (SOUND transformation)

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

(14) Obligation:

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

NUM_IN_G(X) → U2_G(X, p_in_ga(X))
U2_G(X, p_out_ga(X, Y)) → NUM_IN_G(Y)

The TRS R consists of the following rules:

p_in_ga(0) → p_out_ga(0, 0)
p_in_ga(s(X)) → p_out_ga(s(X), X)

The set Q consists of the following terms:

p_in_ga(x0)

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

(15) Narrowing (SOUND transformation)

By narrowing [LPAR04] the rule NUM_IN_G(X) → U2_G(X, p_in_ga(X)) at position [1] we obtained the following new rules [LPAR04]:

NUM_IN_G(0) → U2_G(0, p_out_ga(0, 0))
NUM_IN_G(s(x0)) → U2_G(s(x0), p_out_ga(s(x0), x0))

(16) Obligation:

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

U2_G(X, p_out_ga(X, Y)) → NUM_IN_G(Y)
NUM_IN_G(0) → U2_G(0, p_out_ga(0, 0))
NUM_IN_G(s(x0)) → U2_G(s(x0), p_out_ga(s(x0), x0))

The TRS R consists of the following rules:

p_in_ga(0) → p_out_ga(0, 0)
p_in_ga(s(X)) → p_out_ga(s(X), X)

The set Q consists of the following terms:

p_in_ga(x0)

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

(17) 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.

(18) Obligation:

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

U2_G(X, p_out_ga(X, Y)) → NUM_IN_G(Y)
NUM_IN_G(0) → U2_G(0, p_out_ga(0, 0))
NUM_IN_G(s(x0)) → U2_G(s(x0), p_out_ga(s(x0), x0))

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

p_in_ga(x0)

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

(19) 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].

p_in_ga(x0)

(20) Obligation:

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

U2_G(X, p_out_ga(X, Y)) → NUM_IN_G(Y)
NUM_IN_G(0) → U2_G(0, p_out_ga(0, 0))
NUM_IN_G(s(x0)) → U2_G(s(x0), p_out_ga(s(x0), x0))

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

(21) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule U2_G(X, p_out_ga(X, Y)) → NUM_IN_G(Y) we obtained the following new rules [LPAR04]:

U2_G(0, p_out_ga(0, 0)) → NUM_IN_G(0)
U2_G(s(z0), p_out_ga(s(z0), z0)) → NUM_IN_G(z0)

(22) Obligation:

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

NUM_IN_G(0) → U2_G(0, p_out_ga(0, 0))
NUM_IN_G(s(x0)) → U2_G(s(x0), p_out_ga(s(x0), x0))
U2_G(0, p_out_ga(0, 0)) → NUM_IN_G(0)
U2_G(s(z0), p_out_ga(s(z0), z0)) → NUM_IN_G(z0)

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

(23) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs.

(24) Complex Obligation (AND)

(25) Obligation:

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

U2_G(0, p_out_ga(0, 0)) → NUM_IN_G(0)
NUM_IN_G(0) → U2_G(0, p_out_ga(0, 0))

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

(26) 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 = NUM_IN_G(0) evaluates to t =NUM_IN_G(0)

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:

• Matcher: [ ]
• Semiunifier: [ ]

---

Rewriting sequence

NUM_IN_G(0) → U2_G(0, p_out_ga(0, 0))
with rule NUM_IN_G(0) → U2_G(0, p_out_ga(0, 0)) at position [] and matcher [ ]

U2_G(0, p_out_ga(0, 0)) → NUM_IN_G(0)
with rule U2_G(0, p_out_ga(0, 0)) → NUM_IN_G(0)

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.

(27) FALSE

(28) Obligation:

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

NUM_IN_G(s(x0)) → U2_G(s(x0), p_out_ga(s(x0), x0))
U2_G(s(z0), p_out_ga(s(z0), z0)) → NUM_IN_G(z0)

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

(29) 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:

• U2_G(s(z0), p_out_ga(s(z0), z0)) → NUM_IN_G(z0)
  The graph contains the following edges 1 > 1, 2 > 1

• NUM_IN_G(s(x0)) → U2_G(s(x0), p_out_ga(s(x0), x0))
  The graph contains the following edges 1 >= 1

(30) TRUE

(31) PrologToPiTRSProof (SOUND transformation)

We use the technique of [LOPSTR]. With regard to the inferred argument filtering the predicates were used in the following modes:
num_in: (b)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

num_in_g(0) → U1_g(true_in_)
true_in_ → true_out_
U1_g(true_out_) → num_out_g(0)
num_in_g(X) → U2_g(X, p_in_ga(X, Y))
p_in_ga(0, 0) → p_out_ga(0, 0)
p_in_ga(s(X), X) → p_out_ga(s(X), X)
U2_g(X, p_out_ga(X, Y)) → U3_g(X, num_in_g(Y))
U3_g(X, num_out_g(Y)) → num_out_g(X)

The argument filtering Pi contains the following mapping:
num_in_g(x1)  =  num_in_g(x1)
0  =  0
U1_g(x1)  =  U1_g(x1)
true_in_  =  true_in_
true_out_  =  true_out_
num_out_g(x1)  =  num_out_g
U2_g(x1, x2)  =  U2_g(x2)
p_in_ga(x1, x2)  =  p_in_ga(x1)
p_out_ga(x1, x2)  =  p_out_ga(x2)
s(x1)  =  s(x1)
U3_g(x1, x2)  =  U3_g(x2)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(32) Obligation:

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

num_in_g(0) → U1_g(true_in_)
true_in_ → true_out_
U1_g(true_out_) → num_out_g(0)
num_in_g(X) → U2_g(X, p_in_ga(X, Y))
p_in_ga(0, 0) → p_out_ga(0, 0)
p_in_ga(s(X), X) → p_out_ga(s(X), X)
U2_g(X, p_out_ga(X, Y)) → U3_g(X, num_in_g(Y))
U3_g(X, num_out_g(Y)) → num_out_g(X)

The argument filtering Pi contains the following mapping:
num_in_g(x1)  =  num_in_g(x1)
0  =  0
U1_g(x1)  =  U1_g(x1)
true_in_  =  true_in_
true_out_  =  true_out_
num_out_g(x1)  =  num_out_g
U2_g(x1, x2)  =  U2_g(x2)
p_in_ga(x1, x2)  =  p_in_ga(x1)
p_out_ga(x1, x2)  =  p_out_ga(x2)
s(x1)  =  s(x1)
U3_g(x1, x2)  =  U3_g(x2)

(33) 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:

NUM_IN_G(0) → U1_G(true_in_)
NUM_IN_G(0) → TRUE_IN_
NUM_IN_G(X) → U2_G(X, p_in_ga(X, Y))
NUM_IN_G(X) → P_IN_GA(X, Y)
U2_G(X, p_out_ga(X, Y)) → U3_G(X, num_in_g(Y))
U2_G(X, p_out_ga(X, Y)) → NUM_IN_G(Y)

The TRS R consists of the following rules:

num_in_g(0) → U1_g(true_in_)
true_in_ → true_out_
U1_g(true_out_) → num_out_g(0)
num_in_g(X) → U2_g(X, p_in_ga(X, Y))
p_in_ga(0, 0) → p_out_ga(0, 0)
p_in_ga(s(X), X) → p_out_ga(s(X), X)
U2_g(X, p_out_ga(X, Y)) → U3_g(X, num_in_g(Y))
U3_g(X, num_out_g(Y)) → num_out_g(X)

The argument filtering Pi contains the following mapping:
num_in_g(x1)  =  num_in_g(x1)
0  =  0
U1_g(x1)  =  U1_g(x1)
true_in_  =  true_in_
true_out_  =  true_out_
num_out_g(x1)  =  num_out_g
U2_g(x1, x2)  =  U2_g(x2)
p_in_ga(x1, x2)  =  p_in_ga(x1)
p_out_ga(x1, x2)  =  p_out_ga(x2)
s(x1)  =  s(x1)
U3_g(x1, x2)  =  U3_g(x2)
NUM_IN_G(x1)  =  NUM_IN_G(x1)
U1_G(x1)  =  U1_G(x1)
TRUE_IN_  =  TRUE_IN_
U2_G(x1, x2)  =  U2_G(x2)
P_IN_GA(x1, x2)  =  P_IN_GA(x1)
U3_G(x1, x2)  =  U3_G(x2)

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

(34) Obligation:

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

NUM_IN_G(0) → U1_G(true_in_)
NUM_IN_G(0) → TRUE_IN_
NUM_IN_G(X) → U2_G(X, p_in_ga(X, Y))
NUM_IN_G(X) → P_IN_GA(X, Y)
U2_G(X, p_out_ga(X, Y)) → U3_G(X, num_in_g(Y))
U2_G(X, p_out_ga(X, Y)) → NUM_IN_G(Y)

The TRS R consists of the following rules:

num_in_g(0) → U1_g(true_in_)
true_in_ → true_out_
U1_g(true_out_) → num_out_g(0)
num_in_g(X) → U2_g(X, p_in_ga(X, Y))
p_in_ga(0, 0) → p_out_ga(0, 0)
p_in_ga(s(X), X) → p_out_ga(s(X), X)
U2_g(X, p_out_ga(X, Y)) → U3_g(X, num_in_g(Y))
U3_g(X, num_out_g(Y)) → num_out_g(X)

The argument filtering Pi contains the following mapping:
num_in_g(x1)  =  num_in_g(x1)
0  =  0
U1_g(x1)  =  U1_g(x1)
true_in_  =  true_in_
true_out_  =  true_out_
num_out_g(x1)  =  num_out_g
U2_g(x1, x2)  =  U2_g(x2)
p_in_ga(x1, x2)  =  p_in_ga(x1)
p_out_ga(x1, x2)  =  p_out_ga(x2)
s(x1)  =  s(x1)
U3_g(x1, x2)  =  U3_g(x2)
NUM_IN_G(x1)  =  NUM_IN_G(x1)
U1_G(x1)  =  U1_G(x1)
TRUE_IN_  =  TRUE_IN_
U2_G(x1, x2)  =  U2_G(x2)
P_IN_GA(x1, x2)  =  P_IN_GA(x1)
U3_G(x1, x2)  =  U3_G(x2)

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

(35) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 4 less nodes.

(36) Obligation:

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

NUM_IN_G(X) → U2_G(X, p_in_ga(X, Y))
U2_G(X, p_out_ga(X, Y)) → NUM_IN_G(Y)

The TRS R consists of the following rules:

num_in_g(0) → U1_g(true_in_)
true_in_ → true_out_
U1_g(true_out_) → num_out_g(0)
num_in_g(X) → U2_g(X, p_in_ga(X, Y))
p_in_ga(0, 0) → p_out_ga(0, 0)
p_in_ga(s(X), X) → p_out_ga(s(X), X)
U2_g(X, p_out_ga(X, Y)) → U3_g(X, num_in_g(Y))
U3_g(X, num_out_g(Y)) → num_out_g(X)

The argument filtering Pi contains the following mapping:
num_in_g(x1)  =  num_in_g(x1)
0  =  0
U1_g(x1)  =  U1_g(x1)
true_in_  =  true_in_
true_out_  =  true_out_
num_out_g(x1)  =  num_out_g
U2_g(x1, x2)  =  U2_g(x2)
p_in_ga(x1, x2)  =  p_in_ga(x1)
p_out_ga(x1, x2)  =  p_out_ga(x2)
s(x1)  =  s(x1)
U3_g(x1, x2)  =  U3_g(x2)
NUM_IN_G(x1)  =  NUM_IN_G(x1)
U2_G(x1, x2)  =  U2_G(x2)

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:

NUM_IN_G(X) → U2_G(X, p_in_ga(X, Y))
U2_G(X, p_out_ga(X, Y)) → NUM_IN_G(Y)

The TRS R consists of the following rules:

p_in_ga(0, 0) → p_out_ga(0, 0)
p_in_ga(s(X), X) → p_out_ga(s(X), X)

The argument filtering Pi contains the following mapping:
0  =  0
p_in_ga(x1, x2)  =  p_in_ga(x1)
p_out_ga(x1, x2)  =  p_out_ga(x2)
s(x1)  =  s(x1)
NUM_IN_G(x1)  =  NUM_IN_G(x1)
U2_G(x1, x2)  =  U2_G(x2)

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:

NUM_IN_G(X) → U2_G(p_in_ga(X))
U2_G(p_out_ga(Y)) → NUM_IN_G(Y)

The TRS R consists of the following rules:

p_in_ga(0) → p_out_ga(0)
p_in_ga(s(X)) → p_out_ga(X)

The set Q consists of the following terms:

p_in_ga(x0)

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

(41) UsableRulesReductionPairsProof (EQUIVALENT transformation)

By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

No dependency pairs are removed.

The following rules are removed from R:

p_in_ga(s(X)) → p_out_ga(X)

Used ordering: POLO with Polynomial interpretation [POLO]:

POL(0) = 0   
POL(NUM_IN_G(x₁)) = x₁   
POL(U2_G(x₁)) = x₁   
POL(p_in_ga(x₁)) = x₁   
POL(p_out_ga(x₁)) = 2·x₁   
POL(s(x₁)) = 2·x₁   

(42) Obligation:

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

NUM_IN_G(X) → U2_G(p_in_ga(X))
U2_G(p_out_ga(Y)) → NUM_IN_G(Y)

The TRS R consists of the following rules:

p_in_ga(0) → p_out_ga(0)

The set Q consists of the following terms:

p_in_ga(x0)

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

(43) Narrowing (SOUND transformation)

By narrowing [LPAR04] the rule NUM_IN_G(X) → U2_G(p_in_ga(X)) at position [0] we obtained the following new rules [LPAR04]:

NUM_IN_G(0) → U2_G(p_out_ga(0))

(44) Obligation:

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

U2_G(p_out_ga(Y)) → NUM_IN_G(Y)
NUM_IN_G(0) → U2_G(p_out_ga(0))

The TRS R consists of the following rules:

p_in_ga(0) → p_out_ga(0)

The set Q consists of the following terms:

p_in_ga(x0)

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

(45) 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.

(46) Obligation:

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

U2_G(p_out_ga(Y)) → NUM_IN_G(Y)
NUM_IN_G(0) → U2_G(p_out_ga(0))

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

p_in_ga(x0)

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

(47) 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].

p_in_ga(x0)

(48) Obligation:

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

U2_G(p_out_ga(Y)) → NUM_IN_G(Y)
NUM_IN_G(0) → U2_G(p_out_ga(0))

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

(49) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule U2_G(p_out_ga(Y)) → NUM_IN_G(Y) we obtained the following new rules [LPAR04]:

U2_G(p_out_ga(0)) → NUM_IN_G(0)

(50) Obligation:

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

NUM_IN_G(0) → U2_G(p_out_ga(0))
U2_G(p_out_ga(0)) → NUM_IN_G(0)

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

(51) 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 = U2_G(p_out_ga(0)) evaluates to t =U2_G(p_out_ga(0))

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:

• Matcher: [ ]
• Semiunifier: [ ]

---

Rewriting sequence

U2_G(p_out_ga(0)) → NUM_IN_G(0)
with rule U2_G(p_out_ga(0)) → NUM_IN_G(0) at position [] and matcher [ ]

NUM_IN_G(0) → U2_G(p_out_ga(0))
with rule NUM_IN_G(0) → U2_G(p_out_ga(0))

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.

(52) FALSE