(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(s(0), g(x)) → f(x, g(x))
g(s(x)) → g(x)
Q is empty.
(1) DependencyPairsProof (EQUIVALENT transformation)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
(2) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(s(0), g(x)) → F(x, g(x))
G(s(x)) → G(x)
The TRS R consists of the following rules:
f(s(0), g(x)) → f(x, g(x))
g(s(x)) → g(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(3) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs.
(4) Complex Obligation (AND)
(5) Obligation:
Q DP problem:
The TRS P consists of the following rules:
G(s(x)) → G(x)
The TRS R consists of the following rules:
f(s(0), g(x)) → f(x, g(x))
g(s(x)) → g(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(6) UsableRulesProof (EQUIVALENT transformation)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
(7) Obligation:
Q DP problem:
The TRS P consists of the following rules:
G(s(x)) → G(x)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(8) 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:
- G(s(x)) → G(x)
The graph contains the following edges 1 > 1
(9) YES
(10) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(s(0), g(x)) → F(x, g(x))
The TRS R consists of the following rules:
f(s(0), g(x)) → f(x, g(x))
g(s(x)) → g(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(11) ForwardInstantiation (EQUIVALENT transformation)
By forward instantiating [JAR06] the rule
F(
s(
0),
g(
x)) →
F(
x,
g(
x)) we obtained the following new rules [LPAR04]:
F(s(0), g(s(0))) → F(s(0), g(s(0)))
(12) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(s(0), g(s(0))) → F(s(0), g(s(0)))
The TRS R consists of the following rules:
f(s(0), g(x)) → f(x, g(x))
g(s(x)) → g(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(13) UsableRulesProof (EQUIVALENT transformation)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
(14) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(s(0), g(s(0))) → F(s(0), g(s(0)))
The TRS R consists of the following rules:
g(s(x)) → g(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(15) MRRProof (EQUIVALENT transformation)
By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented rules of the TRS R:
g(s(x)) → g(x)
Used ordering: Knuth-Bendix order [KBO] with precedence:
trivial
and weight map:
s_1=1
F_2=1
g_1=1
0=1
(16) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(s(0), g(s(0))) → F(s(0), g(s(0)))
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(17) 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 =
F(
s(
0),
g(
s(
0))) evaluates to t =
F(
s(
0),
g(
s(
0)))
Thus 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 F(s(0), g(s(0))) to F(s(0), g(s(0))).
(18) NO